How to replace a value of a key in a object
0
Him I am new to the react redux. I have an array of objects which is like,
let result = [ {Id :'a',temp:5,data:'abc (c:1'},{Id :'a',temp:5,data:'pqr (c:1'}{Id :'a',temp:5,data:'xyz (c:1'}]
Now, what I want to do is that there is key called data with some value, I want to take the value which is before that bracket only in that data key.
what I want to look an object ,
{Id :'a',count:5,data:'abc'}
what I have is ,
I have an regex which I use to get that value,
and I have used for loop ,
for (let i = 0; i <= result.length - 1; i++) {
let regex = /^[^(]+/;
let value = data.match(regex)[0].trim();
result[i].data = value
}
fetchQuestions() {
let previous_data = { ...this.props.data };
let result = ;
for (let key in previous_data) {
result = [...result, ...previous_data[key]];
}
for (let i = 0; i <= result.length - 1; i++) {
let regex = /^(.*)(/;
let value = result[i].technology.match(regex)[1].trim();
result[i].technology = value;
}
this.props.fetchQuestions(result);
}
In this this.props.data is the reducer data .Now what I want to do was, I have to get some data from this and do some manipulation. but here, it is directly updating my reducer values itself, But I have used spread operator to get the copy of that data ,but still its updating the reducers values.
But I am not getting the value which I expect. Can any one help me with this ?
javascript reactjs redux react-redux
asked Nov 14 '18 at 6:14
ganeshkganeshk
24219
add a comment |
0
Him I am new to the react redux. I have an array of objects which is like,
let result = [ {Id :'a',temp:5,data:'abc (c:1'},{Id :'a',temp:5,data:'pqr (c:1'}{Id :'a',temp:5,data:'xyz (c:1'}]
Now, what I want to do is that there is key called data with some value, I want to take the value which is before that bracket only in that data key.
what I want to look an object ,
{Id :'a',count:5,data:'abc'}
what I have is ,
I have an regex which I use to get that value,
and I have used for loop ,
for (let i = 0; i <= result.length - 1; i++) {
let regex = /^[^(]+/;
let value = data.match(regex)[0].trim();
result[i].data = value
}
fetchQuestions() {
let previous_data = { ...this.props.data };
let result = ;
for (let key in previous_data) {
result = [...result, ...previous_data[key]];
}
for (let i = 0; i <= result.length - 1; i++) {
let regex = /^(.*)(/;
let value = result[i].technology.match(regex)[1].trim();
result[i].technology = value;
}
this.props.fetchQuestions(result);
}
In this this.props.data is the reducer data .Now what I want to do was, I have to get some data from this and do some manipulation. but here, it is directly updating my reducer values itself, But I have used spread operator to get the copy of that data ,but still its updating the reducers values.
But I am not getting the value which I expect. Can any one help me with this ?
javascript reactjs redux react-redux
asked Nov 14 '18 at 6:14
ganeshkganeshk
24219
try this - let regex =/^(w+)/;
– Nikita R.
Nov 14 '18 at 6:21
add a comment |
0
0
0
Him I am new to the react redux. I have an array of objects which is like,
let result = [ {Id :'a',temp:5,data:'abc (c:1'},{Id :'a',temp:5,data:'pqr (c:1'}{Id :'a',temp:5,data:'xyz (c:1'}]
Now, what I want to do is that there is key called data with some value, I want to take the value which is before that bracket only in that data key.
what I want to look an object ,
{Id :'a',count:5,data:'abc'}
what I have is ,
I have an regex which I use to get that value,
and I have used for loop ,
for (let i = 0; i <= result.length - 1; i++) {
let regex = /^[^(]+/;
let value = data.match(regex)[0].trim();
result[i].data = value
}
fetchQuestions() {
let previous_data = { ...this.props.data };
let result = ;
for (let key in previous_data) {
result = [...result, ...previous_data[key]];
}
for (let i = 0; i <= result.length - 1; i++) {
let regex = /^(.*)(/;
let value = result[i].technology.match(regex)[1].trim();
result[i].technology = value;
}
this.props.fetchQuestions(result);
}
In this this.props.data is the reducer data .Now what I want to do was, I have to get some data from this and do some manipulation. but here, it is directly updating my reducer values itself, But I have used spread operator to get the copy of that data ,but still its updating the reducers values.
