How find a value within an array, with a users input











up vote
2
down vote

favorite












var x1 ='&spades',

    y1 ='&clubs'

    z1 ='&hearts';

var x2 = ' ', y2 = ' ', z2 = ' ';  

var x3 = ' ', y3 = ' ', z3 = ' ';    





var Array = [x1,y1,z1,

             x2,y2,z2,

             x3,y3,z3];


Then I get an input from the user asking them to select a location (ex. x1) store it as input 1, then ask them for another location (ex. x2) and store that as input 2, and I have to swap the locations of the value.



I initially tried something like Array[input1] = Array[input2]



But Array[input1] is undefined even though input1 = x1 because Array[x1] is still undefined.



So how do I make Array[x1] = Array[0] so I can swap the values or is there an easier way?



Sorry if my question is poorly formed.






javascript html arrays







share|improve this question
























  • For starters, 'Array' is a keyword. Probably not a good idea to use it as a variable name. Next, the indices to your array are 0 through 9. JS has no idea how to map 'x2' to 4.
    – Jim B.
    Nov 12 at 4:22












  • use a variable to temporarily store the value of Array[firstValueToSwap] value and store the Array[secondValueToSwap] in Array[firstValueToSwap] and store the value in temp to Array[secondValueToSwap]. It's the easiest way to swap two values. BTW your question is not that clear. hope this is what you're looking for
    – Sanira
    Nov 12 at 4:25












  • @CertainPerformance Yes the input is a string.
    – David
    Nov 12 at 4:28










  • As per my understanding, you have an array with 9 elements in it. Example: [ '&spades', '&clubs', '&hearts', ' ', ' ' , ' ' , ' ' , ' ' , ' ']. Now you want to swap the values at 2 indexes (input1 and input2) for example if the user entered 0 and 2. The new array will be [ '&hearts', '&clubs', '&spades', ' ', ' ' , ' ' , ' ' , ' ' , ' ']. I am writing this comment so that community can understand the question correctly.
    – Suyash Gulati
    Nov 12 at 4:30










  • @JimB. I know this isnt my real code my array is a lot bigger just a simple version, and I know that JS doesnt know how to map x2 to 4 but I was trying to see if there is any way otherwise i have to rewrite my program
    – David
    Nov 12 at 4:30















up vote
2
down vote

favorite












var x1 ='&spades',

    y1 ='&clubs'

    z1 ='&hearts';

var x2 = ' ', y2 = ' ', z2 = ' ';  

var x3 = ' ', y3 = ' ', z3 = ' ';    





var Array = [x1,y1,z1,

             x2,y2,z2,

             x3,y3,z3];


Then I get an input from the user asking them to select a location (ex. x1) store it as input 1, then ask them for another location (ex. x2) and store that as input 2, and I have to swap the locations of the value.



I initially tried something like Array[input1] = Array[input2]



But Array[input1] is undefined even though input1 = x1 because Array[x1] is still undefined.



So how do I make Array[x1] = Array[0] so I can swap the values or is there an easier way?



Sorry if my question is poorly formed.






javascript html arrays







share|improve this question
























  • For starters, 'Array' is a keyword. Probably not a good idea to use it as a variable name. Next, the indices to your array are 0 through 9. JS has no idea how to map 'x2' to 4.
    – Jim B.
    Nov 12 at 4:22












  • use a variable to temporarily store the value of Array[firstValueToSwap] value and store the Array[secondValueToSwap] in Array[firstValueToSwap] and store the value in temp to Array[secondValueToSwap]. It's the easiest way to swap two values. BTW your question is not that clear. hope this is what you're looking for
    – Sanira
    Nov 12 at 4:25












  • @CertainPerformance Yes the input is a string.
    – David
    Nov 12 at 4:28










  • As per my understanding, you have an array with 9 elements in it. Example: [ '&spades', '&clubs', '&hearts', ' ', ' ' , ' ' , ' ' , ' ' , ' ']. Now you want to swap the values at 2 indexes (input1 and input2) for example if the user entered 0 and 2. The new array will be [ '&hearts', '&clubs', '&spades', ' ', ' ' , ' ' , ' ' , ' ' , ' ']. I am writing this comment so that community can understand the question correctly.
    – Suyash Gulati
    Nov 12 at 4:30










