Prepend text to beginning of string

Question

What is the fastest method, to add a new value at the beginning of a string?

score 318 · Accepted Answer · edited Mar 16 '17 at 14:15

318

var mystr = "Doe";
mystr = "John " + mystr;

Wouldn't this work for you?

edited Mar 16 '17 at 14:15

T J

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answered May 23 '11 at 07:30

Thor Jacobsen

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Wait wouldn't using += work as well or would it add it to the end – Aug 27 '19 at 10:04
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@JuicY_Burrito `+=` Appends to the end of the text. – Mark Kramer Nov 26 '19 at 20:29

Gabriele Petrioli · Answer 2 · 2017-06-23T10:36:35.257

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You could do it this way ..

var mystr = 'is my name.';
mystr = mystr.replace (/^/,'John ');

console.log(mystr);

disclaimer: http://xkcd.com/208/

Wait, forgot to escape a space. Wheeeeee[taptaptap]eeeeee.

edited Jun 23 '17 at 10:36

answered May 23 '11 at 07:32

Gabriele Petrioli

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I am tempted to upvote this for the xkcd strip. Great stuff! Your elegant solution is most probably slower because it instantiates the regular expressions processor. – Rolf Nov 10 '17 at 00:40
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@Rolf not most probably. It is **certainly** slower than the `prestring + 'original string';` solution. – Gabriele Petrioli Nov 10 '17 at 07:17
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@GabrielePetrioli The advantage is that you can just add to your dot chain ( pipeline ). You can even append something, like so mystr.replace (/$/,' by Gabriele'). It might be slower but it's exactly what I was looking for. – Mateja Petrovic Aug 09 '18 at 00:23
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Not to worry. https://jsperf.com/prepend-text-to-string/5 It is just around 50-100 times slower than simple string+string. But you still can use your regular expression skills. That outweighs the slowdown, right? – metalim Apr 22 '19 at 11:29

score 82 · Answer 3 · answered Mar 29 '19 at 01:04

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Since the question is about what is the fastest method, I thought I'd throw up add some perf metrics.

TL;DR The winner, by a wide margin, is the + operator, and please never use regex

https://jsperf.com/prepend-text-to-string/1

answered Mar 29 '19 at 01:04

KyleMit

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Since we want to test *prepend*, and not just concatenate, I've updated the tests: https://jsperf.com/prepend-text-to-string/5 – metalim Apr 22 '19 at 11:32
@metalim, were there any material changes of note that you made to the tests? A prepend is just the ordered half of string concatenation. Also, for jsPerf, you don't need to throw the test cases in a loop; jsPerf call each method thousands of times to get a baseline of scores. Perf tests optimize particular scenarios, and I'm not sure how common it prepend a the same string 1,000 times. – KyleMit Apr 22 '19 at 18:08
There can be significant difference if you append short string to long, or long string to short. Since we're interested in *prepending* short string to arbitrary string, that is what is done in tests: prepending short string to increasing longer string. Loops are there to test 1000 different lengths of the suffix string, to avoid testing just short or just long. @KyleMit, if you have better idea, feel free to implement it in next test iteration. – metalim Apr 23 '19 at 08:38

score 17 · Answer 4 · answered Dec 20 '16 at 14:12

17

ES6:

let after = 'something after';
let text = `before text ${after}`;

answered Dec 20 '16 at 14:12

Griffi

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While this code snippet may solve the question, [including an explanation](//meta.stackexchange.com/questions/114762/explaining-entirely-code-based-answers) really helps to improve the quality of your post. Remember that you are answering the question for readers in the future, and those people might not know the reasons for your code suggestion. Please also try not to crowd your code with explanatory comments, this reduces the readability of both the code and the explanations! – kayess Dec 20 '16 at 14:41
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ES6: String Interpolation – Bryan Grace Nov 29 '17 at 19:58

score 8 · Answer 5 · edited Aug 15 '16 at 11:17

8

you could also do it this way

"".concat("x","y")

edited Aug 15 '16 at 11:17

hjpotter92

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answered Aug 15 '16 at 11:15

chirag

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and why not "x".concat("y")? Just to add another string instantiation and make it slower? – metalim Apr 22 '19 at 11:31

score 7 · Answer 6 · edited Nov 11 '19 at 10:58

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If you want to use the version of Javascript called ES 2015 (aka ES6) or later, you can use template strings introduced by ES 2015 and recommended by some guidelines (like Airbnb's style guide):

const after = "test";
const mystr = `This is: ${after}`;

edited Nov 11 '19 at 10:58

Boris Verkhovskiy

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answered Nov 16 '16 at 17:01

Ilan Schemoul

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score 2 · Answer 7 · answered Sep 12 '18 at 14:25

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Another option would be to use join

var mystr = "Matayoshi";
mystr = ["Mariano", mystr].join(' ');

answered Sep 12 '18 at 14:25

MatayoshiMariano

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score 1 · Answer 8 · edited Aug 27 '22 at 16:48

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You can use padStart like this:

'ello'.padStart(5,'h');
//padStart(maxLength, fillString)
//output: hello

See MDN: String.prototype.padStart()

edited Aug 27 '22 at 16:48

som

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answered May 26 '22 at 21:45

Mustafa Yasin

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Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers [in the help center](/help/how-to-answer). – Community May 27 '22 at 07:12
This is actually a nice (functional) answer ... e.g. it allows you to do `a = b?.padStart(b.length + 1, '/')` ... which is equivalent (i think) to the less composable ```a = b && `/${b}` ``` – som Aug 27 '22 at 02:27

score 0 · Answer 9 · edited Aug 12 '12 at 03:40

0

You can use

var mystr = "Doe";
mystr = "John " + mystr;
console.log(mystr)

edited Aug 12 '12 at 03:40

skynet

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answered May 23 '11 at 07:34

Mayank

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Prepend text to beginning of string

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