But I am not getting the value which I expect. Can any one help me with this ?
javascript reactjs redux react-redux
asked Nov 14 '18 at 6:14
ganeshkganeshk
24219
Him I am new to the react redux. I have an array of objects which is like,
let result = [ {Id :'a',temp:5,data:'abc (c:1'},{Id :'a',temp:5,data:'pqr (c:1'}{Id :'a',temp:5,data:'xyz (c:1'}]
Now, what I want to do is that there is key called data with some value, I want to take the value which is before that bracket only in that data key.
what I want to look an object ,
{Id :'a',count:5,data:'abc'}
what I have is ,
I have an regex which I use to get that value,
and I have used for loop ,
for (let i = 0; i <= result.length - 1; i++) {
let regex = /^[^(]+/;
let value = data.match(regex)[0].trim();
result[i].data = value
}
fetchQuestions() {
let previous_data = { ...this.props.data };
let result = ;
for (let key in previous_data) {
result = [...result, ...previous_data[key]];
}
for (let i = 0; i <= result.length - 1; i++) {
let regex = /^(.*)(/;
let value = result[i].technology.match(regex)[1].trim();
result[i].technology = value;
}
this.props.fetchQuestions(result);
}
In this this.props.data is the reducer data .Now what I want to do was, I have to get some data from this and do some manipulation. but here, it is directly updating my reducer values itself, But I have used spread operator to get the copy of that data ,but still its updating the reducers values.
But I am not getting the value which I expect. Can any one help me with this ?
javascript reactjs redux react-redux
javascript reactjs redux react-redux
asked Nov 14 '18 at 6:14
ganeshkganeshk
24219
asked Nov 14 '18 at 6:14
ganeshkganeshk
24219
edited Nov 14 '18 at 8:55
ganeshk
asked Nov 14 '18 at 6:14
ganeshkganeshk
24219
asked Nov 14 '18 at 6:14
ganeshkganeshk
24219
asked Nov 14 '18 at 6:14
ganeshkganeshk
24219
24219
try this - let regex =/^(w+)/;
– Nikita R.
Nov 14 '18 at 6:21
add a comment |
try this - let regex =/^(w+)/;
– Nikita R.
Nov 14 '18 at 6:21
try this - let regex =/^(w+)/;
– Nikita R.
Nov 14 '18 at 6:21
try this - let regex =/^(w+)/;
– Nikita R.
Nov 14 '18 at 6:21
add a comment |
5 Answers
5
active
oldest
votes
2
You can write a regex that creates a group for the value before ( and then get it from the match
let result= [
{
Id: 'a',
temp: 5,
data: 'abc (c:1'
},
{
Id: 'a',
temp: 5,
data: 'pqr (c:1'
},
{
Id: 'a',
temp: 5,
data: 'xyz (c:1'
}
]
for (let i = 0; i <= result.length - 1; i++) {
let regex = /^(.*)(/;
let value = result[i].data.match(regex)[1].trim();
result[i].data = value
}
console.log(result);answered Nov 14 '18 at 6:19
Shubham KhatriShubham Khatri
81.8k1599138
Using this regex I am getting a bracket in the data value
– ganeshk
Nov 14 '18 at 6:24
and If I use 1 then it gives null. I think it should be 0
– ganeshk
Nov 14 '18 at 6:25
Hey @shubham I got so many solutions, Which one will be more efficient
– ganeshk
Nov 14 '18 at 6:29
@ganeshk, the index will be 1 to get the first group match. Also a regex solution will be more efficient than a string manipulation using split. Any regex from the solutions that satisfies your use case should be good enough for you
– Shubham Khatri
Nov 14 '18 at 6:31
I have just updated my question,On reducer value is getting updated , But It should not get update with
– ganeshk
Nov 14 '18 at 8:53
|
show 7 more comments
1
You need to access the data property of each object while you're iterating, in order to use a .match on it:
let result = [{
Id: 'a',
temp: 5,
data: 'abc (c:1'
}, {
Id: 'a',
temp: 5,
data: 'pqr (c:1'
}, {
Id: 'a',
temp: 5,
data: 'xyz (c:1'
}];
for (let i = 0; i <= result.length - 1; i++) {
const { data } = result[i];
let regex = /^[^(]+/;
let value = data.match(regex)[0].trim();
result[i].data = value;
}
console.log(result);But, seeing as you're using react, you might consider using .map instead, which won't mutate the original objects:
let result = [{
Id: 'a',
temp: 5,
data: 'abc (c:1'
}, {
Id: 'a',
temp: 5,
data: 'pqr (c:1'
}, {
Id: 'a',
temp: 5,
data: 'xyz (c:1'
}];
console.log(
result.map(({ data, ...rest }) => ({
data: data.match(/^[^(]+[^( ]/)[0],
...rest
}))
);Note that if you make the final matched character in the regex [^( ] (match anything but a ( or a space), there's no need for trim anymore.