  • @JimB. I know this isnt my real code my array is a lot bigger just a simple version, and I know that JS doesnt know how to map x2 to 4 but I was trying to see if there is any way otherwise i have to rewrite my program
    – David
    Nov 12 at 4:30













up vote
2
down vote

favorite









up vote
2
down vote

favorite











var x1 ='&spades',

    y1 ='&clubs'

    z1 ='&hearts';

var x2 = ' ', y2 = ' ', z2 = ' ';  

var x3 = ' ', y3 = ' ', z3 = ' ';    





var Array = [x1,y1,z1,

             x2,y2,z2,

             x3,y3,z3];


Then I get an input from the user asking them to select a location (ex. x1) store it as input 1, then ask them for another location (ex. x2) and store that as input 2, and I have to swap the locations of the value.



I initially tried something like Array[input1] = Array[input2]



But Array[input1] is undefined even though input1 = x1 because Array[x1] is still undefined.



So how do I make Array[x1] = Array[0] so I can swap the values or is there an easier way?



Sorry if my question is poorly formed.






javascript html arrays







share|improve this question















var x1 ='&spades',

    y1 ='&clubs'

    z1 ='&hearts';

var x2 = ' ', y2 = ' ', z2 = ' ';  

var x3 = ' ', y3 = ' ', z3 = ' ';    





var Array = [x1,y1,z1,

             x2,y2,z2,

             x3,y3,z3];


Then I get an input from the user asking them to select a location (ex. x1) store it as input 1, then ask them for another location (ex. x2) and store that as input 2, and I have to swap the locations of the value.



I initially tried something like Array[input1] = Array[input2]



But Array[input1] is undefined even though input1 = x1 because Array[x1] is still undefined.



So how do I make Array[x1] = Array[0] so I can swap the values or is there an easier way?



Sorry if my question is poorly formed.






javascript html arrays




javascript html arrays






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 12 at 5:46









Andreas

1,7861618




1,7861618










asked Nov 12 at 4:18









David

204




204












  • For starters, 'Array' is a keyword. Probably not a good idea to use it as a variable name. Next, the indices to your array are 0 through 9. JS has no idea how to map 'x2' to 4.
    – Jim B.
    Nov 12 at 4:22












  • use a variable to temporarily store the value of Array[firstValueToSwap] value and store the Array[secondValueToSwap] in Array[firstValueToSwap] and store the value in temp to Array[secondValueToSwap]. It's the easiest way to swap two values. BTW your question is not that clear. hope this is what you're looking for
    – Sanira
    Nov 12 at 4:25












  • @CertainPerformance Yes the input is a string.
    – David
    Nov 12 at 4:28










  • As per my understanding, you have an array with 9 elements in it. Example: [ '&spades', '&clubs', '&hearts', ' ', ' ' , ' ' , ' ' , ' ' , ' ']. Now you want to swap the values at 2 indexes (input1 and input2) for example if the user entered 0 and 2. The new array will be [ '&hearts', '&clubs', '&spades', ' ', ' ' , ' ' , ' ' , ' ' , ' ']. I am writing this comment so that community can understand the question correctly.
    – Suyash Gulati
    Nov 12 at 4:30










  • @JimB. I know this isnt my real code my array is a lot bigger just a simple version, and I know that JS doesnt know how to map x2 to 4 but I was trying to see if there is any way otherwise i have to rewrite my program
    – David
    Nov 12 at 4:30


















  • For starters, 'Array' is a keyword. Probably not a good idea to use it as a variable name. Next, the indices to your array are 0 through 9. JS has no idea how to map 'x2' to 4.
    – Jim B.
    Nov 12 at 4:22












  • use a variable to temporarily store the value of Array[firstValueToSwap] value and store the Array[secondValueToSwap] in Array[firstValueToSwap] and store the value in temp to Array[secondValueToSwap]. It's the easiest way to swap two values. BTW your question is not that clear. hope this is what you're looking for
    – Sanira
    Nov 12 at 4:25












