How to loop through all JSON objects?

Question

I'm trying to add multiple Objects and later to display them all. How would I do this? Every method I googled just uses For-In/For and iterate through characters and I only see the last added object...

for(var i = 0; i < forms.length; i++) 
{
        alert(forms[i].id + " " + 10 + " " + forms[i].value);
        var aObj = new A(forms[i].id, 10, forms[i].value);
        var obj = JSON.stringify(aObj);
}

var str = "";
for(var i = 0; i < jObj.length; i++) 
{
    str += jObj[i]; 
}

alert(str);

Only solution I can come up is to create an array and store them all there?

I don't see where you're adding the object into the jObj array. Can you show that as well? — Joel
– Joel, Commented Jan 30, 2014 at 22:07
Although you aren't adding them to an array in your code (like @Joel already noted), there's nothing wrong in storind them in an array (i.e. with obj[i] in your code above, just declare obj as an Array outside the first loop). — watery
– watery, Commented Jan 30, 2014 at 22:08

score 1 · Accepted Answer · 2014-01-30 22:14:08Z

1

You are reinitalizing the variables aObj and obj and that is causing them to be overwritten every time the loop is executed. They should be arrays with the iteration specified.

var aObj;
var obj = [];
var str = "";

for(var i = 0; i < forms.length; i++) 
{
        alert(forms[i].id + " " + 10 + " " + forms[i].value);
        aObj = new A(forms[i].id, 10, forms[i].value);
        obj[i] = JSON.stringify(aObj);
}

for(var i = 0; i < jObj.length; i++) 
{
    str += jObj[i]; 
}

alert(str);

edited Jan 30, 2014 at 22:14

answered Jan 30, 2014 at 22:03

user3141031

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4 Comments

HelpNeeder Over a year ago

Ok, so array is its. Ok, but I don't think aObj[] need to be an array.

HelpNeeder Over a year ago

aObj refers to A (non-default) constructor.

user3141031 Over a year ago

@HelpNeeder you are right, revised. Btw, what is jObj in the second loop? Did you mean obj?

HelpNeeder Over a year ago

Thats a JSON object which result in stringified output, so I thought. Your answer helps a lot. Thanks.

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