answered Nov 14 '18 at 6:21
CertainPerformanceCertainPerformance
83.3k144168
add a comment |
0
You can use split('(') instead of regex to get the same output:
let result = [{
Id: 'a',
temp: 5,
data: 'abc (c:1'
}, {
Id: 'a',
temp: 5,
data: 'pqr (c:1'
}, {
Id: 'a',
temp: 5,
data: 'xyz (c:1'
}];
for (let i = 0; i <= result.length - 1; i++) {
let value = result[i].data.split('(')[0].trim();
result[i].data = value
}
console.log(result);answered Nov 14 '18 at 6:18
Ankit AgarwalAnkit Agarwal
23.8k52144
add a comment |
0
Try this one.
result.map(el => {
el.data = el.data.match(/^(w+)/)[0];
return el;
})
answered Nov 14 '18 at 6:23
Nikita R.Nikita R.
1456
add a comment |
0
let result = [ {Id :'a',temp:5,data:'abc (c:1'},{Id :'a',temp:5,data:'pqr (c:1'},{Id :'a',temp:5,data:'xyz (c:1'}]
const new_result = result.map((item) => {
let regex = /^[^(]+/;
let value = item.data.match(regex)[0].trim();
item.data = value;
return item;
});
console.log(new_result);
answered Nov 14 '18 at 6:26
BrandonBrandon
111
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
You can write a regex that creates a group for the value before ( and then get it from the match
let result= [
{
Id: 'a',
temp: 5,
data: 'abc (c:1'
},
{
Id: 'a',
temp: 5,
data: 'pqr (c:1'
},
{
Id: 'a',
temp: 5,
data: 'xyz (c:1'
}
]
for (let i = 0; i <= result.length - 1; i++) {
let regex = /^(.*)(/;
let value = result[i].data.match(regex)[1].trim();
result[i].data = value
}
console.log(result);answered Nov 14 '18 at 6:19
Shubham KhatriShubham Khatri
81.8k1599138
Using this regex I am getting a bracket in the data value
– ganeshk
Nov 14 '18 at 6:24
and If I use 1 then it gives null. I think it should be 0
– ganeshk
Nov 14 '18 at 6:25
Hey @shubham I got so many solutions, Which one will be more efficient
– ganeshk
Nov 14 '18 at 6:29
@ganeshk, the index will be 1 to get the first group match. Also a regex solution will be more efficient than a string manipulation using split. Any regex from the solutions that satisfies your use case should be good enough for you
– Shubham Khatri
Nov 14 '18 at 6:31
I have just updated my question,On reducer value is getting updated , But It should not get update with
– ganeshk
Nov 14 '18 at 8:53
|
show 7 more comments
2
You can write a regex that creates a group for the value before ( and then get it from the match
let result= [
{
Id: 'a',
temp: 5,
data: 'abc (c:1'
},
{
Id: 'a',
temp: 5,
data: 'pqr (c:1'
},
{
Id: 'a',
temp: 5,
data: 'xyz (c:1'
}
]
for (let i = 0; i <= result.length - 1; i++) {
let regex = /^(.*)(/;
let value = result[i].data.match(regex)[1].trim();
result[i].data = value
}
console.log(result);answered Nov 14 '18 at 6:19
Shubham KhatriShubham Khatri
81.8k1599138
Using this regex I am getting a bracket in the data value
– ganeshk
Nov 14 '18 at 6:24
and If I use 1 then it gives null. I think it should be 0
– ganeshk
Nov 14 '18 at 6:25
Hey @shubham I got so many solutions, Which one will be more efficient
– ganeshk
