Javascript array into jQuery .post AJAX call

Question

I have a JAVASCRIPT array that looks like this:

postarray['min_price'] = 120000;
postarray['max_price'] = 150000;

I'm trying to pass this to an AJAX call via jQuery .post function so that the .PHP file gets it in this format:

$_REQUEST['query']['min_price'] = 120000;
$_REQUEST['query']['max_price'] = 150000;

So far I've tried:

$.post("ajax_findproperties.php", {query: postarray},
        function(data){
            // processing function with JSON result
        }
    ,'json');

But I've had no luck. I even tried changing the var postarray to query and then tried query.serialize() in place of the bracketed variable block, but with no luck either.

When I check my status on Firebug, the AJAX call has absolutely no POST vars set whatsoever - complete blank.

It is probably possible, but why not just query_min_price, query_max_price? — sdfadfaasd
– sdfadfaasd, Commented Jan 7, 2012 at 21:54
Because it's easier on my PHP side if all this is unified within an array. The original form had name="query[min_price]" in there... Trying to not change too much. — jeffkee
– jeffkee, Commented Jan 7, 2012 at 21:56
Also an odd note: when I alert postarray.length, it returns 0. But when I do a for(x in postarray) loop and alert the x and postarray[x], the variables & array keys come up just fine. What's up with that? — jeffkee
– jeffkee, Commented Jan 7, 2012 at 22:01

scessor · Accepted Answer · 2012-01-07 22:07:45Z

2

The javascript array is not an array, it's an object. Define it before:

var postarray = {};
postarray['min_price'] = 120000;
postarray['max_price'] = 150000;

or replace with:

var postarray = {
    min_price: 120000,
    max_price: 150000
};

Now the JSON.stringify works:

alert(JSON.stringify(postarray));

Also see this example.

But this object should also be send without JSON.stringify():

$.post("ajax_findproperties.php", {query: postarray}, ... );

answered Jan 7, 2012 at 22:07

scessor

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1 Comment

jeffkee Over a year ago

Best answer so far! I'm trying it, so far it shows positive signs.. just a few more tweaks to make sure it's all correct...

Milad Naseri · Accepted Answer · 2012-01-07 21:54:30Z

0

Have you tried converting it with JSON.stringify(); and then doing a json_decode(...); in the PHP script?

answered Jan 7, 2012 at 21:54

Milad Naseri

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5 Comments

jeffkee Over a year ago

I tried JSON.stringify(postarray) and that returned [] only, no strings or vars whatsoever.

Milad Naseri Over a year ago

You should probably change it into a proper object rather than an array: query.min_price = 120000; and query.max_price = 150000;. If you insist on keeping it in array format, rather than object format, you have to tell jQuery about it by adding brackets to the variable name: {'query[]': postarray}, although I'm not sure that will handle non-integer indices properly.

jeffkee Over a year ago

In that case, how do I generate the query.min_price part from the form data? They are not pre-set. They are determined by the name of the input field being used, and more or less may be added dynamically. query.min_price, query.max_price, query.other_fieldname, query.more_fieldnames etc. etc. How do I loop that based on the <input> field attributes?

Milad Naseri Over a year ago

You can use var inputs = document.getElementById('your-form-id').getElementsByTagName('input'); and then perform a for loop with this check: if (inputs[i].name && /^query\[/.test(inputs[i].name)) { ... }.

jeffkee Over a year ago

Yes I knkow how to do a loop, but if I'm using an object style like query.[name], how do I assign it? Let say fieldname = min_price is set, and I want to automatically translate this into query.min_price.. do I just do query.'fieldname' and this becomes query.min_price object?

Julien Lafont · Accepted Answer · 2012-01-07 21:56:55Z

0

Try this solution : add [] to your query key

$.post("ajax_findproperties.php", { 'query[]': postarray },
    function(data) { },
    'json');

Source : http://api.jquery.com/jQuery.post/#example-2

answered Jan 7, 2012 at 21:56

Julien Lafont

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