Is there a way to get PHP to return an AJAX error code if the PHP script fails somewhere? I was following a tutorial and typed this in to my PHP:
$return['error'] = true; $return['msg'] = "Could not connect to DB";
And all was well, until I realised it was JSON data. Is there a way to return errors using standard $_POST and returned HTML data (as in, trigger jQuery's AJAX error: event?
I don't know about jQuery, but if it distinguishes between successful and unsuccessful (HTTP 200 OK vs. HTTP != 200) Ajax requests, you might want your PHP script respond with an HTTP code not equal to 200:
if ($everything_is_ok)
{
header('Content-Type: application/json');
print json_encode($result);
}
else
{
header('HTTP/1.1 500 Internal Server Booboo');
header('Content-Type: application/json; charset=UTF-8');
die(json_encode(array('message' => 'ERROR', 'code' => 1337)));
}
Booboo official, or can it be anything?
Booboo should be the standard. Either that or Foofd
<httpErrors errorMode="Custom" existingResponse="PassThrough"/>$return = array();
$return['msg'] = "Could not connect to DB";
if(isset($_SERVER['HTTP_X_REQUESTED_WITH']) && $_SERVER['HTTP_X_REQUESTED_WITH'] == 'XMLHttpRequest')
{
header('Cache-Control: no-cache, must-revalidate');
header('Expires: Mon, 26 Jul 1997 05:00:00 GMT');
header('Content-type: application/json');
die(json_encode($return));
}
try this out. Hope it helps.
<!-- This is index.php --->
<html>
<head>
<script src="js/jquery.js" type="text/javascript"></script><!-- link to jQuery -->
<script language="javascript">
$(document).ready(function () {
$('input#send').click(function(){
/* Empty div#error and div#result incase they contain info from the last submission */
$('#error').empty('');
$('#result').empty('');
/* Client-side validation */
var name = $("input#name").val();
var age = $("input#age").val();
if (name==''){
alert('Insert your name.');
}
else if (age==''){
alert('Insert your age.');
} else {
var params = 'name=' + name + '&age=' + age;
$.ajax({
url:'b.php',
type:'post',
dataType:'html',
data:params,
cache: false,
success:data
});
function data (html) {
var $html = $( html ); // create DOM elements in a jQuery object
/* Return errors from 'b.php' */
$html.filter('#err').appendTo("#error");
/* Return valid Post */
$html.filter('#res').appendTo("#result");
/* Clear name input */
$('input#name').val('');
/* Clear age input */
$('input#age').val('');
}
}
});
});
</script>
<style type='text/css'>
#error{
color:maroon;
font-weight:bold;
}
#result{
color:green;
font-weight:bold;
}
</style>
</head>
<body>
<div id="error"></div><!-- View Errors -->
<div id="result"></div><!-- View valid Post -->
<input type='text' name='name' id="name" value='' /><br/ >
<input type='text' name='age' id="age" value='' /><br/ >
<input type='submit' name='send' id="send" value='Send' />
</body>
<?php
/* This is b.php */
if(($_POST['name']=='')||($_POST['age']=='')){
$error = '<div id="err">Error: Fill in ALL fields.</div>';
} elseif(is_numeric($_POST['name'])){
$error = '<div id="err">Error: Name should NOT contain numbers.</div>';
} elseif(!is_numeric($_POST['age'])){
$error = '<div id="err">Error: Age should ONLY contain numbers.</div>';
} else{
$result = '<div id="res">Hi '.$_POST['name'].', you are '.$_POST['age'].' years old.</div>';
}
echo $error;
echo $result;
?>
For simple transfer of data from PHP to AJAX, Json encoding or key value pairs are not always necessary. It can be simply done using string handling.
A sample code explaining this functionality is below.
$query = "SELECT email FROM login WHERE email = '". mysqli_real_escape_string($conn,$email) ."' AND password = '". mysqli_real_escape_string($conn,$pass) ."'" ;
$result = mysqli_query($conn,$query);
$row=mysqli_fetch_row($result);
$row_cnt = mysqli_num_rows($result);
$s_email = $row[0];
if ($row_cnt == 1) {
$response="success";
} else {
$response = "Invalid Credentials";
}
$conn->close();
$_SESSION["email"]= $s_email;
echo $response;
This code shows how a 'success' response is sent back to ajax calling the php code. Further in ajax the following could be done to retrieve the response.
$.ajax({ type : 'POST',
data : {email: $('#email').val(), password: $('#pass').val()},
url : 'login.php',
success: function (data) {
if(data == 'success'){ //the value echoed from php can be stored in the data variable of ajax
window.location.assign('upload.php');
}
else{
alert(data);
}
},
error: function ( xhr ) {
alert("error");
}
});
The above concept can be extended to return MULTIPLE VALUES also. This method allows simple tranfer of data from PHP to AJAX in a string format.
We have to follow a simple step of echoing everything we need to send as a response from the php to ajax, seperated by a unique seperator.
echo $response.#;
echo $email.#;
echo $password.#;
echo "Hello".#;
echo "World";
Then the variable data in ajax could be simply retrieved as in the above example and a simple function like,
var res = data.split("#");
data, being the variable within ajax. The res array can then be used within js for whatever purpose we need.
Seeing such complicated answers pains me.; your answer is probably the most unnecessarily complex by creating your own data format. You should also be using PDO (or better alternative; I haven't been keeping up with the PHP ecosystem) instead of mysqli_* for better security.