Jquery和Ajax內部服務器錯誤500

Question

我在ajax調用上收到內部服務器錯誤500。 我使用LAMP堆棧作為我的服務器，我已經檢查了權限，因此不能成為我的錯誤。

這是我的javascript代碼：

//$('#addAthleteForm').trigger("reset");

//$(document).ready(function()  {

$(function () {
    $("#dialog").dialog({
        autoOpen: false,
        maxWidth: 600,
        maxHeight: 500,
        width: 500,
        height: 460,
        close: function () {
            $('#addAthleteForm').trigger("reset");
        }

    });
    $("#addAthlete").on("click", function () {
        $("#dialog").dialog("open");
    });

    $("#addAthleteForm").submit(function (e) {
        e.preventDefault();
        var postData = jQuery(this).serialize();
        $("#dialog").dialog("close")
        $.ajax({
            type: "POST",
            url: "AddAthletes.php",
            dataType: 'json',
            data: postData,
            success: function (data) {
                alert(data);
            },
            error: function (jqXHR, exception) {
                if (jqXHR.status === 0) {
                    alert('Not connect.\n Verify Network.');
                } else if (jqXHR.status == 404) {
                    alert('Requested page not found. [404]');
                } else if (jqXHR.status == 500) {
                    alert('Internal Server Error [500].');
                } else if (exception === 'parsererror') {
                    alert('Requested JSON parse failed.');
                } else if (exception === 'timeout') {
                    alert('Time out error.');
                } else if (exception === 'abort') {
                    alert('Ajax request aborted.');
                } else {
                    alert('Uncaught Error.\n' + jqXHR.responseText);
                }
            }



        });
    });


    $("#editAthlete").submit(function (e) {
        e.preventDefault();
        var editData = jQuery(this).serialize();
        $.ajax({
            type: "POST",
            url: "GetAthlete.php",
            dataType: 'json',
            data: editData,
            success: function (data) {

                var form = document.forms['addAthleteForm'];
                form.fname.value = data.fname;
                form.lname.value = data.lname;
                form.school.value = data.school;
                form.agegrp.value = data.agegrp;
            }

        });

        $("#dialog").dialog("open");

    });
})

這是我的PHP代碼：

<?php
$fname = $_POST['fname'];
$lname = $_POST['lname'];
$school = $_POST['school'];
$agegrp = $_POST['agegrp'];


$db = mysqli_connect("localhost", "root", "passwrd","site");
    if(!$db){
      exit("Error in database connection");
    }

else{

$result = mysqli_query($db, "SELECT * FROM `School` WHERE `SchoolLong`='$school'");
$row = mysqli_fetch_array($result));
$SchoolID = $row['SchoolID'];

$result = mysqli_query($db,"SELECT * FROM `AgeGroup` WHERE `AgeGroupLong`='$agegrp'");
$row = mysqli_fetch_array($result));
$AgeGroupID = $row['AgeGroupID'];


mysqli_query($db, "INSERT INTO `Athlete` (`NameFirst`,`NameLast`, `SchoolID`, `AgeGroupID`) VALUES ('$fname', '$lname', $schoolID, $agegrpID)");


}

echo json_encode($fname);
?>

我不確定這個錯誤來自哪里。 用來工作的代碼。

Answer 1

下面一行中的語法錯誤： -

$row = mysqli_fetch_array($result));

它應該是 ：-

$row = mysqli_fetch_array($result);

同樣的錯誤，以下是mysqli_fetch_array()

您更正后的代碼塊是： -

$result = mysqli_query($db, "SELECT * FROM `School` WHERE `SchoolLong`='$school'");
$row = mysqli_fetch_array($result);
$SchoolID = $row['SchoolID'];

$result = mysqli_query($db,"SELECT * FROM `AgeGroup` WHERE `AgeGroupLong`='$agegrp'");
$row = mysqli_fetch_array($result);
$AgeGroupID = $row['AgeGroupID'];