Nov 14 '18 at 6:29
@ganeshk, the index will be 1 to get the first group match. Also a regex solution will be more efficient than a string manipulation using split. Any regex from the solutions that satisfies your use case should be good enough for you
– Shubham Khatri
Nov 14 '18 at 6:31
I have just updated my question,On reducer value is getting updated , But It should not get update with
– ganeshk
Nov 14 '18 at 8:53
|
show 7 more comments
2
2
2
You can write a regex that creates a group for the value before ( and then get it from the match
let result= [
{
Id: 'a',
temp: 5,
data: 'abc (c:1'
},
{
Id: 'a',
temp: 5,
data: 'pqr (c:1'
},
{
Id: 'a',
temp: 5,
data: 'xyz (c:1'
}
]
for (let i = 0; i <= result.length - 1; i++) {
let regex = /^(.*)(/;
let value = result[i].data.match(regex)[1].trim();
result[i].data = value
}
console.log(result);answered Nov 14 '18 at 6:19
Shubham KhatriShubham Khatri
81.8k1599138
You can write a regex that creates a group for the value before ( and then get it from the match
let result= [
{
Id: 'a',
temp: 5,
data: 'abc (c:1'
},
{
Id: 'a',
temp: 5,
data: 'pqr (c:1'
},
{
Id: 'a',
temp: 5,
data: 'xyz (c:1'
}
]
for (let i = 0; i <= result.length - 1; i++) {
let regex = /^(.*)(/;
let value = result[i].data.match(regex)[1].trim();
result[i].data = value
}
console.log(result);let result= [
{
Id: 'a',
temp: 5,
data: 'abc (c:1'
},
{
Id: 'a',
temp: 5,
data: 'pqr (c:1'
},
{
Id: 'a',
temp: 5,
data: 'xyz (c:1'
}
]
for (let i = 0; i <= result.length - 1; i++) {
let regex = /^(.*)(/;
let value = result[i].data.match(regex)[1].trim();
result[i].data = value
}
console.log(result);let result= [
{
Id: 'a',
temp: 5,
data: 'abc (c:1'
},
{
Id: 'a',
temp: 5,
data: 'pqr (c:1'
},
{
Id: 'a',
temp: 5,
data: 'xyz (c:1'
}
]
for (let i = 0; i <= result.length - 1; i++) {
let regex = /^(.*)(/;
let value = result[i].data.match(regex)[1].trim();
result[i].data = value
}
console.log(result);answered Nov 14 '18 at 6:19
Shubham KhatriShubham Khatri
81.8k1599138
edited Nov 14 '18 at 6:33
answered Nov 14 '18 at 6:19
Shubham KhatriShubham Khatri
81.8k1599138
answered Nov 14 '18 at 6:19
Shubham KhatriShubham Khatri
81.8k1599138
answered Nov 14 '18 at 6:19
Shubham KhatriShubham Khatri
81.8k1599138
81.8k1599138
Using this regex I am getting a bracket in the data value
– ganeshk
Nov 14 '18 at 6:24
and If I use 1 then it gives null. I think it should be 0
– ganeshk
Nov 14 '18 at 6:25
Hey @shubham I got so many solutions, Which one will be more efficient
– ganeshk
Nov 14 '18 at 6:29
@ganeshk, the index will be 1 to get the first group match. Also a regex solution will be more efficient than a string manipulation using split. Any regex from the solutions that satisfies your use case should be good enough for you
– Shubham Khatri
Nov 14 '18 at 6:31
I have just updated my question,On reducer value is getting updated , But It should not get update with
– ganeshk
Nov 14 '18 at 8:53
|
show 7 more comments
