AngularJS Sending Objects as Array For PHP

Question

I'm getting use to AngularJS ng-resource but I am having a problem serieliazing an array. My ng-resource is like this

app.factory('MyModel', ['$resource', 'api_domain',
function($resource, api_domain) {

        return $resource(api_domain + 'adsizes/:id', {
            id : '@id'
        }, {
            get : {method: 'GET', isArray: true }
        })

}]);

And my controller looks like this:

    app.controller("MyCtrl", ['$scope', 'MyModel',
    function($scope, MyModel) {

        MyModel.get({
                    id : id,
                    'conditions': { 'join' : 'table2'}
                }, function() {

                });
});

The problem is it passes the conditions like so:

conditions:{"join":"table2"}

Which passes into php as a string that has to get decoded. My question is how can I pass the conditions as associative array for php?

Wouldn't you just use json_encode for that in PHP, as you can't send anything other than strings with a GET or POST request anyway ? — adeneo
– adeneo, Commented Nov 23, 2013 at 22:59
It would, I of course meant json_decode, converting the string to an object or array. — adeneo
– adeneo, Commented Nov 23, 2013 at 23:06
I could do that, but that would require modifying the PHP to suite that javascript, something I want to avoid — Devin Dixon
– Devin Dixon, Commented Nov 23, 2013 at 23:20

Michael J. Calkins · Accepted Answer · 2013-11-24 10:37:27Z

1

Append this to your get request and var dump $_GET.

?conditions[join]=table2

answered Nov 24, 2013 at 10:37

Michael J. Calkins

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Comments

Jeff Hubbard · Accepted Answer · 2013-11-24 00:31:36Z

0

Your parameter needs to be an array, rather than an object:

...
conditions: ['join', 'table2']
...

edited Nov 24, 2013 at 0:31

answered Nov 23, 2013 at 23:06

Jeff Hubbard

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3 Comments

Devin Dixon Over a year ago

Not quite. The parameters are being passed through $GET so there is no need for php://input

Devin Dixon Over a year ago

The reason why I'm not doing it PHP is because I do not want to have to change all my php to accomdate this one change

Jeff Hubbard Over a year ago

In that case, you need to use an array--rather than an object--as the input. I'll update my answer accordingly.

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