CODEWITHSUNDEEP
phpjavascriptjqueryajax
GeekedOut
GeekedOut

Reputation: 17185

How to get the PHP AJAX error to get to show up in my jQuery code?

I have some PHP AJAX code that is supposed to validate some parameters sent by jQuery and return some values. Currently, it consistently returns invokes the jQuery error case, and I am not sure why.

Here is my jQuery code:

$('.vote_up').click(function() 
{        
    alert ( "test: " + $(this).attr("data-problem_id") );
    problem_id = $(this).attr("data-problem_id");

    var dataString = 'problem_id='+ problem_id + '&vote=+';

    $.ajax({
                type: "POST",
                url: "/problems/vote.php",
                dataType: "json",
                data: dataString,
                success: function(json)
                {           
                    // ? :)
                    alert (json);


                },
                error : function(json) 
                {
                alert("ajax error, json: " + json);

                //for (var i = 0, l = json.length; i < l; ++i) 
                    //{
                    //  alert (json[i]);
                    //}
                }
            });


    //Return false to prevent page navigation
    return false;
});

and here is the PHP code. The validation errors in PHP do occur, but I see no sign that the error that is happening on the php side, is the one that is invoking the jQuery error case.

This is the snippet that gets invoked:

if ( empty ( $member_id ) || !isset ( $member_id ) )
{
    error_log ( ".......error validating the problem - no member id");
    $error = "not_logged_in";
    echo json_encode ($error);
}

But how do I get the "not_logged_in" to show up in my JavaScript of the jQuery so that I know it is the bit that got returned? And if it isn't, how do I make it that that error is what comes back to the jQuery?

Thanks!

Upvotes: 9

Views: 21281

Answers (4)

jxwd
jxwd

Reputation: 187

Easiest way to debug PHP in this case, is to modify .htaccess to create an error log.

php_flag log_errors on

Upvotes: 0

veritas
veritas

Reputation: 2052

jQuery uses the .success(...) method when the response status is 200 (OK) any other status like 404 or 500 is considered an error so jQuery would use .error(...).

Upvotes: 2

jeroen
jeroen

Reputation: 91792

You must handle all output returned from the php script in the success handler in javascript. So a not-logged in user in php can still (should normally...) result in a successful ajax call.

If you are consistently getting the error handler in your javascript call, your php script was not run or is returning a real error instead of a json object.

According to the manual, you have 3 variables available in the error handler, so just checking these will tell you exactly what the problem is:

// success
success: function(data)
{
  if (data == 'not_logged_in') {
    // not logged in
  } else {
    // data contains some json object
  }
},
// ajax error
error: function(jqXHR, textStatus, errorThrown)
{
  console.log(jqXHR);
  console.log(textStatus);
  console.log(errorThrown);
}
//

Upvotes: 1

comu
comu

Reputation: 921

Don't echo $error in the json_encode() method just simply echo $error like so. Also, don't use the variable json, use the variable data. Edited code below:

PHP

if ( empty ( $member_id ) || !isset ( $member_id ) )
{
    error_log ( ".......error validating the problem - no member id");
    $error = "not_logged_in";
    echo $error;
}

jQuery

$('.vote_up').click(function() 
{        
    alert ( "test: " + $(this).attr("data-problem_id") );
    problem_id = $(this).attr("data-problem_id");

    var dataString = 'problem_id='+ problem_id + '&vote=+';

    $.ajax({
                type: "POST",
                url: "/problems/vote.php",
                dataType: "json",
                data: dataString,
                success: function(data)
                {           
                    // ? :)
                    alert (data);


                },
                error : function(data) 
                {
                alert("ajax error, json: " + data);

                //for (var i = 0, l = json.length; i < l; ++i) 
                    //{
                    //  alert (json[i]);
                    //}
                }
            });


    //Return false to prevent page navigation
    return false;
});

Upvotes: 8

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