Extracting a part of simple string using Javascript

Question

Let's say I have these strings (each new line being a separate string):

EducationLink
BioLink
InterestsLink

And I wanted to extract the part that's not "Link" using Javascript. How would I do this? The expected results are

Education
Bio
Interests

I tried a Regex, but I'm not very experienced with them, and it failed:

/^ (^Link) $/

Using string functions such as slice() and substr(), I only got the values to the right of the selected text, not to the left as desired.

Thanks for your help.

Darko · Accepted Answer · 2012-02-23 21:45:24Z

2

you can use .replace()

'EducationLink'.replace('Link', ''); //returns Education

https://developer.mozilla.org/en/JavaScript/Reference/Global_Objects/String/replace

The thing is you CAN use regex to do this replace, but being such a simple scenario (unless there is more to it) why would you overcomplicate things?

answered Feb 23, 2012 at 21:45

Darko

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3 Comments

Niet the Dark Absol Over a year ago

Because if there's more than one line to work on. Because there might be Link in the middle of the string somewhere. Because there's just so much that could go wrong with plain string substitution. Apart from that, no reason.

LonelyWebCrawler Over a year ago

Neither of these scenarios are true, so this seems like the best answer.

Darko Over a year ago

@Kolink no need to get your knickers in a twist. The asker plainly states he has several strings and goes on to list them.

Niet the Dark Absol · Accepted Answer · 2012-02-23 21:45:20Z

2

text = input.replace(/Link(\n|$)/g,"\n");

answered Feb 23, 2012 at 21:45

Niet the Dark Absol

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3 Comments

benekastah Over a year ago

If you assume that each line is an individual string, then it should go like this instead: text = input.replace(/Link$/, '');. I'm not sure which the OP meant, though.

LonelyWebCrawler Over a year ago

Each line is a separate string, sorry about that.

Niet the Dark Absol Over a year ago

The updated regex will still work, but you can simplify it to /Link$/g if you want.

Mike Christensen · Accepted Answer · 2012-02-23 21:46:37Z

1

You would need to put the Link into the end of your RegEx, such as:

^([A-Z]+)Link$

This matches:

One or more characters between A and Z
Then the word Link
Then the end of the string

The part in parentheses will be returned as a group.

Furthermore, if all you need to do is remove the suffix Link, you can use the replace() method in Javascript:

var x = myStr.replace('Link', '');

answered Feb 23, 2012 at 21:46

Mike Christensen

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4 Comments

LonelyWebCrawler Over a year ago

Oh, that was rather obvious. However, why did my RegEx not work?

Mike Christensen Over a year ago

Your RegEx doesn't make much sense. The ^ character is used to match the beginning of the string, or to negate characters in a bracket group. Such as [^Link] would mean any character that is neither L, nor i nor n nor k.

LonelyWebCrawler Over a year ago

Ok, but I used normal parentheses (), not square brackets [].

Mike Christensen Over a year ago

I can't think of anything that would match your expression.

calebds · Accepted Answer · 2012-02-23 21:46:53Z

0

One way is to use String.split()^MDN

var txt = 'EducationLink'.split('Link')[0]; // txt == 'Education'

answered Feb 23, 2012 at 21:46

calebds

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Comments

Dan Kanze · Accepted Answer · 2012-02-23 21:48:52Z

0

Try This:

<script>

var str = "EducationLink";
index = str.search(/link/i);
str = str.substr(0,index);

</script>

answered Feb 23, 2012 at 21:48

Dan Kanze

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Comments

pete · Accepted Answer · 2012-02-23 22:01:00Z

As others have pointed out, if all the strings end in 'Link', just use a replace. Otherwise, any of these methods will work fine:

var i = 0;
var strings = [
    'EducationLink',
    'BioLink',
    'InterestsLink'
];
var s = '';
var r = /^(.+)Link$/
for (i = 0; i < strings.length; i += 1) {
    s = strings[i];
    //s = s.substring(0, s.indexOf('Link')); //Uncomment for substring
    //s = s.match(r)[1]; //Uncomment for regex
    s = s.replace('Link', '');
    strings[i] = s;
}
console.log(strings);

Copy and paste into a browser console window to run.

To answer your question ("However, why did my RegEx not work?") in a comment on another answer, your RegEx does the following:

/^ (^Link) $/
//Each regex term on own line, commented below
^   //start of string
    //Literal space
(^Link) //Capturing group, string starting with 'Link'.
    //Literal space
$   //end of string

While you can have two "start of string" identifiers in a Regex, there will only ever be one "start of string". The ^ character is only a negation in a character class which is enclosed in square brackets, not curved. If you want to capture a literal caret (^), put a backslash in front of it (\^).

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