Finding if next value in object exists or not in javascript

Question

I want to find if the next value in a object exists or not, if it exists, add '&' else add nothing .

serializeForQueryString: function(obj) {
    var queryString ='';

    for (var key in obj) {
        if (obj.hasOwnProperty(key)) {
            queryString=queryString+key+'='+ obj[key];

            if (obj.hasOwnProperty(key)!=null){
                queryString=queryString+'&';
            }
            alert(queryString);
        }
    }

    //document.write(obj.name);
    var queryString = '';

    //jquery
    //$.each(obj, function(key, value) {});

    //$.isArray

    return queryString;
},

Make sure you use "encodeURIComponent" on keys and values if you use your own solution — Ian
– Ian, Commented May 1, 2013 at 3:50

dave · Accepted Answer · 2013-05-01 03:56:29Z

2

Just build an array and then join it using &:

for (var key in obj) {
            if (obj.hasOwnProperty(key)) {

                 queryString.push(key + '=' + obj[key]);


            }
 }

queryString.join('&');

edited May 1, 2013 at 3:56

answered May 1, 2013 at 3:48

dave

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5 Comments

Ian Over a year ago

I think you mean: querystring.push(key + "=" + obj[key]);

JLRishe Over a year ago

It would also be wise to url escape at least obj[key].

dave Over a year ago

I agree, but I figured as the question was specifically about how to get the amperstand in between key value pairs, I would just show that.

Joseph Myers Over a year ago

This is a bad answer because it does not encode the data. Try to decode the query string that it produces if there is anything complicated contained in obj[key]. What if your object contains a value of 'x = 5 & y = 7' for example?

dave Over a year ago

Then he should definitely URL encode it. However, the problem was to "find if the next value in a object exists or not, if it exists, add '&' else add nothing." Url-encoding is outside the scope of the question asked. Anyways - he has your comment, and other people mentioning the same thing, so I'm sure he and anyone else can figure it out.

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