  • @CertainPerformance Yes the input is a string.
    – David
    Nov 12 at 4:28










  • As per my understanding, you have an array with 9 elements in it. Example: [ '&spades', '&clubs', '&hearts', ' ', ' ' , ' ' , ' ' , ' ' , ' ']. Now you want to swap the values at 2 indexes (input1 and input2) for example if the user entered 0 and 2. The new array will be [ '&hearts', '&clubs', '&spades', ' ', ' ' , ' ' , ' ' , ' ' , ' ']. I am writing this comment so that community can understand the question correctly.
    – Suyash Gulati
    Nov 12 at 4:30










  • @JimB. I know this isnt my real code my array is a lot bigger just a simple version, and I know that JS doesnt know how to map x2 to 4 but I was trying to see if there is any way otherwise i have to rewrite my program
    – David
    Nov 12 at 4:30
















For starters, 'Array' is a keyword. Probably not a good idea to use it as a variable name. Next, the indices to your array are 0 through 9. JS has no idea how to map 'x2' to 4.
– Jim B.
Nov 12 at 4:22






For starters, 'Array' is a keyword. Probably not a good idea to use it as a variable name. Next, the indices to your array are 0 through 9. JS has no idea how to map 'x2' to 4.
– Jim B.
Nov 12 at 4:22














use a variable to temporarily store the value of Array[firstValueToSwap] value and store the Array[secondValueToSwap] in Array[firstValueToSwap] and store the value in temp to Array[secondValueToSwap]. It's the easiest way to swap two values. BTW your question is not that clear. hope this is what you're looking for
– Sanira
Nov 12 at 4:25






use a variable to temporarily store the value of Array[firstValueToSwap] value and store the Array[secondValueToSwap] in Array[firstValueToSwap] and store the value in temp to Array[secondValueToSwap]. It's the easiest way to swap two values. BTW your question is not that clear. hope this is what you're looking for
– Sanira
Nov 12 at 4:25














@CertainPerformance Yes the input is a string.
– David
Nov 12 at 4:28




@CertainPerformance Yes the input is a string.
– David
Nov 12 at 4:28












As per my understanding, you have an array with 9 elements in it. Example: [ '&spades', '&clubs', '&hearts', ' ', ' ' , ' ' , ' ' , ' ' , ' ']. Now you want to swap the values at 2 indexes (input1 and input2) for example if the user entered 0 and 2. The new array will be [ '&hearts', '&clubs', '&spades', ' ', ' ' , ' ' , ' ' , ' ' , ' ']. I am writing this comment so that community can understand the question correctly.
– Suyash Gulati
Nov 12 at 4:30




As per my understanding, you have an array with 9 elements in it. Example: [ '&spades', '&clubs', '&hearts', ' ', ' ' , ' ' , ' ' , ' ' , ' ']. Now you want to swap the values at 2 indexes (input1 and input2) for example if the user entered 0 and 2. The new array will be [ '&hearts', '&clubs', '&spades', ' ', ' ' , ' ' , ' ' , ' ' , ' ']. I am writing this comment so that community can understand the question correctly.
– Suyash Gulati
Nov 12 at 4:30












@JimB. I know this isnt my real code my array is a lot bigger just a simple version, and I know that JS doesnt know how to map x2 to 4 but I was trying to see if there is any way otherwise i have to rewrite my program
– David
Nov 12 at 4:30




@JimB. I know this isnt my real code my array is a lot bigger just a simple version, and I know that JS doesnt know how to map x2 to 4 but I was trying to see if there is any way otherwise i have to rewrite my program
– David
Nov 12 at 4:30












2 Answers
2






active

oldest

votes

















up vote
2
down vote



accepted










Assuming the inputs are strings ('x1' and 'x2'), you need a way to represent the variable names in your data structure. One way would be to have, rather than an array of strings, an array of objects, containing a label and a value property. Then, just find the indicies of both labels for both inputs, and reassign those indicies.



Also, better not to assign to a variable named Array, because that will shadow the global Array object - name it something else instead, like arr:






const arr = [

  { label: 'x1', value: '&spades' },

  { label: 'y1', value: '&clubs' },

  { label: 'z1', value: '&hearts' }

  // ...

