How to access php session data inside template files jquery code

Question

I have ajax that calls a php file called "fetch_quarter_limit.php" from template file.

     $.ajax({

         type: 'POST',
         url: 'fetch_quarter_limit.php',
         data: { 
             leavefrom: from,
             leaveto: to,
             empid: emp
         },
         success: function(data) {
            var quarter_limit = data['myVar'];
            alert(quarter_limit);
         }
     });

In my .php file i have tried to return the session data as an array. Fetched the required data, stored in session and formed an array.

$_SESSION['quarter_limit_mend'] = $quarterLimit;
$returnVal = array('myVar' => $_SESSION['quarter_limit_mend']);
echo $returnVal;

As shown in above ajax code part, i tried to fetch it, but all i am getting is "undefined" when i output the variable using alert.

Please help.

Ajax code updated, p2 :

Adding dataType is making code not to work.

$.ajax({

         type: 'POST',
         url: 'fetch_quarter_limit.php',
         dataType: 'json',
         data: { 
             leavefrom: from,
             leaveto: to,
             empid: emp
         },
         success: function(data) {
             alert(data);
    
         }
 });

As @Tim mentioned i have added custom json encode function to my .php file. It returns as expected {"myVar": 2}

echo array_to_json($returnVal);

This is returned from php file. But not able to access in ajax code.

@ADyson php version in < 5, so json_encode is not available! — Akhil paul
– Akhil paul, Commented Feb 24, 2021 at 9:26
@ADyson I tried adding session_start();, but still it outputs undefined. — Akhil paul
– Akhil paul, Commented Feb 24, 2021 at 9:28
php version in < 5...then you need to upgrade urgently. You are years out of support. php.net/supported-versions.php . I guess you'll have to make the JSON manually then, or find a polyfill. Echoing an array directly in PHP will just produce a warning (or nothing, if warnings are switched off, or set to log to a file) because it can't directly convert an array to a string (and even if it could, it still wouldn't be valid JSON that Javascript could read). — ADyson
– ADyson, Commented Feb 24, 2021 at 9:31

Dharman · Accepted Answer · 2021-02-24 09:51:15Z

1

You're using echo on an array, which is not possible. As described in the PHP manual echo outputs one or more strings.

Usually you'd use json_encode() on your array and then output it to the screen. But as you've commented you are using php < 5. First of all, if possible, you should consider to upgrade to PHP > 7, as this not only improves performance, it also improves security.

If you can't upgrade to a PHP version above PHP 5, then you can use workarounds. On this question there is already an answer for the workaround, and the workaround can be found on the PHP manual itself.

You should be able to use the returned JSON data.

edited Feb 24, 2021 at 9:51

Dharman♦

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answered Feb 24, 2021 at 9:45

Tim

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4 Comments

Akhil paul Over a year ago

I have added the method from the discussion. echo array_to_json($returnVal); Passing this way. But there is no success repsonse. When i remove dataType: 'json', from ajax code, the alert returns as undefined. Am i missing something.

Tim Over a year ago

Try using console.log() on your data instead of an alert and look in the console what it returns. console.log(data); @Akhilpaul

Tim Over a year ago

In jQuery dataType: 'json' automatically converts JSON to an object. So to get the value of myVar you'd do console.log(data.myVar) or alert(data.myVar)

Tim Over a year ago

If the succes is not being called, then there could be an error. You can use error: function() {} to get the error response

Roboroads · Accepted Answer · 2021-02-24 16:16:09Z

0

Reply for after your edit

So you have your data as JSON now, but JS will still see it as a string. In your success function you still have to parse it.

success: function(data) {
    jsonData = JSON.parse(data);
    alert(jsonData.myVar);
}

I do have to add too that it is very insecure to continue using php < 7.3 (as of today, 24-02-2021)

answered Feb 24, 2021 at 16:16

Roboroads

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