split string only on first instance of specified character

Question

In my code I split a string based on _ and grab the second item in the array.

var element = $(this).attr('class');
var field = element.split('_')[1];

Takes good_luck and provides me with luck. Works great!

But, now I have a class that looks like good_luck_buddy. How do I get my javascript to ignore the second _ and give me luck_buddy?

I found this var field = element.split(new char [] {'_'}, 2); in a c# stackoverflow answer but it doesn't work. I tried it over at jsFiddle...

Answer 1

Use capturing parentheses:

'good_luck_buddy'.split(/_(.*)/s)
['good', 'luck_buddy', ''] // ignore the third element

They are defined as

If separator contains capturing parentheses, matched results are returned in the array.

So in this case we want to split at _.* (i.e. split separator being a sub string starting with _) but also let the result contain some part of our separator (i.e. everything after _).

In this example our separator (matching _(.*)) is _luck_buddy and the captured group (within the separator) is lucky_buddy. Without the capturing parenthesis the luck_buddy (matching .*) would've not been included in the result array as it is the case with simple split that separators are not included in the result.

We use the s regex flag to make . match on newline (\n) characters as well, otherwise it would only split to the first newline.

Answer 2

What do you need regular expressions and arrays for?

myString = myString.substring(myString.indexOf('_')+1)

var myString= "hello_there_how_are_you"
myString = myString.substring(myString.indexOf('_')+1)
console.log(myString)

Answer 3

With help of destructuring assignment it can be more readable:

let [first, ...rest] = "good_luck_buddy".split('_')
rest = rest.join('_')

Answer 4

I avoid RegExp at all costs. Here is another thing you can do:

"good_luck_buddy".split('_').slice(1).join('_')

Answer 5

A simple ES6 way to get both the first key and remaining parts in a string would be:

 const [key, ...rest] = "good_luck_buddy".split('_')
 const value = rest.join('_')
 console.log(key, value) // good, luck_buddy

Answer 6

Nowadays String.prototype.split does indeed allow you to limit the number of splits.

str.split([separator[, limit]])

...

limit Optional

A non-negative integer limiting the number of splits. If provided, splits the string at each occurrence of the specified separator, but stops when limit entries have been placed in the array. Any leftover text is not included in the array at all.

The array may contain fewer entries than limit if the end of the string is reached before the limit is reached. If limit is 0, no splitting is performed.

caveat

It might not work the way you expect. I was hoping it would just ignore the rest of the delimiters, but instead, when it reaches the limit, it splits the remaining string again, omitting the part after the split from the return results.

let str = 'A_B_C_D_E'
const limit_2 = str.split('_', 2)
limit_2
(2) ["A", "B"]
const limit_3 = str.split('_', 3)
limit_3
(3) ["A", "B", "C"]

I was hoping for:

let str = 'A_B_C_D_E'
const limit_2 = str.split('_', 2)
limit_2
(2) ["A", "B_C_D_E"]
const limit_3 = str.split('_', 3)
limit_3
(3) ["A", "B", "C_D_E"]

Answer 7

This solution worked for me

var str = "good_luck_buddy";
var index = str.indexOf('_');
var arr = [str.slice(0, index), str.slice(index + 1)];

//arr[0] = "good"
//arr[1] = "luck_buddy"

OR

var str = "good_luck_buddy";
var index = str.indexOf('_');
var [first, second] = [str.slice(0, index), str.slice(index + 1)];

//first = "good"
//second = "luck_buddy"

Answer 8

You can use the regular expression like:

var arr = element.split(/_(.*)/)

You can use the second parameter which specifies the limit of the split. i.e: var field = element.split('_', 1)[1];

Answer 9

Replace the first instance with a unique placeholder then split from there.

"good_luck_buddy".replace(/\_/,'&').split('&')

["good","luck_buddy"]

This is more useful when both sides of the split are needed.

Answer 10

I need the two parts of string, so, regex lookbehind help me with this.

const full_name = 'Maria do Bairro';
const [first_name, last_name] = full_name.split(/(?<=^[^ ]+) /);
console.log(first_name);
console.log(last_name);

Answer 11

Non-regex solution

I ran some benchmarks, and this solution won hugely:¹

str.slice(str.indexOf(delim) + delim.length)

// as function
function gobbleStart(str, delim) {
    return str.slice(str.indexOf(delim) + delim.length);
}

// as polyfill
String.prototype.gobbleStart = function(delim) {
    return this.slice(this.indexOf(delim) + delim.length);
};

Performance comparison with other solutions

The only close contender was the same line of code, except using substr instead of slice.

Other solutions I tried involving split or RegExps took a big performance hit and were about 2 orders of magnitude slower. Using join on the results of split, of course, adds an additional performance penalty.

Why are they slower? Any time a new object or array has to be created, JS has to request a chunk of memory from the OS. This process is very slow.

Here are some general guidelines, in case you are chasing benchmarks:

• New dynamic memory allocations for objects {} or arrays [] (like the one that split creates) will cost a lot in performance.
• RegExp searches are more complicated and therefore slower than string searches.
• If you already have an array, destructuring arrays is about as fast as explicitly indexing them, and looks awesome.

