regex replace with associative array (of regex patterns)

Question

Essentially this code searches through a string, finds and replaces according to the each of these regex patterns before spitting the updated variable back.

I'm having difficulty translating it from PHP code to Javascript:

in PHP I have achieved it as such:

$dataPreg = array(
'/(.*)\s§/' => '$1 <span class="suppSmall">quasi</span>',
'/(\bor|and\b)/'=>'<span class="orChar">$1</span>',
'/\:(\s*see[\:]?)/'=>'<span class="orChar"> $1: </span>'
);

$root = preg_replace(array_keys($dataPreg),array_values($dataPreg),$root);

return $root;

in Javascript I attempted to port it as such:

var dataPreg = {
'/(.*)\s§/':'$1 <span class="suppSmall">quasi</span>',
'/(\bor|and\b)/':'<span class="orChar">$1</span>',
'/\:(\s*see[\:]?)/':'<span class="orChar"> $1: </span>'
};

for (var val in dataPreg)
$string = $string.replace(new RegExp(val, "g"), dataPreg[val]);
return $string;

Needless to say, its not working in Javascript and I can't figure out why. It doesn't trigger any JS errors, it simply returns the string back..unchanged.

Paul S. · Accepted Answer · 2014-04-16 17:48:05Z

Keys of Objects should be Strings, so I suggest you saving effort and cycles by using an Array of Objects, i.e.

var dataPreg = [
    {re: /(.*)\s§/g        , sub: '$1 <span class="suppSmall">quasi</span>'},
    {re: /(\bor|and\b)/g   , sub: '<span class="orChar">$1</span>'},
    {re: /\:(\s*see[\:]?)/g, sub: '<span class="orChar"> $1: </span>'}
];

Then

(function () {
    var i; // don't pollute namespace ;)
    for (i = 0; i < dataPreg.length; ++i)
        $string = $string.replace(dataPreg[i].re, dataPreg[i].sub);
}());

e.g.

var $string = 'foo § bar or then and you :see: here';
/* becomes
foo <span class="suppSmall">quasi</span> bar <span class="orChar">or</span> then <span class="orChar">and</span> you <span class="orChar"> see:: </span> here
*/

Thanks Paul very clever, worked like a charm and very easy to read.

Sabuj Hassan · Accepted Answer · 2014-04-16 17:29:15Z

1

The RegExp takes the pattern as parameter without the delimiter. For example it takes [a-z]+ as parameter, and not take /[a-z]+/ like you do it in PHP. So remove the // delimiters from the both end of your regex. One example:

'(.*)\s§':'$1 <span class="suppSmall">quasi</span>',
 ^ the / is gone

answered Apr 16, 2014 at 17:29

Sabuj Hassan

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1 Comment

Tourshi Over a year ago

Thanks--I did try that, however, it appears as though it has no effect--something else is also to blame..

anubhava · Accepted Answer · 2014-04-16 17:47:05Z

1

You array strings aren't right for creating new RegExp object. (Using delimiters and single escaped propertied e.g. \s instead of \\s).

Try this instead:

dataPreg = {
'/(.*)\\s§/g':'$1 <span class="suppSmall">quasi</span>',
'/(\\bor|and\\b)/g':'<span class="orChar">$1</span>',
'/:(\\s*see:?)/g':'<span class="orChar"> $1: </span>'
};
for (var val in dataPreg)
    $string = $string.replace(val, dataPreg[val]);
return $string;

edited Apr 16, 2014 at 17:47

answered Apr 16, 2014 at 17:33

anubhava

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4 Comments

Tourshi Over a year ago

I just tested it, it didn't work, but the opposite did! I escaped everything with double slashes.. so instead of \s I put it as \\s --I'm guessing that's what you meant? Thanks

anubhava Over a year ago

Whether regex is right or not I can't tell since you haven't provided value of $string

Tourshi Over a year ago

I'm referring to your advice (to use single escaped..)--I double escaped and it worked :) and I'm guessing that's what you meant to tell me

Paul S. Over a year ago

' '.replace('/\\s/', '') !== ''

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