Ajax不执行成功函数


Ajax does not execute success function

我有一个jquery ajax函数,它将表单数据提交到一个php文件,在该文件中对数据库进行检查并返回响应。但是我的成功函数并没有被执行,而是通过error函数返回响应。

以下是我的login.js代码:

$(document).ready(function(){
    $("form#loginForm").submit(function() { // loginForm is submitted
        var username = $('#username').attr('value'); // get username
        var password = $('#password').attr('value'); // get password
    if (username && password) { // values are not empty
        $.ajax({
            type: "POST",
            url: "url", // URL of the Php script
            contentType: "application/json; charset=utf-8",
            dataType: "json", //expected from server in response
            // send username and password to the Php script
            data: "username=" + username + "&password=" + password,

            error: function(XMLHttpRequest, textStatus, errorThrown) { // script call was *not* successful
                $('div#loginResult').text("responseText: " + XMLHttpRequest.responseText
                + ", textStatus: " + textStatus
                + ", errorThrown: " + errorThrown);
                $('div#loginResult').addClass("error");
            },
            success: function(data){ //script was successful, data contains response from mysql database
                if (data.error) { // login was not successful
                    $('div#loginResult').text("data.error: " + data.error);
                    $('div#loginResult').addClass("error");
                } // if
                else { // login was successful
                    $('form#loginForm').hide();
                    $('div#loginResult').text("data.success: " + data.success);
                    $('div#loginResult').addClass("success");
                } //else
            } // success
        }); // ajax
    } // if
    else {
        $('div#loginResult').text("enter username and password");
        $('div#loginResult').addClass("error");
    } // else
    $('div#loginResult').fadeIn();
    return false;
});
});

这是我的Login.php代码:

 <?php

$username = $_POST["username"];
$password = $_POST["password"];

$dbhost = '*****';
$dbuser = '*****';
$dbpass = '*****';
$conn = mysql_connect($dbhost, $dbuser, $dbpass); // connect to server
if(! $conn )
{
    die('Could not connect: ' . mysql_error());
}
echo 'Connected to database successfully,';
mysql_select_db('sl493',$conn); //pick sl493 database

$result = mysql_query("SELECT *
        FROM metauser
        WHERE metauser.username = '$username'
        AND metauser.password = '$password'") or die(mysql_error()); //select data from metauser table
$row = mysql_fetch_assoc($result);
if($row['username'] == $username) //If username and password not accepted
{$result = 'true login';
 $arr = array('success' => "login is successful");
 echo json_encode($arr);}
else //if username and password are not accepted
{$result = 'login failed';
 $arr = array('error' => "username or password is wrong");
 echo json_encode($arr);}
?>

现在,无论我键入的凭据是正确的还是不正确的,我都会得到以下响应:

responseText:成功连接到数据库,{"成功":"登录成功"},text状态:parsererror,errorThrow:无效JSON:已成功连接到数据库,{"success":"login is successful"}https://i.stack.imgur.com/4xynj.jpg

它返回"Connected to database successfully",这意味着它成功地转到了login.php,但它也返回了responseText,这是ajax的一部分:error,由于函数执行成功,它不应该转到那里。哈尔普?

我发现了问题所在:

  1. 首先,我将ContentType指定为json,同时将其作为字符串,所以我不得不改变它。

  2. 其次,我使用$_POST的Is_set andis_empty函数我肯定得到了POST字段。

  3. 最后,由于某种原因,返回的ajax变量数据不会显示出来by键,因此data.success未定义,但data返回所有输出从php文件

我仍在试图弄清楚为什么我不能按键访问返回的JSON。如果你有任何建议,请告诉我,我不想再提出另一个问题。以下是工作代码:

