javascript - split string by specifying starting and ending characters

Question

I have a string (100*##G. Mobile Dashboard||Android App ( Practo.com )||# of new installs@@-##G. Mobile Dashboard||Android App ( Practo.com )||# of uninstalls@@

I want to split the string in such a way that it returns the following result (i.e it matches all characters that starts with ## and ends with @@ and splits the string with matched characters)

["(100*", "G. Mobile Dashboard||Android App ( Practo.com )||# of new installs", '-', 'G. Mobile Dashboard||Android App ( Practo.com )||# of uninstalls'

The initial string will have hyphen between those 2 characters. So when splitting the string with this specified characters, it needs to give hyphen in ouput — Saravanan S
– Saravanan S, Commented Aug 7, 2016 at 10:31

Community · Accepted Answer · 2017-05-23 10:28:43Z

8

Use String.prototype.split() passing a regex.

var str = "(100*##G. Mobile Dashboard||Android App ( Practo.com )||# of new installs@@-##G. Mobile Dashboard||Android App ( Practo.com )||# of uninstalls@@";

var re = /##(.*?)@@/;

var result = str.split(re);
console.log(result);

When you use Capturing parentheses in the regex, the captured text is also returned in the array.

Note that this will have an ending "" entry, because your string ends with @@. If you don't want that, just remove it.

If you always assume a well-formed string, the following regex yields the same result:
```
/##|@@/
```
^{*Commented by T.J. Crowder}
If you expect newlines in between ## and @@, change the expression to:
```
/##([\s\S]*?)@@/
```
If you need it to perform better, specially failing faster with longer strings:
```
/##([^@]*(?:@[^@]+)*)@@/
```
^*Benchmark

edited May 23, 2017 at 10:28

CommunityBot

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answered Aug 7, 2016 at 10:30

Mariano

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4 Comments

T.J. Crowder Over a year ago

Seems like the same result you get with /##|@@/, which would be simpler. (Assuming a well-formed string.)

Mariano Over a year ago

@T.J.Crowder Except when you have unclosed structures, e.g. aaa##zzz. So it depends on what you expect in the input.

T.J. Crowder Over a year ago

Heh, was just adding "...assuming a well-formed string." :-)

Mariano Over a year ago

Haha... And I was adding the "...depends on what you expect..." :-)

Lazar Ljubenović · Accepted Answer · 2016-08-07 10:30:27Z

4

You can first split by ##, then split each of the results by @@, and flatten the resulting array, like the following.

s.split('##').map(el => el.split('@@')).reduce((acc, curr) => acc.concat(curr))

Note that the last element of the resulting array will be empty string if you original string ends with @@, so you might need to remove it, depending on your use-case.

answered Aug 7, 2016 at 10:30

Lazar Ljubenović

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Comments

anubhava · Accepted Answer · 2016-08-07 10:31:22Z

4

You can use:

var s = '(100*##G. Mobile Dashboard||Android App ( Practo.com )||# of new installs@@-##G. Mobile Dashboard||Android App ( Practo.com )||# of uninstalls@@'

var arr = s.split(/(##.*?@@)/).filter(Boolean)

//=> ["(100*", "##G. Mobile Dashboard||Android App ( Practo.com )||# of new installs@@", "-", "##G. Mobile Dashboard||Android App ( Practo.com )||# of uninstalls@@"]

Use a capturing group to get split text in resulting array
filter(Boolean) is needed to remove empty result from array

answered Aug 7, 2016 at 10:31

anubhava

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