Using this regex I am getting a bracket in the data value
– ganeshk
Nov 14 '18 at 6:24
and If I use 1 then it gives null. I think it should be 0
– ganeshk
Nov 14 '18 at 6:25
Hey @shubham I got so many solutions, Which one will be more efficient
– ganeshk
Nov 14 '18 at 6:29
@ganeshk, the index will be 1 to get the first group match. Also a regex solution will be more efficient than a string manipulation using split. Any regex from the solutions that satisfies your use case should be good enough for you
– Shubham Khatri
Nov 14 '18 at 6:31
I have just updated my question,On reducer value is getting updated , But It should not get update with
– ganeshk
Nov 14 '18 at 8:53
Using this regex I am getting a bracket in the data value
– ganeshk
Nov 14 '18 at 6:24
Using this regex I am getting a bracket in the data value
– ganeshk
Nov 14 '18 at 6:24
and If I use 1 then it gives null. I think it should be 0
– ganeshk
Nov 14 '18 at 6:25
and If I use 1 then it gives null. I think it should be 0
– ganeshk
Nov 14 '18 at 6:25
Hey @shubham I got so many solutions, Which one will be more efficient
– ganeshk
Nov 14 '18 at 6:29
Hey @shubham I got so many solutions, Which one will be more efficient
– ganeshk
Nov 14 '18 at 6:29
@ganeshk, the index will be 1 to get the first group match. Also a regex solution will be more efficient than a string manipulation using split. Any regex from the solutions that satisfies your use case should be good enough for you
– Shubham Khatri
Nov 14 '18 at 6:31
@ganeshk, the index will be 1 to get the first group match. Also a regex solution will be more efficient than a string manipulation using split. Any regex from the solutions that satisfies your use case should be good enough for you
– Shubham Khatri
Nov 14 '18 at 6:31
I have just updated my question,On reducer value is getting updated , But It should not get update with
– ganeshk
Nov 14 '18 at 8:53
I have just updated my question,On reducer value is getting updated , But It should not get update with
– ganeshk
Nov 14 '18 at 8:53
|
show 7 more comments
1
You need to access the data property of each object while you're iterating, in order to use a .match on it:
let result = [{
Id: 'a',
temp: 5,
data: 'abc (c:1'
}, {
Id: 'a',
temp: 5,
data: 'pqr (c:1'
}, {
Id: 'a',
temp: 5,
data: 'xyz (c:1'
}];
for (let i = 0; i <= result.length - 1; i++) {
const { data } = result[i];
let regex = /^[^(]+/;
let value = data.match(regex)[0].trim();
result[i].data = value;
}
console.log(result);But, seeing as you're using react, you might consider using .map instead, which won't mutate the original objects:
let result = [{
Id: 'a',
temp: 5,
data: 'abc (c:1'
}, {
Id: 'a',
temp: 5,
data: 'pqr (c:1'
}, {
Id: 'a',
temp: 5,
data: 'xyz (c:1'
}];
console.log(
result.map(({ data, ...rest }) => ({
data: data.match(/^[^(]+[^( ]/)[0],
...rest
}))
);Note that if you make the final matched character in the regex [^( ] (match anything but a ( or a space), there's no need for trim anymore.
