String to jQuery ajax function object

Question

Is there any way to convert this string to jQuery function/object?

var str = '';
str=str+"$.ajax({ ";
str=str+"url: 'index.php', ";
str=str+"type: 'post', ";
str=str+"data: 'somevar=' + somevar, ";
str=str+"dataType: 'json', ";
str=str+"success: function(json) { ";
str=str+"alert('test'); ";
str=str+"} ";
str=str+"}); ";

I want to get the same result as...

var myFunction = $.ajax({
    url: 'index.php',
    type: 'post',
    data: 'somevar=' + somevar,
    dataType: 'json',
    success: function(json) { 
        alert('test');
    }
});

Cherniv, I don't think so. It doen't work - 'not a function'. Can u provide a demo? Arun P Johny, I've unparse function... stackoverflow.com/questions/17014735/… — Yevgen
– Yevgen, Commented Jun 10, 2013 at 6:43
@Cherniv, okey... No surprise - you've executed the string. That was not the question. Tnx anyway! Do you know how to get jquery object from this string? — Yevgen
– Yevgen, Commented Jun 10, 2013 at 7:06
so you need this: var myAjaxParamObj = { url: 'index.php', type: 'post', data: 'somevar=' + somevar, dataType: 'json', success: function(json) { alert('test'); } } ; — Ivan Chernykh
– Ivan Chernykh, Commented Jun 10, 2013 at 7:25

jcubic · Accepted Answer · 2013-06-10 06:52:34Z

2

you can use Function constructor.

var your_function = new Function(str);

but you have 2 errors in your string, you can't put // because you will comment everything after that (you don't have new lines) and you have 1 reduntant closing curly brace.

EDIT: to get jquery object, you can execute that function

var somevar = 'something';
var str = '';
str=str+"return $.ajax({ ";
str=str+"url: 'index.php', ";
str=str+"type: 'post', ";
str=str+"data: 'somevar=' + somevar, ";
str=str+"dataType: 'json', ";
str=str+"success: function(json) { ";
str=str+"} ";
str=str+"}); ";

var ajax = new Function(str)();

edited Jun 10, 2013 at 6:52

answered Jun 10, 2013 at 6:46

jcubic

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2 Comments

Yevgen Over a year ago

Thx, sorry, it is simple code for demo. Fixed. Anyway, in this case I'll get function, but not jQuery object.

Yevgen Over a year ago

thx a lot, but i don't need to execute this function. Updated question to be more specific.

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