];

// Inputs:

const label1 = 'x1';

const label2 = 'y1';



// Calculate indicies:

const [index1, index2] = [label1, label2].map(

  findLabel => arr.findIndex(({ label }) => label === findLabel)

);

([arr[index1], arr[index2]] = [arr[index2], arr[index1]]);

console.log(arr);





The destructuring line at the end there allows for swapping variables in two positions at once, without having to resort to an intermediate variable:



([arr[index1], arr[index2]] = [arr[index2], arr[index1]]);


If you find that confusing, an alternative is to actually use an intermediate variable:



const orig1 = arr[index1];

arr[index1] = arr[index2];

arr[index2] = orig1;





share|improve this answer





















  • I will try this as soon as I get back home, I see what you mean, I used a very very simple version of my code as an example of what I was trying to do I know not to name it array was just example but I will try to implement this method into my own code will let you know how it goes, thank you
    – David
    Nov 12 at 4:44










  • Yes, thank you @CertainPerformance, by making my array of objects now allows me to make the connection between the array and users input so now I can switch them, at the bottom I get a bit fuzzy on how the whole switching works, I just used some if statements with an intermediate var because I didnt understand what the bottom half was doing. Though I will try to play around to try and understand it. Thank you for your help.
    – David
    Nov 12 at 5:52


















up vote
0
down vote













Here's what your code might look like if you use an object:






    let arr = {};

    arr['x1'] = '&spades';

    arr['y1'] = '&clubs';

    arr['z1'] = '&hearts';

    arr['x2'] = ' ';

    arr['y2'] = ' ';

    arr['z2'] = ' ';

    arr['x3'] = ' ';

    arr['y3'] = ' ';

    arr['z3'] = ' ';

    

    console.log(arr['z1']);

    arr['z1'] = 'foo';

    console.log(arr['z1']);








share|improve this answer





















  • I see what you mean now CertainPerformace was saying something about the same will try as soon as I can, thank you
    – David
    Nov 12 at 4:45










  • Yes, @JimB. you were right the answer was that I had to make an array of objects. Thank you for your help.
    – David
    Nov 12 at 5:54











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










Assuming the inputs are strings ('x1' and 'x2'), you need a way to represent the variable names in your data structure. One way would be to have, rather than an array of strings, an array of objects, containing a label and a value property. Then, just find the indicies of both labels for both inputs, and reassign those indicies.



Also, better not to assign to a variable named Array, because that will shadow the global Array object - name it something else instead, like arr:






const arr = [

  { label: 'x1', value: '&spades' },

  { label: 'y1', value: '&clubs' },

  { label: 'z1', value: '&hearts' }

  // ...

];

// Inputs:

const label1 = 'x1';

const label2 = 'y1';



// Calculate indicies:

const [index1, index2] = [label1, label2].map(

  findLabel => arr.findIndex(({ label }) => label === findLabel)

);

([arr[index1], arr[index2]] = [arr[index2], arr[index1]]);

console.log(arr);





The destructuring line at the end there allows for swapping variables in two positions at once, without having to resort to an intermediate variable:



([arr[index1], arr[index2]] = [arr[index2], arr[index1]]);


If you find that confusing, an alternative is to actually use an intermediate variable:



const orig1 = arr[index1];

arr[index1] = arr[index2];

arr[index2] = orig1;





share|improve this answer





















  • I will try this as soon as I get back home, I see what you mean, I used a very very simple version of my code as an example of what I was trying to do I know not to name it array was just example but I will try to implement this method into my own code will let you know how it goes, thank you
    – David
    Nov 12 at 4:44










  • Yes, thank you @CertainPerformance, by making my array of objects now allows me to make the connection between the array and users input so now I can switch them, at the bottom I get a bit fuzzy on how the whole switching works, I just used some if statements with an intermediate var because I didnt understand what the bottom half was doing. Though I will try to play around to try and understand it. Thank you for your help.
    – David
    Nov 12 at 5:52















up vote
2
down vote



accepted










Assuming the inputs are strings ('x1' and 'x2'), you need a way to represent the variable names in your data structure. One way would be to have, rather than an array of strings, an array of objects, containing a label and a value property. Then, just find the indicies of both labels for both inputs, and reassign those indicies.



