jQuery AJAX：成功返回request.fail

Question

Have the following jQuery AJAX call. It functions correctly and pulls off the PHP/MySQL operations, but returns the request.fail function instead of the normal done/success. Works fine if you comment/remove the MySQL query.

Did a lot of searching and believe it is something to do with the dataType , which I've tried as text , html , and "" according to other posts. Tried using html as the dataType and just changing the echo "Success."; for standard HTML/text output after the PHP code, but no luck. Any help would be appreciated.

This code is a small portion of a much larger project that has been isolated for the sake of the question. The $variable is just an example and is within scope (and not causing an issue).

Calling File:

<script type="text/javascript">
        function ajaxCall() {
            var request = $.ajax({
                url: '<?php echo "ajax.php?variable=$value"; ?>',
                type: "GET",            
                dataType: "html"
            });

            request.done(function(msg) {
                alert(msg);

            });

            request.fail(function(jqXHR, textStatus) {
                alert( "Request failed: " + textStatus );
            });
        }
</script>

AJAX File:

<?php
    $variable = $_GET['variable'];
    $sign = mysqli_query(mysqli_connect('localhost', 'user', 'password', 'database'), "INSERT INTO `Table` (`Column`) VALUES ('$variable')");
    echo "Success.";
?>

Answer 1

Like I said above, you need to work on error reporting for this all to work as you want it. I would start by building errors coming back from your PHP , then building the result back into JSON like this:

<?php
ini_set('display_errors', false);

$variable = $_GET['variable'];

$servername = "localhost";
$username = "root";
$password = "root";
$dbname = "test";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    $json_arr['response'] = FALSE;
    $json_arr['message'] = "Connection failed: " . $conn->connect_error;
} else {
    $sql = "INSERT INTO `tbl1` (`col1`) VALUES ('$variable')";
    $result = $conn->query($sql);
    if ($result) {
        $json_arr['response'] = TRUE;
        $json_arr['message'] = "success";
    }else{
        $json_arr['response'] = FALSE;
        $json_arr['message'] = $conn->error;
    }
}
$conn->close();

header('Content-Type: application/json');
echo json_encode($json_arr);

Once that is set, you can then use AJAX to obtain your results in the following manner:

function ajaxCall() {
    var request = $.ajax({
        url: '<?php echo "success.php?variable=$value"; ?>',
        type: "GET",
        dataType: "json"
    });

    request.done(function (result) {
        if (result.response) {
            alert(result.message);
            // your success code
        } else {
            alert(result.message);
            // your fail code
        }

    });

    request.fail(function (jqXHR, textStatus, error) {
        alert("FAIL");
        console.log(textStatus);
        console.log(error);
    });
}

ajaxCall();

Answer 2

/* js */

function callFunction() {
    $.ajax({
        url: <url>
        type: 'GET',
        data: {},
        dataType: 'json',
        success: function(response) {
            alert(response.msg)
        },
        error: function() {
            alert('Error!');
        }
    });
    return false;
}

/* php */

echo json_encode([
    'msg' => 'Message'
]);
exit