Jquery AJAX return from PHP

Question

New to using this Jquery AJAX method. So i'm making an AJAX call with Jquery, In my PHP script i'm checking a condition to be true from my Database, before I pass in my datastring and bind the variables, then execute that query. Both the script and Database insert work correctly.

My question is, how can I get it to display the error message in my AJAX call on return from my PHP script?

JS

$.ajax({
type: "POST",
url: 'submit_form.php',
data: dataString,
error: function() { alert('Error'); },
success: function() { alert('Success'); }
});

SUBMIT_FORM.PHP

if ( count of rows in db < x  )
 {

return false;
exit();

}

if this condition is true I want to stop execution of my script and execute the error condition in my AJAX function.

elseif ($stmt = $mysqli->prepare("INSERT INTO blah (stuff, morestuff) values (?, ?)")) {

 /* Bind our params */
  $stmt->bind_param("ss", $param1, $param1);

        /* Set our params */
$param1= $_POST["name"];
$param2= $_POST["email"];


/* Execute the prepared Statement */
$stmt->execute();
/* Close the statement */
$stmt->close(); 
}

if the first condition is false then I want to return the success function in my ajax call which it is currently doing.

This can be helpful, http://stackoverflow.com/questions/4417690/return-errors-from-php-run-via-ajax — MJQ, Oct 24 '12 at 06:46

score 7 · Accepted Answer · answered Oct 24 '12 at 06:51

To invoke the error handler of your $.ajax, you can write the following code in PHP (for FastCGI):

header('Status: 404 Not found');
exit;

Alternatively, use header('HTTP/1.0 404 Not Found') if you're not using FastCGI.

Personally I wouldn't use this approach, because it blurs the line between web server errors and application errors.

You're better off with an error mechanism like:

echo json_encode(array('error' => array(
    'message' => 'Error message',
    'code' => 123,
)));
exit;

Inside your success handler:

if (data.error) {
    alert(data.error.message);
}

score 2 · Answer 2 · answered Oct 24 '12 at 06:45

2

If you want to execute error command in jquery you can try using: throw new Exception("Invalid"); inside your php code (where error should invoke).

This will cause http server to return error code 500 to your javascript code.

answered Oct 24 '12 at 06:45

Arkadiusz 'flies' Rzadkowolski

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score 2 · Answer 3 · answered Oct 24 '12 at 06:46

YOur concept is wrong. error condtion means error in ajax call Change your code to this

$.ajax({
type: "POST",
url: 'submit_form.php',
data: dataString,
error: function() {
 alert('Error'); //here the error is the error in ajax call.eg failing to call or a 404..etc
 },
success: function(msg) {
if(msg=='error'){
   alert('Error'); //this is the error you are looking for
}else{
 alert('Success'); 
}

}
});

Your PHP

if ( count of rows in db < x  )
 {
echo 'error'; //this is  enough
exit;

}

score 0 · Answer 4 · answered Oct 24 '12 at 06:47

i think you are little confused about jquery error setting, according to jquery

error : A function to be called if the request fails.

so your error code will be fired only because of the following reasons

404 Error – ajax file not found error

500 Error – ajax internal system error

Jquery AJAX return from PHP

SUBMIT_FORM.PHP

4 Answers4

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