Also, better not to assign to a variable named Array, because that will shadow the global Array object - name it something else instead, like arr:






const arr = [

  { label: 'x1', value: '&spades' },

  { label: 'y1', value: '&clubs' },

  { label: 'z1', value: '&hearts' }

  // ...

];

// Inputs:

const label1 = 'x1';

const label2 = 'y1';



// Calculate indicies:

const [index1, index2] = [label1, label2].map(

  findLabel => arr.findIndex(({ label }) => label === findLabel)

);

([arr[index1], arr[index2]] = [arr[index2], arr[index1]]);

console.log(arr);





The destructuring line at the end there allows for swapping variables in two positions at once, without having to resort to an intermediate variable:



([arr[index1], arr[index2]] = [arr[index2], arr[index1]]);


If you find that confusing, an alternative is to actually use an intermediate variable:



const orig1 = arr[index1];

arr[index1] = arr[index2];

arr[index2] = orig1;





share|improve this answer





















  • I will try this as soon as I get back home, I see what you mean, I used a very very simple version of my code as an example of what I was trying to do I know not to name it array was just example but I will try to implement this method into my own code will let you know how it goes, thank you
    – David
    Nov 12 at 4:44










  • Yes, thank you @CertainPerformance, by making my array of objects now allows me to make the connection between the array and users input so now I can switch them, at the bottom I get a bit fuzzy on how the whole switching works, I just used some if statements with an intermediate var because I didnt understand what the bottom half was doing. Though I will try to play around to try and understand it. Thank you for your help.
    – David
    Nov 12 at 5:52













up vote
2
down vote



accepted







up vote
2
down vote



accepted






Assuming the inputs are strings ('x1' and 'x2'), you need a way to represent the variable names in your data structure. One way would be to have, rather than an array of strings, an array of objects, containing a label and a value property. Then, just find the indicies of both labels for both inputs, and reassign those indicies.



Also, better not to assign to a variable named Array, because that will shadow the global Array object - name it something else instead, like arr:






const arr = [

  { label: 'x1', value: '&spades' },

  { label: 'y1', value: '&clubs' },

  { label: 'z1', value: '&hearts' }

  // ...

];

// Inputs:

const label1 = 'x1';

const label2 = 'y1';



// Calculate indicies:

const [index1, index2] = [label1, label2].map(

  findLabel => arr.findIndex(({ label }) => label === findLabel)

);

([arr[index1], arr[index2]] = [arr[index2], arr[index1]]);

console.log(arr);





The destructuring line at the end there allows for swapping variables in two positions at once, without having to resort to an intermediate variable:



([arr[index1], arr[index2]] = [arr[index2], arr[index1]]);


If you find that confusing, an alternative is to actually use an intermediate variable:



const orig1 = arr[index1];

arr[index1] = arr[index2];

arr[index2] = orig1;





share|improve this answer












Assuming the inputs are strings ('x1' and 'x2'), you need a way to represent the variable names in your data structure. One way would be to have, rather than an array of strings, an array of objects, containing a label and a value property. Then, just find the indicies of both labels for both inputs, and reassign those indicies.



Also, better not to assign to a variable named Array, because that will shadow the global Array object - name it something else instead, like arr:






const arr = [

  { label: 'x1', value: '&spades' },

  { label: 'y1', value: '&clubs' },

  { label: 'z1', value: '&hearts' }

  // ...

];

// Inputs:

const label1 = 'x1';

const label2 = 'y1';



// Calculate indicies:

const [index1, index2] = [label1, label2].map(

  findLabel => arr.findIndex(({ label }) => label === findLabel)

);

([arr[index1], arr[index2]] = [arr[index2], arr[index1]]);

console.log(arr);





The destructuring line at the end there allows for swapping variables in two positions at once, without having to resort to an intermediate variable:



([arr[index1], arr[index2]] = [arr[index2], arr[index1]]);


If you find that confusing, an alternative is to actually use an intermediate variable:



const orig1 = arr[index1];

arr[index1] = arr[index2];

arr[index2] = orig1;







const arr = [

  { label: 'x1', value: '&spades' },

  { label: 'y1', value: '&clubs' },

  { label: 'z1', value: '&hearts' }

  // ...

