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rated 0 times [  0] [ 7]  / answers: 1 / hits: 24593  / 13 Years ago, fri, september 9, 2011, 12:00:00

I would like to use create a object that contains regular expressions as the key value. I tried to use the following syntax:



var kv = {

    /key/g : value

};


But it fails according JavaScript lint:



SyntaxError: invalid property id


How can I fix it?



Update



Background: The reason why I want to do this is to implement a workaround that fixes wrong unicode in a HTTP API result. I know this is very hackish, but since I have no control over the API server code I think this is the best I can do.



Currently I implemented the mapping by having a keys array and a values array:



function fixUnicode(text) {



    var result = text;

    var keys = [];

    var values = [];

    keys.push(/é/g); values.push(é);

    keys.push(/è/g); values.push(è);

    keys.push(/ê/g); values.push(ê);

    keys.push(/ë/g); values.push(ë);

    keys.push(/à/g); values.push(à);

    keys.push(/ä/g); values.push(ä);

    keys.push(/â/g); values.push(â);

    keys.push(/ù/g); values.push(ù);

    keys.push(/û/g); values.push(û);

    keys.push(/ü/g); values.push(ü);

    keys.push(/ô/g); values.push(ô);

    keys.push(/ö/g); values.push(ö);

    keys.push(/î/g); values.push(î);

    keys.push(/ï/g); values.push(ï);

    keys.push(/ç/g); values.push(ç);



    for (var i = 0; i < keys.length; ++i) {

        result = result.replace(keys[i], values[i]);

    }

    return result;

}


But I want to implement to use a JavaScript object to map keys as values:



function fixUnicode(text) {



    var result = text;

    var keys = {

        /&Atilde;&copy;/g : &eacute;,

        /&Atilde;&uml;/g :  &egrave;

        // etc...

    };



    for (var key in keys) {

        result = result.replace(key, keys[key]);

    }

    return result;

}

More From » javascript
 Answers
62

This can be done, but not using object literal syntax. You'll need to do it like this:



var kv = {};

kv[/key/g] = value;

console.log(kv[/key/g]); // value



Edit: this could probably use some explaining. As xanatos commented below, what's really happening here is that the key, /key/g in this case, is being toString()'d to create the key. This is important to know, because it has an effect on key uniqueness. Consider the following:



var x = {},

    reg = /foo/;



x[reg] = 'bar';

console.log(x[reg]); // bar

console.log(x[reg.toString()]); // bar

console.log(x['/foo/']); // bar'


In summary, I'm semi-scared to ask why you need to do this, but assuming you have your reasons, be careful and make sure you understand what is really happening :)




Edit 2: So in response to your updated question, you should be able to achieve something pretty close to what you want. You can use the object literal syntax as long as you wrap the regular expression in quotes. Unfortunately that means you'll have to manually reconstruct an actually RegExp object out of that key though. For example:



var result = abcdef,

    replacements = {

        /a/g: FOO,

        /d/i: BAR

    };



for (var key in replacements) {

    var parts = key.split('/');

    result = result.replace(new RegExp(parts[1], parts[2]), replacements[key]);

}



console.log(result); //FOObcBARef



Edit 3: Because, why not. I was so stuck on making your object-literal syntax work, that I didn't consider the fact that you never need to actually look up the replacement by the pattern itself (i.e., there's no need for object keys at all). Here's a more efficient approach using arrays that doesn't require the RegExp reconstruction:



var result = abcdef,

    replacements = [

        [/a/g, FOO],

        [/d/i, BAR]

    ];



for (var i = 0, len = replacements.length; i < len; i++) {

    var replacement = replacements[i];

    result = result.replace(replacement[0], replacement[1]);

}



console.log(result); //FOObcBARef



Edit 4: Because I'm bored and I like this question. Here's the ninja version:



var result = abcdef,

    replacements = [

        [/a/g, FOO],

        [/d/i, BAR]

    ], r;



while ((r = replacements.shift()) && (result = String.prototype.replace.apply(result, r))) {}



console.log(result); //FOObcBARef

[#90179] Thursday, September 8, 2011, 13 Years  [reply] [flag answer]
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