Loop through JSON Array

Question

I have a JSON array that I'm passing to PHP. Eventually I'll be passing and receiving alot more from my PHP file but for the moment this is it. Right now it's recieving 1 array and sending back that same array, no problem. I can loop through the data in Javascript but how do I loop through the array I pass to my PHP file? I get errors with foreach and a for loop didn't seem to help any. Suggestions?

Javascript

var fullName = ["John Doe", "Jane Doe"];

$(window).load(function(){
    getList();
});

function getList(){
    $.getJSON(
        "names.php",
        {names : JSON.stringify(fullName)},
        function(data) 
        {
            for(var i = 0; i < data.test.length; i++)
            {
                window.alert(data.test[i]);
            }
        }
      );
}

PHP

<?php
    $names=json_decode($_REQUEST['names']);

foreach($names as $name)
{
    echo $name;
}

    $data['test'] = $names;
    echo json_encode($data);

The foreach errors out on the foreach line telling me "Warning: Invalid argument supplied for foreach()"

Which error do you get with that for-loop? And also please post the received JSON (inspect it at the network panel of your dev tools) — Bergi
– Bergi, Commented Feb 26, 2013 at 21:11
In your PHP, $names should be an array. What's the problem with reading it? — gen_Eric
– gen_Eric, Commented Feb 26, 2013 at 21:13
Invalid argument? To what? Is it a PHP or a JS error? Please post the exact error message, including the line it points to. — Bergi
– Bergi, Commented Feb 26, 2013 at 21:14
success in jquery just means that the script was called without any problems. It does not mean there are no errors in the script itself. — jeroen
– jeroen, Commented Feb 26, 2013 at 21:29

James C · Accepted Answer · 2013-02-27 00:38:29Z

1

json_decode() doesn't return an array. To get it to do so you'd need to do json_decode($_REQUEST['names'], true)

http://php.net/manual/en/function.json-decode.php

answered Feb 27, 2013 at 0:38

James C

14.2k1 gold badge37 silver badges45 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Martin Zeitler Over a year ago

that's true, but as the sample data looks like it's already an array.

jeroen · Accepted Answer · 2013-02-26 22:03:05Z

0

There seems to be something strange going on when using:

{names : JSON.stringify(fullName)}

As you are sending key - value pairs using GET or POST, you probably need to encode the value correctly.

You could do that using:

{names : encodeURIComponent(JSON.stringify(fullName))}

answered Feb 26, 2013 at 22:03

jeroen

91.8k22 gold badges117 silver badges132 bronze badges

1 Comment

Howdy_McGee Over a year ago

Still no luck with that. My var_dumps still show NULL. Maybe I need to use a different ajax method.

Deathspike · Accepted Answer · 2013-02-26 22:31:32Z

0

EDIT: jQuery is being insane again.

$.getJSON("names.php?names=" + encodeURIComponent(JSON.stringify(fullName)), function(data)
        {
                for(var i = 0; i < data.test.length; i++)
                {
                        window.alert(data.test[i]);
                }
        }
);

The problem was it was adding each item in the array as seperate value to URI.

edited Feb 26, 2013 at 22:31

answered Feb 26, 2013 at 22:23

Deathspike

8,8206 gold badges49 silver badges83 bronze badges

1 Comment

Howdy_McGee Over a year ago

No luck with this either. It even breaks the alerts that would otherwise work (even though it shouldn't)

apoq · Accepted Answer · 2013-02-26 22:33:20Z

0

Try

$names=json_decode($_POST['names']); // or $_GET

instead of

$names=json_decode($_REQUEST['names']);

Sometimes $_REQUEST can be empty because of php.ini conf. SEE request_order

answered Feb 26, 2013 at 22:33

apoq

1,45413 silver badges14 bronze badges

Comments

Shridhar · Accepted Answer · 2013-02-26 22:51:12Z

0

Try this, In your javascript function replace below line with

 {names : JSON.stringify(fullName)},

with

"{names :"+ JSON.stringify(fullName) +"}",

debug test

var fullName = ['John Doe', 'Jane Doe'];
console.log({names : JSON.stringify(fullName)});
// gives Object { names="["John Doe","Jane Doe"]"}

but with

console.log("{names :"+ JSON.stringify(fullName) +"}");
// gives {names :["John Doe","Jane Doe"]}

// so i guess second is correct json string to pass

answered Feb 26, 2013 at 22:51

Shridhar

1661 silver badge5 bronze badges

3 Comments

Howdy_McGee Over a year ago

When I use the console.log as suggested I do get {names :["John Doe","Jane Doe"]} but in my PHP i'm still getting Nulls when I var_dump.

Shridhar Over a year ago

Ok, try one step back, we will remove full name array for a time being and just pass {names : "my test name1"}, and check if you get correct value in php script. if you get correct value that means we have identified an issue with array to json string. otherwise there some other issues

Howdy_McGee Over a year ago

Still nothing in PHP, all nulls. There gotta be something really wrong but I've used a similar script before. I'm just going to try from scratch with the ajax method mentioned below...

Martin Zeitler · Accepted Answer · 2013-02-27 00:38:35Z

0

/* client-side */
$.ajax({

   /* the request's method: */
   type: 'GET',

   /* the request's location: */
   url:'/names.php',

   /* the request's fields: */
   data: JSON.stringify({names: fullName}),

   /* the request's content-type : */
   contentType: 'application/json; charset=utf-8',

   /* the response's content-type: */
   dataType:'json',

   /* the callback function */ 
   success: function(json){

     if(json.length > 0){
         $.each(json, function(i, v){

            console.info(v);

         });
      }
      else {
         alert('wtf?!');
      }
});

/* server-side */
$req=array_merge($_GET, $_POST);

// die(print_r($req));

$names=json_decode($req['names']);
header('content-type: application/json; charset=utf8;');
echo json_encode($names);

Alternatively one could set the request-method for the request, I'm just a little lazy and with that array_merge() it just doesn't matter if the request was a POST or GET (well, function $.ajax defaults to GET). Best practice is to use FireBug and checkout it's net-panel, especially the XHR tab. Probably the question should have been: "How to use a XHR debugger?" because there's nothing tricky about this at all.

It seems that the jQuery API has been slightly updated (see the Deprecation Notice):

http://api.jquery.com/jQuery.ajax/

https://getfirebug.com

edited Feb 27, 2013 at 0:38

answered Feb 26, 2013 at 22:58

Martin Zeitler

79.6k20 gold badges169 silver badges243 bronze badges

6 Comments

Martin Zeitler Over a year ago

If FireBug doesn't work ... you must be missing something fundamental. This code and above posted codes should both "work" - at least log anything to JS console.

Martin Zeitler Over a year ago

besides "doesn't work" isn't considered an error description... while you're not telling where exactly the request bugs out - this is only guessing around. at least it's an example how to create the proper headers for request & response... and FireBug (should) display in detail what happens.

Martin Zeitler Over a year ago

Have you ever used a JS debugger, seriously?? It's all somewhat obvious when setting watches for certain variables and some breakpoints, e.g. on callback.

Martin Zeitler Over a year ago

just set a watch on $ or jQuery ... and you can inspect jQuery itself

Howdy_McGee Over a year ago

I suppose I must admit that the problem doesn't seem to be with the Jquery at all and with the PHP. Good rant through.

|

Collectives™ on Stack Overflow

Loop through JSON Array

6 Answers 6

1 Comment

1 Comment

1 Comment

Comments

3 Comments

6 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

6 Answers 6

1 Comment

1 Comment

1 Comment

Comments

3 Comments

6 Comments

Your Answer

Sign up or log in

Post as a guest

Related