];

// Inputs:

const label1 = 'x1';

const label2 = 'y1';



// Calculate indicies:

const [index1, index2] = [label1, label2].map(

  findLabel => arr.findIndex(({ label }) => label === findLabel)

);

([arr[index1], arr[index2]] = [arr[index2], arr[index1]]);

console.log(arr);





const arr = [

  { label: 'x1', value: '&spades' },

  { label: 'y1', value: '&clubs' },

  { label: 'z1', value: '&hearts' }

  // ...

];

// Inputs:

const label1 = 'x1';

const label2 = 'y1';



// Calculate indicies:

const [index1, index2] = [label1, label2].map(

  findLabel => arr.findIndex(({ label }) => label === findLabel)

);

([arr[index1], arr[index2]] = [arr[index2], arr[index1]]);

console.log(arr);






share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 12 at 4:25









CertainPerformance

71.5k143453




71.5k143453












  • I will try this as soon as I get back home, I see what you mean, I used a very very simple version of my code as an example of what I was trying to do I know not to name it array was just example but I will try to implement this method into my own code will let you know how it goes, thank you
    – David
    Nov 12 at 4:44










  • Yes, thank you @CertainPerformance, by making my array of objects now allows me to make the connection between the array and users input so now I can switch them, at the bottom I get a bit fuzzy on how the whole switching works, I just used some if statements with an intermediate var because I didnt understand what the bottom half was doing. Though I will try to play around to try and understand it. Thank you for your help.
    – David
    Nov 12 at 5:52


















  • I will try this as soon as I get back home, I see what you mean, I used a very very simple version of my code as an example of what I was trying to do I know not to name it array was just example but I will try to implement this method into my own code will let you know how it goes, thank you
    – David
    Nov 12 at 4:44










  • Yes, thank you @CertainPerformance, by making my array of objects now allows me to make the connection between the array and users input so now I can switch them, at the bottom I get a bit fuzzy on how the whole switching works, I just used some if statements with an intermediate var because I didnt understand what the bottom half was doing. Though I will try to play around to try and understand it. Thank you for your help.
    – David
    Nov 12 at 5:52
















I will try this as soon as I get back home, I see what you mean, I used a very very simple version of my code as an example of what I was trying to do I know not to name it array was just example but I will try to implement this method into my own code will let you know how it goes, thank you
– David
Nov 12 at 4:44




I will try this as soon as I get back home, I see what you mean, I used a very very simple version of my code as an example of what I was trying to do I know not to name it array was just example but I will try to implement this method into my own code will let you know how it goes, thank you
– David
Nov 12 at 4:44












Yes, thank you @CertainPerformance, by making my array of objects now allows me to make the connection between the array and users input so now I can switch them, at the bottom I get a bit fuzzy on how the whole switching works, I just used some if statements with an intermediate var because I didnt understand what the bottom half was doing. Though I will try to play around to try and understand it. Thank you for your help.
– David
Nov 12 at 5:52




Yes, thank you @CertainPerformance, by making my array of objects now allows me to make the connection between the array and users input so now I can switch them, at the bottom I get a bit fuzzy on how the whole switching works, I just used some if statements with an intermediate var because I didnt understand what the bottom half was doing. Though I will try to play around to try and understand it. Thank you for your help.
– David
Nov 12 at 5:52












up vote
0
down vote













Here's what your code might look like if you use an object:






    let arr = {};

    arr['x1'] = '&spades';

    arr['y1'] = '&clubs';

    arr['z1'] = '&hearts';

    arr['x2'] = ' ';

    arr['y2'] = ' ';

    arr['z2'] = ' ';

    arr['x3'] = ' ';

    arr['y3'] = ' ';

    arr['z3'] = ' ';

    

    console.log(arr['z1']);

    arr['z1'] = 'foo';

    console.log(arr['z1']);








share|improve this answer





















  • I see what you mean now CertainPerformace was saying something about the same will try as soon as I can, thank you
    – David
    Nov 12 at 4:45










