JavaScript let if else statement args
0
I'm trying to define a let in JavaScript so that if there's no arg[2], then it will just use arg[1], but if there is arg[1] and arg[2] it will use them both.
ie. command: !hello world
then: let cmd = arg[1]
but also
command: !hello world one
then: let cmd = arg[1] + " " + arg[2]
I have this, however it only works if arg[2] exists, and doesn't if only arg[1] is supplied:
let cmd = args[1] + " " + args[2];
Is there anyway you can specify an arg to pass through spaces? So !hello world one, would be arg[1] arg[2] (arg[2] being: world one)
javascript let
edited Nov 14 '18 at 1:58
Pedro Lobito
48.4k14133164
asked Nov 14 '18 at 1:51
ah1ah1
168
add a comment |
0
I'm trying to define a let in JavaScript so that if there's no arg[2], then it will just use arg[1], but if there is arg[1] and arg[2] it will use them both.
ie. command: !hello world
then: let cmd = arg[1]
but also
command: !hello world one
then: let cmd = arg[1] + " " + arg[2]
I have this, however it only works if arg[2] exists, and doesn't if only arg[1] is supplied:
let cmd = args[1] + " " + args[2];
Is there anyway you can specify an arg to pass through spaces? So !hello world one, would be arg[1] arg[2] (arg[2] being: world one)
javascript let
edited Nov 14 '18 at 1:58
Pedro Lobito
48.4k14133164
asked Nov 14 '18 at 1:51
ah1ah1
168
1
You seem to be asking two very different questions: (1) how to deal with an optional second argument and (2) how to pass an argument that contains spaces. It's not clear if you're asking the second question as a way of dealing with the first or if it's a separate issue. The first issue can be handled with a ternary expression:let cmd = arg[2] === undefined ? arg[1] + " " + arg[2] : arg[1];.
– Ted Hopp
Nov 14 '18 at 1:58
1
let cmd = args.join(" ");would suffice if you didn't mind them possibly supplying more than two arguments.
– Tyler Roper
Nov 14 '18 at 2:02
Thank you. @TedHopp, how would I go about the second, where args[3] would contain everything from args[3] and onward?
– ah1
Nov 14 '18 at 18:58
1
You can use what @Tyler suggested and collect all remaining arguments with something likeargs.slice(3).join(" ")(which would create a single string containing all elements ofargfrom index 3 onward, separated by a space).
– Ted Hopp
Nov 14 '18 at 19:36
add a comment |
0
0
0
I'm trying to define a let in JavaScript so that if there's no arg[2], then it will just use arg[1], but if there is arg[1] and arg[2] it will use them both.
ie. command: !hello world
then: let cmd = arg[1]
but also
command: !hello world one
then: let cmd = arg[1] + " " + arg[2]
I have this, however it only works if arg[2] exists, and doesn't if only arg[1] is supplied:
let cmd = args[1] + " " + args[2];
Is there anyway you can specify an arg to pass through spaces? So !hello world one, would be arg[1] arg[2] (arg[2] being: world one)
javascript let
edited Nov 14 '18 at 1:58
Pedro Lobito
48.4k14133164
asked Nov 14 '18 at 1:51
ah1ah1
168
I'm trying to define a let in JavaScript so that if there's no arg[2], then it will just use arg[1], but if there is arg[1] and arg[2] it will use them both.
ie. command: !hello world
then: let cmd = arg[1]
but also
command: !hello world one
then: let cmd = arg[1] + " " + arg[2]
I have this, however it only works if arg[2] exists, and doesn't if only arg[1] is supplied:
let cmd = args[1] + " " + args[2];
Is there anyway you can specify an arg to pass through spaces? So !hello world one, would be arg[1] arg[2] (arg[2] being: world one)
javascript let
javascript let
edited Nov 14 '18 at 1:58
Pedro Lobito
48.4k14133164
asked Nov 14 '18 at 1:51
ah1ah1
168
edited Nov 14 '18 at 1:58
Pedro Lobito
48.4k14133164
asked Nov 14 '18 at 1:51
ah1ah1
168
edited Nov 14 '18 at 1:58
Pedro Lobito
48.4k14133164
edited Nov 14 '18 at 1:58
Pedro Lobito
48.4k14133164
edited Nov 14 '18 at 1:58
Pedro Lobito
48.4k14133164
48.4k14133164
asked Nov 14 '18 at 1:51
ah1ah1
168
asked Nov 14 '18 at 1:51
ah1ah1
168
asked Nov 14 '18 at 1:51
ah1ah1
168
168
1
You seem to be asking two very different questions: (1) how to deal with an optional second argument and (2) how to pass an argument that contains spaces. It's not clear if you're asking the second question as a way of dealing with the first or if it's a separate issue. The first issue can be handled with a ternary expression:let cmd = arg[2] === undefined ? arg[1] + " " + arg[2] : arg[1];.
– Ted Hopp
Nov 14 '18 at 1:58
1
let cmd = args.join(" ");would suffice if you didn't mind them possibly supplying more than two arguments.
