Jquery object vs string problems

Question

I deleted my old post as it contained a lot of spam. And I created a jsfiddle to reproduce the issue I am having.

console.clear();
var args = '{ action: downline }';

showObject(args);

showObject({ action: 'downline' });


function showObject(args) {

    console.log(args);        
}

How do I pass args as an object? I'm currently building args into a string dynamically and need to pass it as an object for the function to work correctly.

You could parse args explicitly to json: showObject(JSON.parse(args)). — Dominik Schreiber
– Dominik Schreiber, Commented Mar 30, 2012 at 18:27

devstruck · Accepted Answer · 2012-03-30 18:26:18Z

2

It is not your passing, but rather your declaration that is the problem. Define args as

var args = { action: 'downline' };

answered Mar 30, 2012 at 18:26

devstruck

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1 Comment

James Wilson Over a year ago

Ahh, I'm glad someone understood what I was trying to say. I don't explain myself the best sometimes. This worked like a charm, thank you very much!

Frenchi In LA · Accepted Answer · 2012-03-30 18:26:16Z

0

args.action, or args[action] will show downline

you can make also your object like: args = {}; args.action = 'downline'; etc

What's the problem?

answered Mar 30, 2012 at 18:26

Frenchi In LA

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Comments

Naftali · Accepted Answer · 2012-03-30 18:26:23Z

0

I am not sure what the issue is.

Your second way is how you pass an object to a function.

This fiddle shows the log: http://jsfiddle.net/maniator/mLxSP/1/

answered Mar 30, 2012 at 18:26

Naftali

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user400654 · Accepted Answer · 2012-03-30 18:27:27Z

0

If you generated it as valid JSON, you could parse the string as json and display that.

console.clear();
var args = '{ "action": "downline" }';

showObject($.parseJSON(args));

showObject({ action: 'downline' });


function showObject(args) {

    console.log(args);        
}

answered Mar 30, 2012 at 18:27

user400654

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1 Comment

James Wilson Over a year ago

Ahh, that is interesting. I will have to play around with that some.

jeewiya · Accepted Answer · 2012-03-30 18:48:33Z

0

Jquery "parseJSON" will help you.

$(document).ready(function() {

    var args = $.parseJSON('{"action": "downline" }');
    showObject(args);
    showObject({
        action: 'downline'
    });

    function showObject(args) {
        console.log(args);
    }
});

answered Mar 30, 2012 at 18:48

jeewiya

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