  • Yes, @JimB. you were right the answer was that I had to make an array of objects. Thank you for your help.
    – David
    Nov 12 at 5:54















up vote
0
down vote













Here's what your code might look like if you use an object:






    let arr = {};

    arr['x1'] = '&spades';

    arr['y1'] = '&clubs';

    arr['z1'] = '&hearts';

    arr['x2'] = ' ';

    arr['y2'] = ' ';

    arr['z2'] = ' ';

    arr['x3'] = ' ';

    arr['y3'] = ' ';

    arr['z3'] = ' ';

    

    console.log(arr['z1']);

    arr['z1'] = 'foo';

    console.log(arr['z1']);








share|improve this answer





















  • I see what you mean now CertainPerformace was saying something about the same will try as soon as I can, thank you
    – David
    Nov 12 at 4:45










  • Yes, @JimB. you were right the answer was that I had to make an array of objects. Thank you for your help.
    – David
    Nov 12 at 5:54













up vote
0
down vote










up vote
0
down vote









Here's what your code might look like if you use an object:






    let arr = {};

    arr['x1'] = '&spades';

    arr['y1'] = '&clubs';

    arr['z1'] = '&hearts';

    arr['x2'] = ' ';

    arr['y2'] = ' ';

    arr['z2'] = ' ';

    arr['x3'] = ' ';

    arr['y3'] = ' ';

    arr['z3'] = ' ';

    

    console.log(arr['z1']);

    arr['z1'] = 'foo';

    console.log(arr['z1']);








share|improve this answer












Here's what your code might look like if you use an object:






    let arr = {};

    arr['x1'] = '&spades';

    arr['y1'] = '&clubs';

    arr['z1'] = '&hearts';

    arr['x2'] = ' ';

    arr['y2'] = ' ';

    arr['z2'] = ' ';

    arr['x3'] = ' ';

    arr['y3'] = ' ';

    arr['z3'] = ' ';

    

    console.log(arr['z1']);

    arr['z1'] = 'foo';

    console.log(arr['z1']);








    let arr = {};

    arr['x1'] = '&spades';

    arr['y1'] = '&clubs';

    arr['z1'] = '&hearts';

    arr['x2'] = ' ';

    arr['y2'] = ' ';

    arr['z2'] = ' ';

    arr['x3'] = ' ';

    arr['y3'] = ' ';

    arr['z3'] = ' ';

    

    console.log(arr['z1']);

    arr['z1'] = 'foo';

    console.log(arr['z1']);





    let arr = {};

    arr['x1'] = '&spades';

    arr['y1'] = '&clubs';

    arr['z1'] = '&hearts';

    arr['x2'] = ' ';

    arr['y2'] = ' ';

    arr['z2'] = ' ';

    arr['x3'] = ' ';

    arr['y3'] = ' ';

    arr['z3'] = ' ';

    

    console.log(arr['z1']);

    arr['z1'] = 'foo';

    console.log(arr['z1']);






share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 12 at 4:37









Jim B.

2,6411928




2,6411928












  • I see what you mean now CertainPerformace was saying something about the same will try as soon as I can, thank you
    – David
    Nov 12 at 4:45










  • Yes, @JimB. you were right the answer was that I had to make an array of objects. Thank you for your help.
    – David
    Nov 12 at 5:54


















  • I see what you mean now CertainPerformace was saying something about the same will try as soon as I can, thank you
    – David
    Nov 12 at 4:45










  • Yes, @JimB. you were right the answer was that I had to make an array of objects. Thank you for your help.
    – David
    Nov 12 at 5:54
















I see what you mean now CertainPerformace was saying something about the same will try as soon as I can, thank you
– David
Nov 12 at 4:45




I see what you mean now CertainPerformace was saying something about the same will try as soon as I can, thank you
– David
Nov 12 at 4:45












Yes, @JimB. you were right the answer was that I had to make an array of objects. Thank you for your help.
– David
Nov 12 at 5:54




Yes, @JimB. you were right the answer was that I had to make an array of objects. Thank you for your help.
– David
Nov 12 at 5:54


















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