CODEWITHSUNDEEP
javascriptphpajaxjson
user2301506
user2301506

Reputation: 300

Not getting JSON array back from Ajax call

I know this has been asked before but I've reviewed all the previous posts I can find and still cannot get this to work. Hopefully the problem is simple and you guys can help me solve it.

I am not able to get an array, which I have json_encoded, from the PHP script back into a Javascript array. If I remove the condition that dataType should be json then the success function complains about unexpected data.

I have a Javascript function which calls a PHP script with POST method. In the PHP script I write some debugging messages to a log file and to display the content of the json_encoded array. When I check that using JSONLint the content is JSON compliant but my javascript function always goes to error.

The PHP function looks like:

<?php

// Include the php logger class so that we can write log messages to a log file
require_once('/myscripts/phplogger.php');

// Set up php log file
$log = new Logging();
$log->lfile('/mylogfiles/php-debug.log');

// Log status
$log->lwrite('Starting debug');
// ...
// (there's some more stuff here to get the data from a MySQL database)
// ...

// The following line when read in the log file shows a valid JSON array
$log->lwrite('array '.json_encode($dataVnvMethods));

$json_dataVnvMethods = json_encode($dataVnvMethods);

$log->lwrite('Ending debug');

?>

The Javascript function looks like:

function jsgetvnvfilters(projid,repid,weekid)
{
    $.ajax(
            {
            url: './getvnvfilter.php',
            data: {'projid':projid, 'weekid':weekid, 'repid':repid},
            type: 'post',
            dataType: 'json',
            success: function()
              {
                    alert('success');
                    var prevnvlist = '<?php echo $json_dataVnvMethods ?>';
                    var vnvlist = JSON.parse(prevnvlist);
                    for (var x = 1; x <= vnvlist.length; x++) {
                            var vnv = vnvlist[x]['VnVMethod'];
                            vnvSel.options[vnvSel.options.length] = new Option(vnv, vnv);
                    }
              },
            error: function()
              {
                    alert('failure');
              }

            }
    );

}

Upvotes: 3

Views: 384

Answers (3)

Bart Van Den Boomen
Bart Van Den Boomen

Reputation: 53

use data: JSON.stringify(value[, replacer [, space]])

Upvotes: 0

Branimir Đurek
Branimir Đurek

Reputation: 632

Press f12 and open developer tools. Go to section network. And then try. If you got 404 error it means that you enter wrong URI or something. Basically, if Ajax return you error it means you didn't access your script for some reason. Maybe you entered wrong URI, or you don't have permission.

Upvotes: 0

jeroen
jeroen

Reputation: 91792

You misunderstand how the javascript and php are related; the php will be rendered only once on page-load so you cannot echo the returned php data in your javascript success handler.

Instead, everything that is outputted by your php script will be available in a javascript variable:

success: function(vnvlist) {
           //     ^^^^^^^ this is what has been outputted by php, the
           //             json already parsed

           // your data is available in `vnvlist`
           // var prevnvlist = '<?php echo $json_dataVnvMethods ?>';

           // with dataType='json' you don't need to parse anything
           // as jQuery will parse it for you
           // var vnvlist = JSON.parse(prevnvlist);

           // so all you need is this...
           alert('success');
           for (var x = 1; x <= vnvlist.length; x++) {
             var vnv = vnvlist[x]['VnVMethod'];
             vnvSel.options[vnvSel.options.length] = new Option(vnv, vnv);
           }
         },

And you need to make sure that php outputs only your json string:

...
// The following line when read in the log file shows a valid JSON array
$log->lwrite('array '.json_encode($dataVnvMethods));

// output your data so that it is available on the client-side
echo json_encode($dataVnvMethods);

$log->lwrite('Ending debug');
...

Upvotes: 5

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