– Tyler Roper
Nov 14 '18 at 2:02
Thank you. @TedHopp, how would I go about the second, where args[3] would contain everything from args[3] and onward?
– ah1
Nov 14 '18 at 18:58
1
You can use what @Tyler suggested and collect all remaining arguments with something likeargs.slice(3).join(" ")(which would create a single string containing all elements ofargfrom index 3 onward, separated by a space).
– Ted Hopp
Nov 14 '18 at 19:36
add a comment |
1
You seem to be asking two very different questions: (1) how to deal with an optional second argument and (2) how to pass an argument that contains spaces. It's not clear if you're asking the second question as a way of dealing with the first or if it's a separate issue. The first issue can be handled with a ternary expression:let cmd = arg[2] === undefined ? arg[1] + " " + arg[2] : arg[1];.
– Ted Hopp
Nov 14 '18 at 1:58
1
let cmd = args.join(" ");would suffice if you didn't mind them possibly supplying more than two arguments.
– Tyler Roper
Nov 14 '18 at 2:02
Thank you. @TedHopp, how would I go about the second, where args[3] would contain everything from args[3] and onward?
– ah1
Nov 14 '18 at 18:58
1
You can use what @Tyler suggested and collect all remaining arguments with something likeargs.slice(3).join(" ")(which would create a single string containing all elements ofargfrom index 3 onward, separated by a space).
– Ted Hopp
Nov 14 '18 at 19:36
1
1
You seem to be asking two very different questions: (1) how to deal with an optional second argument and (2) how to pass an argument that contains spaces. It's not clear if you're asking the second question as a way of dealing with the first or if it's a separate issue. The first issue can be handled with a ternary expression:
let cmd = arg[2] === undefined ? arg[1] + " " + arg[2] : arg[1];.– Ted Hopp
Nov 14 '18 at 1:58
You seem to be asking two very different questions: (1) how to deal with an optional second argument and (2) how to pass an argument that contains spaces. It's not clear if you're asking the second question as a way of dealing with the first or if it's a separate issue. The first issue can be handled with a ternary expression:
let cmd = arg[2] === undefined ? arg[1] + " " + arg[2] : arg[1];.– Ted Hopp
Nov 14 '18 at 1:58
1
1
let cmd = args.join(" "); would suffice if you didn't mind them possibly supplying more than two arguments.– Tyler Roper
Nov 14 '18 at 2:02
let cmd = args.join(" "); would suffice if you didn't mind them possibly supplying more than two arguments.– Tyler Roper
Nov 14 '18 at 2:02
Thank you. @TedHopp, how would I go about the second, where args[3] would contain everything from args[3] and onward?
– ah1
Nov 14 '18 at 18:58
Thank you. @TedHopp, how would I go about the second, where args[3] would contain everything from args[3] and onward?
– ah1
Nov 14 '18 at 18:58
1
1
You can use what @Tyler suggested and collect all remaining arguments with something like
args.slice(3).join(" ") (which would create a single string containing all elements of arg from index 3 onward, separated by a space).– Ted Hopp
Nov 14 '18 at 19:36
You can use what @Tyler suggested and collect all remaining arguments with something like
args.slice(3).join(" ") (which would create a single string containing all elements of arg from index 3 onward, separated by a space).– Ted Hopp
Nov 14 '18 at 19:36
add a comment |
3 Answers
3
active
oldest
votes
0
I usually do something like:
const myFunction = (argA, argB) => {
if (argB === undefined) argB = "";
console.log(argA + " " + argB);
return "" + argA + " " + argB
}
answered Nov 14 '18 at 1:55
FallenreaperFallenreaper
3,95883483
add a comment |
0
You mean like:
let test = arg[1] && arg[2] ? arg[1]+' '+arg[1] : arg[1];
answered Nov 14 '18 at 2:01
Yoannes GeisslerYoannes Geissler
81118
add a comment |
0
If you are using ES6 you can use the rest operator like so:
const myFunction = (...args) => {
console.log(args[0])
}
if you don't use the fat arrow syntax you can refer to the arguments key word:
function func1(a, b, c) {
if(arguments && arguments.length == 2){
console.log('do a thing')
}else{
console.log('do a different thing')
}
};
answered Nov 14 '18 at 2:06
ruby_newbieruby_newbie
2,30521120
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
I usually do something like:
const myFunction = (argA, argB) => {
if (argB === undefined) argB = "";
console.log(argA + " " + argB);
return "" + argA + " " + argB
}
answered Nov 14 '18 at 1:55
FallenreaperFallenreaper
3,95883483
add a comment |
0
I usually do something like:
const myFunction = (argA, argB) => {
if (argB === undefined) argB = "";