$(document).ready(function(){
    $("form#loginForm").submit(function() {
    // loginForm is submitted
        var username = $('#username').attr('value'); // get username
        var password = $('#password').attr('value'); // get password
        console.log(username + password) ;
        if (username && password) { // values are not empty
            $.ajax({
                type: "POST",
                url: "https://xxx/login.php", // URL of the Php script
                contentType: "application/x-www-form-urlencoded; charset=utf-8",
                dataType: "application/json", //expected from server in response
                // send username and password to the Php script
                //data: "username=" + username + "&password=" + password,
                data:'username='+ username+'&password='+ password,
                error: function(XMLHttpRequest, textStatus, errorThrown) { // script call was *not* successful
                    $('div#loginResult').text("responseText: " + XMLHttpRequest.responseText
                    + ", textStatus: " + textStatus
                    + ", errorThrown: " + errorThrown);
                    $('div#loginResult').addClass("error");
                },
                success: function(data){

                    $('form#loginForm').hide();
                    $('div#loginResult').text("Login: " + data );
                    $('div#loginResult').addClass("success");
                                                                }

            });
        }
        else {
            $('div#loginResult').text("enter username and password");
            $('div#loginResult').addClass("error");
        } // else
        $('div#loginResult').fadeIn();
        return false;
    });
})

Login.php

<?php
if (isset($_POST["username"]) && !empty($_POST["username"])) {
    $username = $_POST["username"];}
if (isset($_POST["password"]) && !empty($_POST["password"])) {
    $password = $_POST["password"];}
$dbhost = 'xxx';
$dbuser = 'xxxxxx';
$dbpass = 'xxxxxxxx';
$conn = mysql_connect($dbhost, $dbuser, $dbpass); // connect to server
if(! $conn )
{
    die('Could not connect: ' . mysql_error());
}
mysql_select_db('sl493',$conn);
$query = mysql_query("SELECT *FROM metauser WHERE metauser.username = '$username'AND metauser.password = '$password'") or die(mysql_error()); //select data from metauser table
$row = mysql_fetch_assoc($query);

if($row['username'] == $username)
{
        if($row['usertype']== 'student')
              { $type = 'student'; }
                    else{ $type = 'admin'; }
    //$arr = array('result' => 'loginOK', 'usertype' => $type);
    $associativeArray = array();
    $associativeArray ['result'] = 'success';
    $associativeArray ['usertype'] = $type;
    //$arr = '{"success":"login is successful"}';
    //echo $type;
    echo json_encode($associativeArray); }

else
{$arr = '{"error":"username or password is wrong"}';
    echo json_encode($arr);}
?>

试试这个:

var data="";
data=['username':username,'password':password];//Add this
 $.ajax({
            type: "POST",
            url: "url", // URL of the Php script
            contentType: "application/json; charset=utf-8",
            dataType: "json", //expected from server in response
            // send username and password to the Php script
            data: JSON.stringify(data),//Stringify your data while sending
            success: function(data){ //script was successful, data contains response from mysql database
                if (data.error) { // login was not successful
                    $('div#loginResult').text("data.error: " + data.error);
                    $('div#loginResult').addClass("error");
                } // if
                else { // login was successful
                    $('form#loginForm').hide();
                    $('div#loginResult').text("data.success: " + data.success);
                    $('div#loginResult').addClass("success");
                } //else
            },
            error: function(XMLHttpRequest, textStatus, errorThrown) { // script call was *not* successful
                $('div#loginResult').text("responseText: " + XMLHttpRequest.responseText
                + ", textStatus: " + textStatus
                + ", errorThrown: " + errorThrown);
                $('div#loginResult').addClass("error");
            }, // success
        }); // ajax
请将您的ajax数据类型从dataType: "json"更改为
datatype: "application/json"

为了在未经授权的响应情况下解决响应标头问题,请在代码的其他部分添加响应标头作为header('HTTP/1.1 401 Unauthorized', true, 401);,如下所示。

else //if username and password are not accepted
{$result = 'login failed';
 $arr = array('error' => "username or password is wrong");
header('HTTP/1.1 401 Unauthorized', true, 401);
 echo json_encode($arr);}