answered Nov 14 '18 at 6:21
CertainPerformanceCertainPerformance
83.3k144168
add a comment |
1
You need to access the data property of each object while you're iterating, in order to use a .match on it:
let result = [{
Id: 'a',
temp: 5,
data: 'abc (c:1'
}, {
Id: 'a',
temp: 5,
data: 'pqr (c:1'
}, {
Id: 'a',
temp: 5,
data: 'xyz (c:1'
}];
for (let i = 0; i <= result.length - 1; i++) {
const { data } = result[i];
let regex = /^[^(]+/;
let value = data.match(regex)[0].trim();
result[i].data = value;
}
console.log(result);But, seeing as you're using react, you might consider using .map instead, which won't mutate the original objects:
let result = [{
Id: 'a',
temp: 5,
data: 'abc (c:1'
}, {
Id: 'a',
temp: 5,
data: 'pqr (c:1'
}, {
Id: 'a',
temp: 5,
data: 'xyz (c:1'
}];
console.log(
result.map(({ data, ...rest }) => ({
data: data.match(/^[^(]+[^( ]/)[0],
...rest
}))
);Note that if you make the final matched character in the regex [^( ] (match anything but a ( or a space), there's no need for trim anymore.
answered Nov 14 '18 at 6:21
CertainPerformanceCertainPerformance
83.3k144168
add a comment |
1
1
1
You need to access the data property of each object while you're iterating, in order to use a .match on it:
let result = [{
Id: 'a',
temp: 5,
data: 'abc (c:1'
}, {
Id: 'a',
temp: 5,
data: 'pqr (c:1'
}, {
Id: 'a',
temp: 5,
data: 'xyz (c:1'
}];
for (let i = 0; i <= result.length - 1; i++) {
const { data } = result[i];
let regex = /^[^(]+/;
let value = data.match(regex)[0].trim();
result[i].data = value;
}
console.log(result);But, seeing as you're using react, you might consider using .map instead, which won't mutate the original objects:
let result = [{
Id: 'a',
temp: 5,
data: 'abc (c:1'
}, {
Id: 'a',
temp: 5,
data: 'pqr (c:1'
}, {
Id: 'a',
temp: 5,
data: 'xyz (c:1'
}];
console.log(
result.map(({ data, ...rest }) => ({
data: data.match(/^[^(]+[^( ]/)[0],
...rest
}))
);Note that if you make the final matched character in the regex [^( ] (match anything but a ( or a space), there's no need for trim anymore.
answered Nov 14 '18 at 6:21
CertainPerformanceCertainPerformance
83.3k144168
You need to access the data property of each object while you're iterating, in order to use a .match on it:
let result = [{
Id: 'a',
temp: 5,
data: 'abc (c:1'
}, {
Id: 'a',
temp: 5,
data: 'pqr (c:1'
}, {
Id: 'a',
temp: 5,
data: 'xyz (c:1'
}];
for (let i = 0; i <= result.length - 1; i++) {
const { data } = result[i];
let regex = /^[^(]+/;
let value = data.match(regex)[0].trim();
result[i].data = value;
}
console.log(result);But, seeing as you're using react, you might consider using .map instead, which won't mutate the original objects:
let result = [{
Id: 'a',
temp: 5,
data: 'abc (c:1'
}, {
Id: 'a',
temp: 5,
data: 'pqr (c:1'
}, {
Id: 'a',
temp: 5,
data: 'xyz (c:1'
}];
console.log(
result.map(({ data, ...rest }) => ({
data: data.match(/^[^(]+[^( ]/)[0],
...rest
}))
);Note that if you make the final matched character in the regex [^( ] (match anything but a ( or a space), there's no need for trim anymore.