console.log(argA + " " + argB);
return "" + argA + " " + argB
}
answered Nov 14 '18 at 1:55
FallenreaperFallenreaper
3,95883483
add a comment |
0
0
0
I usually do something like:
const myFunction = (argA, argB) => {
if (argB === undefined) argB = "";
console.log(argA + " " + argB);
return "" + argA + " " + argB
}
answered Nov 14 '18 at 1:55
FallenreaperFallenreaper
3,95883483
I usually do something like:
const myFunction = (argA, argB) => {
if (argB === undefined) argB = "";
console.log(argA + " " + argB);
return "" + argA + " " + argB
}
answered Nov 14 '18 at 1:55
FallenreaperFallenreaper
3,95883483
answered Nov 14 '18 at 1:55
FallenreaperFallenreaper
3,95883483
answered Nov 14 '18 at 1:55
FallenreaperFallenreaper
3,95883483
answered Nov 14 '18 at 1:55
FallenreaperFallenreaper
3,95883483
3,95883483
add a comment |
add a comment |
0
You mean like:
let test = arg[1] && arg[2] ? arg[1]+' '+arg[1] : arg[1];
answered Nov 14 '18 at 2:01
Yoannes GeisslerYoannes Geissler
81118
add a comment |
0
You mean like:
let test = arg[1] && arg[2] ? arg[1]+' '+arg[1] : arg[1];
answered Nov 14 '18 at 2:01
Yoannes GeisslerYoannes Geissler
81118
add a comment |
0
0
0
You mean like:
let test = arg[1] && arg[2] ? arg[1]+' '+arg[1] : arg[1];
answered Nov 14 '18 at 2:01
Yoannes GeisslerYoannes Geissler
81118
You mean like:
let test = arg[1] && arg[2] ? arg[1]+' '+arg[1] : arg[1];
answered Nov 14 '18 at 2:01
Yoannes GeisslerYoannes Geissler
81118
answered Nov 14 '18 at 2:01
Yoannes GeisslerYoannes Geissler
81118
answered Nov 14 '18 at 2:01
Yoannes GeisslerYoannes Geissler
81118
answered Nov 14 '18 at 2:01
Yoannes GeisslerYoannes Geissler
81118
81118
add a comment |
add a comment |
0
If you are using ES6 you can use the rest operator like so:
const myFunction = (...args) => {
console.log(args[0])
}
if you don't use the fat arrow syntax you can refer to the arguments key word:
function func1(a, b, c) {
if(arguments && arguments.length == 2){
console.log('do a thing')
}else{
console.log('do a different thing')
}
};
answered Nov 14 '18 at 2:06
ruby_newbieruby_newbie
2,30521120
add a comment |
0
If you are using ES6 you can use the rest operator like so:
const myFunction = (...args) => {
console.log(args[0])
}
if you don't use the fat arrow syntax you can refer to the arguments key word:
function func1(a, b, c) {
if(arguments && arguments.length == 2){
console.log('do a thing')
}else{
console.log('do a different thing')
}
};
answered Nov 14 '18 at 2:06
ruby_newbieruby_newbie
2,30521120
add a comment |
0
0
0
If you are using ES6 you can use the rest operator like so:
const myFunction = (...args) => {
console.log(args[0])
}
if you don't use the fat arrow syntax you can refer to the arguments key word:
function func1(a, b, c) {
if(arguments && arguments.length == 2){
console.log('do a thing')
}else{
console.log('do a different thing')
}
};
answered Nov 14 '18 at 2:06
ruby_newbieruby_newbie
2,30521120
If you are using ES6 you can use the rest operator like so:
const myFunction = (...args) => {
console.log(args[0])
}
if you don't use the fat arrow syntax you can refer to the arguments key word:
function func1(a, b, c) {
if(arguments && arguments.length == 2){
console.log('do a thing')
}else{
console.log('do a different thing')
}
};
answered Nov 14 '18 at 2:06
ruby_newbieruby_newbie
2,30521120
edited Nov 14 '18 at 2:11
answered Nov 14 '18 at 2:06
ruby_newbieruby_newbie
2,30521120
answered Nov 14 '18 at 2:06
ruby_newbieruby_newbie
2,30521120
answered Nov 14 '18 at 2:06
ruby_newbieruby_newbie
2,30521120
2,30521120
add a comment |
add a comment |
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1
You seem to be asking two very different questions: (1) how to deal with an optional second argument and (2) how to pass an argument that contains spaces. It's not clear if you're asking the second question as a way of dealing with the first or if it's a separate issue. The first issue can be handled with a ternary expression:
let cmd = arg[2] === undefined ? arg[1] + " " + arg[2] : arg[1];.– Ted Hopp
Nov 14 '18 at 1:58
1
let cmd = args.join(" ");would suffice if you didn't mind them possibly supplying more than two arguments.– Tyler Roper
Nov 14 '18 at 2:02
Thank you. @TedHopp, how would I go about the second, where args[3] would contain everything from args[3] and onward?
– ah1
Nov 14 '18 at 18:58
1
You can use what @Tyler suggested and collect all remaining arguments with something like
args.slice(3).join(" ")(which would create a single string containing all elements ofargfrom index 3 onward, separated by a space).– Ted Hopp
Nov 14 '18 at 19:36