Return errors from PHP run via. AJAX?

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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\

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4条回答

花落未央

2020-12-02 20:43

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)));
    }

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执念已碎

2020-12-02 20:46

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;
?>

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野趣味

2020-12-02 20:47

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.

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不要未来只要你来

2020-12-02 20:50

$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));
}

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