let result = [{
Id: 'a',
temp: 5,
data: 'abc (c:1'
}, {
Id: 'a',
temp: 5,
data: 'pqr (c:1'
}, {
Id: 'a',
temp: 5,
data: 'xyz (c:1'
}];
for (let i = 0; i <= result.length - 1; i++) {
const { data } = result[i];
let regex = /^[^(]+/;
let value = data.match(regex)[0].trim();
result[i].data = value;
}
console.log(result);let result = [{
Id: 'a',
temp: 5,
data: 'abc (c:1'
}, {
Id: 'a',
temp: 5,
data: 'pqr (c:1'
}, {
Id: 'a',
temp: 5,
data: 'xyz (c:1'
}];
for (let i = 0; i <= result.length - 1; i++) {
const { data } = result[i];
let regex = /^[^(]+/;
let value = data.match(regex)[0].trim();
result[i].data = value;
}
console.log(result);let result = [{
Id: 'a',
temp: 5,
data: 'abc (c:1'
}, {
Id: 'a',
temp: 5,
data: 'pqr (c:1'
}, {
Id: 'a',
temp: 5,
data: 'xyz (c:1'
}];
console.log(
result.map(({ data, ...rest }) => ({
data: data.match(/^[^(]+[^( ]/)[0],
...rest
}))
);let result = [{
Id: 'a',
temp: 5,
data: 'abc (c:1'
}, {
Id: 'a',
temp: 5,
data: 'pqr (c:1'
}, {
Id: 'a',
temp: 5,
data: 'xyz (c:1'
}];
console.log(
result.map(({ data, ...rest }) => ({
data: data.match(/^[^(]+[^( ]/)[0],
...rest
}))
);answered Nov 14 '18 at 6:21
CertainPerformanceCertainPerformance
83.3k144168
answered Nov 14 '18 at 6:21
CertainPerformanceCertainPerformance
83.3k144168
answered Nov 14 '18 at 6:21
CertainPerformanceCertainPerformance
83.3k144168
answered Nov 14 '18 at 6:21
CertainPerformanceCertainPerformance
83.3k144168
83.3k144168
add a comment |
add a comment |
0
You can use split('(') instead of regex to get the same output:
let result = [{
Id: 'a',
temp: 5,
data: 'abc (c:1'
}, {
Id: 'a',
temp: 5,
data: 'pqr (c:1'
}, {
Id: 'a',
temp: 5,
data: 'xyz (c:1'
}];
for (let i = 0; i <= result.length - 1; i++) {
let value = result[i].data.split('(')[0].trim();
result[i].data = value
}
console.log(result);answered Nov 14 '18 at 6:18
Ankit AgarwalAnkit Agarwal
23.8k52144
add a comment |
0
You can use split('(') instead of regex to get the same output:
let result = [{
Id: 'a',
temp: 5,
data: 'abc (c:1'
}, {
Id: 'a',
temp: 5,
data: 'pqr (c:1'
}, {
Id: 'a',
temp: 5,
data: 'xyz (c:1'
}];
for (let i = 0; i <= result.length - 1; i++) {
let value = result[i].data.split('(')[0].trim();
result[i].data = value
}
console.log(result);answered Nov 14 '18 at 6:18
Ankit AgarwalAnkit Agarwal
23.8k52144
add a comment |
0
0
0
You can use split('(') instead of regex to get the same output:
let result = [{
Id: 'a',
temp: 5,
data: 'abc (c:1'
}, {
Id: 'a',
temp: 5,
data: 'pqr (c:1'
}, {
Id: 'a',
temp: 5,
data: 'xyz (c:1'
}];
for (let i = 0; i <= result.length - 1; i++) {
let value = result[i].data.split('(')[0].trim();
result[i].data = value
}
console.log(result);answered Nov 14 '18 at 6:18
Ankit AgarwalAnkit Agarwal
23.8k52144
You can use split('(') instead of regex to get the same output:
let result = [{
Id: 'a',
temp: 5,
data: 'abc (c:1'
}, {
Id: 'a',
temp: 5,
data: 'pqr (c:1'
}, {
Id: 'a',
temp: 5,
data: 'xyz (c:1'
}];
for (let i = 0; i <= result.length - 1; i++) {
let value = result[i].data.split('(')[0].trim();
result[i].data = value
}
console.log(result);let result = [{
Id: 'a',
temp: 5,
data: 'abc (c:1'
}, {
Id: 'a',
temp: 5,
data: 'pqr (c:1'
}, {
Id: 'a',
temp: 5,
data: 'xyz (c:1'
}];
for (let i = 0; i <= result.length - 1; i++) {
let value = result[i].data.split('(')[0].trim();
result[i].data = value
}
console.log(result);let result = [{
Id: 'a',
temp: 5,
data: 'abc (c:1'
}, {
Id: 'a',
temp: 5,
data: 'pqr (c:1'
}, {
Id: 'a',
temp: 5,
data: 'xyz (c:1'
}];
for (let i = 0; i <= result.length - 1; i++) {
let value = result[i].data.split('(')[0].trim();
result[i].data = value
}
console.log(result);answered Nov 14 '18 at 6:18
Ankit AgarwalAnkit Agarwal
23.8k52144
answered Nov 14 '18 at 6:18
Ankit AgarwalAnkit Agarwal
23.8k52144
answered Nov 14 '18 at 6:18
Ankit AgarwalAnkit Agarwal
23.8k52144
answered Nov 14 '18 at 6:18
Ankit AgarwalAnkit Agarwal
23.8k52144
23.8k52144
add a comment |
add a comment |
0
Try this one.
