PHP parse array posted from jQuery

Question

I have this jQuery code that adds to an array inside an $each loop:

new_edit = {};
new_edit['file_id'] = $file_id;
new_edit['file_realname'] = $new_file_realname;
new_edit['file_folder'] = $new_file_folder;
new_file_edits.push(JSON.stringify(new_edit));

jQuery new_file_edits array output

{"file_id":"1857","file_realname":"dddd[1].mp4","file_folder":"/"}, 
{"file_id":"1856","file_realname":"aaaa[1].jpg","file_folder":"/"},
{"file_id":"1855","file_realname":"ssss[1].jpg","file_folder":"/"}

and im trying to post it to call.php like this:

$.ajax({
    type: "POST",
    url: "/call.php",
    data: "edit_files="+ arrayVarHere,
    contentType: "application/x-www-form-urlencoded;charset=UTF-8",
        success: function(msg){
            alert(+msg);
        }
    });

In the PHP file call.php i have this code:

if(isset($_POST['edit_files'])){
    $edit_array = $_POST['edit_files'];
    $file_array = array();
    foreach($edit_array as $key => $value){
        $file_array[$key] = $value;
    }
print_r($file_array);
    die();
}

but i get this error and i cant figure out how to fix it, been googling it for a while now...

error:

Warning: Invalid argument supplied for foreach() in /home2/dddd/public_html/call.php on line 237

Line 237: foreach($edit_array as $key => $value){

Any help is much appreciated!

-Morten

EDIT 1: I changed $edit_array = $_POST['edit_files']; to $edit_array = array($_POST['edit_files']);

And now it outputs:

{"file_id":"1857","file_realname":"dddd[1].mp4","file_folder":"/"}, 
{"file_id":"1856","file_realname":"aaaa[1].jpg","file_folder":"/"},
{"file_id":"1855","file_realname":"ssss[1].jpg","file_folder":"/"}

How do i go from here with the foreach($edit_array as $key => $value){ part?

Edit 2: i build my arrayVarHere like this:

$.each( $selected_files_array, function( index, value ){
//i get $file_id, $new_file_realname and $new_file_folder with some code here
        new_edit = {};
        new_edit['file_id'] = $file_id;
        new_edit['file_realname'] = $new_file_realname;
        new_edit['file_folder'] = $new_file_folder;
        arrayVarHere.push(JSON.stringify(new_edit));
    });

Possible duplicate of PHP Warning: Invalid argument supplied for foreach() — Thamilhan
– Thamilhan, Commented Jun 8, 2016 at 5:05

Toney · Accepted Answer · 2016-06-08 05:15:13Z

0

change your code like this:

data: "edit_files="+ arrayVarHere,

answered Jun 8, 2016 at 5:15

Toney

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Comments

Jack jdeoel · Accepted Answer · 2016-06-08 05:17:36Z

0

call json data by json_decode in call.php

if(isset($_POST['edit_files'])){
$edit_array = json_decode($_POST['edit_files'],true); //return array 
$file_array = array();
foreach($edit_array as $key => $value){
    $file_array[$key] = $value;
}
print_r($file_array);
die();
}

in your ajax alert should be

 success: function(msg){
            alert(msg); //remove plus sign ,this can output `NaN`
        }

answered Jun 8, 2016 at 5:17

Jack jdeoel

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2 Comments

mlolweb Over a year ago

How can i output file_realname for each file_id in the array?

Jack jdeoel Over a year ago

what will output $key and $value in foreach

swap · Accepted Answer · 2016-06-08 05:28:18Z

0

check edit_array is array print_r($edit_array);

check contentType is used form data then correct otherwise used json type

if form data then check data: edit_files[]: arrayVarHere,

answered Jun 8, 2016 at 5:28

swap

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Comments

mlolweb · Accepted Answer · 2016-06-08 06:08:23Z

0

It returns this array

Array ( [0] => [{"file_id":"1857","file_realname":"aq53pop_460sv[1].mp4","file_folder":"/"},{"file_id":"1859","file_realname":"aKqbD8Q_460sv[1].mp4","file_folder":"/"},{"file_id":"1856","file_realname":"aDopymK_700b[1].jpg","file_folder":"/"}] )

how can i loop the array and access file_id value etc?

answered Jun 8, 2016 at 6:08

mlolweb

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Comments

Poiz · Accepted Answer · 2016-06-08 06:24:10Z

The way you are passing Data to PHP may not be that ideal. Here is what might get you started:

PHP: call.php

    <?php

        $data0      = isset($_POST['data_0']) ? json_decode($_POST['data_0'], true) : null;
        $data1      = isset($_POST['data_1']) ? json_decode($_POST['data_1'], true) : null;
        $data2      = isset($_POST['data_2']) ? json_decode($_POST['data_2'], true) : null;

        $arrData    = array(
            'data0' => $data0,
            'data1' => $data1,
            'data2' => $data2,
        );

        foreach($arrData as $dataKey=>$dataObject){
            $tempFileID     = $dataObject->file_id;
            $tempFileName   = $dataObject->file_realname;
            $tempFileFolder = $dataObject->file_folder;

            // DO SOMETHING WITH THESE DATA AND THEN BUILD UP A RESPONSE FOR AJAX + SEND IT...

            // UNCOMMENT TO SEE THE DATA IN YOUR NETWORKS CONSOLE BUT 
            // IF YOU DO THIS (SINCE YOU EXPECT JSON RESPONSE) YOUR AJAX WILL RESPOND WITH AN ERROR...
            // var_dump($dataObject);
        }

        // RESPOND WITH THE SAME DATA YOU RECEIVED FROM AJAX: JUST FOR TESTING PURPOSES... 
        die( json_encode($arrData));

    ?>

JAVASCRIPT: BUILDING THE DATA FOR PHP

    <script type="text/javascript">
        var dataForPHP              = {};
        var iCount                  = 0;

        $.each( $selected_files_array, function( index, value ){
            var keyName                 = "data_" + iCount;
            new_edit                    = {};
            new_edit['file_id']         = $file_id;
            new_edit['file_realname']   = $new_file_realname;
            new_edit['file_folder']     = $new_file_folder;
            dataForPHP[keyName]         = new_edit;
            iCount++;
        });
    </script>

JAVASCRIPT: AJAX REQUEST

    <script type="text/javascript">
        $.ajax({
            type        : "POST",
            url         : "/call.php",
            datatype    : "JSON",
            data        : dataForPHP,
            contentType: "application/x-www-form-urlencoded;charset=UTF-8",

            success: function(msg){
                if(msg){
                    alert("Ajax Succeeded");
                    console.log(msg);
                }
            }
        });
    </script>

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