Fetch specific values from string

Question

I have following string

var a = "{gallery: 'gal', smallimage: '/uploads/photo/image/3256/big_1303-1.jpg', largeimage: '/uploads/photo/image/3256/zoom_1303-1.jpg'}";

What is the best way to fetch largeimage value which is "/uploads/photo/image/3256/zoom_1303-1.jpg" Let me tell you guyz currently i am using this.(which is very basic)

var a = "{gallery: 'gal', smallimage: '/uploads/photo/image/3256/big_1303-1.jpg', largeimage: '/uploads/photo/image/3256/zoom_1303-1.jpg'}";
var b = a.split('largeimage:')[1].split("'")[1];

@E_net4 are you just commenting to make him feel bad, or do you want to be helpful at all? — Joe, Sep 18 '14 at 10:23
@error-404 I suggested it before actually looking at the JSON, which isn't valid. — George, Sep 18 '14 at 10:25
Does it have to be a string? Remove the "" and you have a valid object notation — Johan, Sep 18 '14 at 10:28
@Joe I apologize for that interpretation. I seriously started imagining the question asking for a few steps further, such as retrieving the image data. And only a few moments after, I noticed the lack of quotes in the attributes. — E_net4, Sep 18 '14 at 10:30
@johan : it has to be string, i am fetching this string from somewhere, dont have control over that — error-404, Sep 18 '14 at 10:34
Just posted a working object parser for you. It may not work with way-complicated objects, but should handle your sample case — Joe, Sep 18 '14 at 10:43

score 0 · Answer 1 · answered Sep 18 '14 at 10:33

Have only this idea.

var a = "{gallery: 'gal', smallimage: '/uploads/photo/image/3256/big_1303-1.jpg', largeimage: '/uploads/photo/image/3256/zoom_1303-1.jpg'}";

var b = a.split(',');

var c = b[2].split("'");

var img = c[1];

console.log(img););     // "/uploads/photo/image/3256/zoom_1303-1.jpg"

JSFIDDLE.

Scription · Answer 2 · 2014-09-21T09:47:18.563

0

You have two choice here:

Make your string a valid JSON - to do this you will need to wrap gallery, largeimage, smallimage with apostrophe. so it would look like this:

var validJSONstring = '{"gallery": "gal", "smallimage": "/uploads/photo/image/3256/big_1303-1.jpg", "largeimage": "/uploads/photo/image/3256/zoom_1303-1.jpg"}';

var validJSONObject = JSON.parse(validJSONstring);
console.log(validJSONObject.largeimage);

Then you have to parse it to JSON and get the property name.

Split the string where separator sign is "," comma, loop through results and split each result when separator sign is ":" and check for the property you are looking for.

edited Sep 21 '14 at 09:47

answered Sep 18 '14 at 10:33

Scription

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i cant change the string, no control over it – error-404 Sep 18 '14 at 10:38
What you posted is not valid JSON either. – Felix Kling Sep 18 '14 at 15:11
@Felix, you are correct, I have changed my answer and now the it's a valid JSON. the change I made was replacing the out apostrophe with Quote – Scription Sep 21 '14 at 09:48

score 0 · Answer 3 · answered Sep 18 '14 at 10:39

Here's a rough implementation of parsing a valid JS object. It doesn't support nested objects, but for the sample string it will work correctly.

function parseObject(objString) {
    var obj = {},
        parts,
        part,
        splitPart;

    // Remove leading and trailing {}s
    objString = objString.trim().replace(/^{|}$/g, '');
    parts = objString.split(','); // split properties into an array

    for (part in parts) {
        splitPart = parts[part].split(':'); // split key/value

        // .join for values containing ':', then the regex removes leading and trailing quotes and spaces
        obj[ splitPart[0] ] = splitPart.slice(1).join(':').replace(/^\s*['|"]\s*|\s*['|"]\s*$/g, '');
    }

    return obj;
}

score -1 · Answer 4 · edited May 23 '17 at 10:26

-1

var a = "{gallery: 'gal', smallimage: '/uploads/photo/image/3256/big_1303-1.jpg', largeimage: '/uploads/photo/image/3256/zoom_1303-1.jpg'}";

var jsontemp = a.replace((/([\w]+)(:)/g), "\"$1\"$2");
var correctjson = jsontemp.replace((/'/g), "\"");
var obj = JSON.parse(correctjson);
var c = obj.largeimage;

JSFiddle

For turning into valid json, I used the solution from this post: https://stackoverflow.com/a/24462870/888177

edited May 23 '17 at 10:26

Community

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answered Sep 18 '14 at 10:25

Stefan

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`a` is a string. It does not have a `.largeImage` property – Joe Sep 18 '14 at 10:26
Working on it, hang on! – Stefan Sep 18 '14 at 10:28

Fetch specific values from string

4 Answers4