result.map(el => {
el.data = el.data.match(/^(w+)/)[0];
return el;
})
answered Nov 14 '18 at 6:23
Nikita R.Nikita R.
1456
add a comment |
0
Try this one.
result.map(el => {
el.data = el.data.match(/^(w+)/)[0];
return el;
})
answered Nov 14 '18 at 6:23
Nikita R.Nikita R.
1456
add a comment |
0
0
0
Try this one.
result.map(el => {
el.data = el.data.match(/^(w+)/)[0];
return el;
})
answered Nov 14 '18 at 6:23
Nikita R.Nikita R.
1456
Try this one.
result.map(el => {
el.data = el.data.match(/^(w+)/)[0];
return el;
})
answered Nov 14 '18 at 6:23
Nikita R.Nikita R.
1456
answered Nov 14 '18 at 6:23
Nikita R.Nikita R.
1456
answered Nov 14 '18 at 6:23
Nikita R.Nikita R.
1456
answered Nov 14 '18 at 6:23
Nikita R.Nikita R.
1456
1456
add a comment |
add a comment |
0
let result = [ {Id :'a',temp:5,data:'abc (c:1'},{Id :'a',temp:5,data:'pqr (c:1'},{Id :'a',temp:5,data:'xyz (c:1'}]
const new_result = result.map((item) => {
let regex = /^[^(]+/;
let value = item.data.match(regex)[0].trim();
item.data = value;
return item;
});
console.log(new_result);
answered Nov 14 '18 at 6:26
BrandonBrandon
111
add a comment |
0
let result = [ {Id :'a',temp:5,data:'abc (c:1'},{Id :'a',temp:5,data:'pqr (c:1'},{Id :'a',temp:5,data:'xyz (c:1'}]
const new_result = result.map((item) => {
let regex = /^[^(]+/;
let value = item.data.match(regex)[0].trim();
item.data = value;
return item;
});
console.log(new_result);
answered Nov 14 '18 at 6:26
BrandonBrandon
111
add a comment |
0
0
0
let result = [ {Id :'a',temp:5,data:'abc (c:1'},{Id :'a',temp:5,data:'pqr (c:1'},{Id :'a',temp:5,data:'xyz (c:1'}]
const new_result = result.map((item) => {
let regex = /^[^(]+/;
let value = item.data.match(regex)[0].trim();
item.data = value;
return item;
});
console.log(new_result);
answered Nov 14 '18 at 6:26
BrandonBrandon
111
let result = [ {Id :'a',temp:5,data:'abc (c:1'},{Id :'a',temp:5,data:'pqr (c:1'},{Id :'a',temp:5,data:'xyz (c:1'}]
const new_result = result.map((item) => {
let regex = /^[^(]+/;
let value = item.data.match(regex)[0].trim();
item.data = value;
return item;
});
console.log(new_result);
answered Nov 14 '18 at 6:26
BrandonBrandon
111
answered Nov 14 '18 at 6:26
BrandonBrandon
111
answered Nov 14 '18 at 6:26
BrandonBrandon
111
answered Nov 14 '18 at 6:26
BrandonBrandon
111
111
add a comment |
add a comment |
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try this - let regex =/^(w+)/;
– Nikita R.
Nov 14 '18 at 6:21