Removing beyond the first instance

Here's a solution that will slice up to and including the nth instance. It's not quite as fast, but on the OP's question, gobble(element, '_', 1) is still >2x faster than a RegExp or split solution and can do more:

/*
`gobble`, given a positive, non-zero `limit`, deletes
characters from the beginning of `haystack` until `needle` has
been encountered and deleted `limit` times or no more instances
of `needle` exist; then it returns what remains. If `limit` is
zero or negative, delete from the beginning only until `-(limit)`
occurrences or less of `needle` remain.
*/
function gobble(haystack, needle, limit = 0) {
  let remain = limit;
  if (limit <= 0) { // set remain to count of delim - num to leave
    let i = 0;
    while (i < haystack.length) {
      const found = haystack.indexOf(needle, i);
      if (found === -1) {
        break;
      }
      remain++;
      i = found + needle.length;
    }
  }

  let i = 0;
  while (remain > 0) {
    const found = haystack.indexOf(needle, i);
    if (found === -1) {
      break;
    }
    remain--;
    i = found + needle.length;
  }
  return haystack.slice(i);
}

With the above definition, gobble('path/to/file.txt', '/') would give the name of the file, and gobble('prefix_category_item', '_', 1) would remove the prefix like the first solution in this answer.

---

1. Tests were run in Chrome 70.0.3538.110 on macOSX 10.14.

Answer 12

Use the string replace() method with a regex:

var result = "good_luck_buddy".replace(/.*?_/, "");
console.log(result);

This regex matches 0 or more characters before the first _, and the _ itself. The match is then replaced by an empty string.

Answer 13

Javascript's String.split unfortunately has no way of limiting the actual number of splits. It has a second argument that specifies how many of the actual split items are returned, which isn't useful in your case. The solution would be to split the string, shift the first item off, then rejoin the remaining items::

var element = $(this).attr('class');
var parts = element.split('_');

parts.shift(); // removes the first item from the array
var field = parts.join('_');

Answer 14

Here's one RegExp that does the trick.

'good_luck_buddy' . split(/^.*?_/)[1] 

First it forces the match to start from the start with the '^'. Then it matches any number of characters which are not '_', in other words all characters before the first '_'.

The '?' means a minimal number of chars that make the whole pattern match are matched by the '.*?' because it is followed by '_', which is then included in the match as its last character.

Therefore this split() uses such a matching part as its 'splitter' and removes it from the results. So it removes everything up till and including the first '_' and gives you the rest as the 2nd element of the result. The first element is "" representing the part before the matched part. It is "" because the match starts from the beginning.

There are other RegExps that work as well like /_(.*)/ given by Chandu in a previous answer.

The /^.*?_/ has the benefit that you can understand what it does without having to know about the special role capturing groups play with replace().

Answer 15

This should be quite fast

function splitOnFirst (str, sep) {
  const index = str.indexOf(sep);
  return index < 0 ? [str] : [str.slice(0, index), str.slice(index + sep.length)];
}

console.log(splitOnFirst('good_luck', '_')[1])
console.log(splitOnFirst('good_luck_buddy', '_')[1])

Answer 16

if you are looking for a more modern way of doing this:

let raw = "good_luck_buddy"

raw.split("_")
    .filter((part, index) => index !== 0)
    .join("_")

Answer 17

Mark F's solution is awesome but it's not supported by old browsers. Kennebec's solution is awesome and supported by old browsers but doesn't support regex.

So, if you're looking for a solution that splits your string only once, that is supported by old browsers and supports regex, here's my solution:

String.prototype.splitOnce = function(regex)
{
    var match = this.match(regex);
    if(match)
    {
        var match_i = this.indexOf(match[0]);
        
        return [this.substring(0, match_i),
        this.substring(match_i + match[0].length)];
    }
    else
    { return [this, ""]; }
}

var str = "something/////another thing///again";

alert(str.splitOnce(/\/+/)[1]);

Answer 18

For beginner like me who are not used to Regular Expression, this workaround solution worked:

   var field = "Good_Luck_Buddy";
   var newString = field.slice( field.indexOf("_")+1 );

slice() method extracts a part of a string and returns a new string and indexOf() method returns the position of the first found occurrence of a specified value in a string.

Answer 19

This worked for me on Chrome + FF:

"foo=bar=beer".split(/^[^=]+=/)[1] // "bar=beer"
"foo==".split(/^[^=]+=/)[1] // "="
"foo=".split(/^[^=]+=/)[1] // ""
"foo".split(/^[^=]+=/)[1] // undefined

If you also need the key try this:

"foo=bar=beer".split(/^([^=]+)=/) // Array [ "", "foo", "bar=beer" ]
"foo==".split(/^([^=]+)=/) // [ "", "foo", "=" ]
"foo=".split(/^([^=]+)=/) // [ "", "foo", "" ]
"foo".split(/^([^=]+)=/) // [ "foo" ]

//[0] = ignored (holds the string when there's no =, empty otherwise)
//[1] = hold the key (if any)
//[2] = hold the value (if any)

Answer 20

a simple es6 one statement solution to get the first key and remaining parts

let raw = 'good_luck_buddy'

raw.split('_')
   .reduce((p, c, i) => i === 0 ? [c] : [p[0], [...p.slice(1), c].join('_')], [])

Answer 21

You could also use non-greedy match, it's just a single, simple line:

a = "good_luck_buddy"
const [,g,b] = a.match(/(.*?)_(.*)/)
console.log(g,"and also",b)