Question

我有一个html表单，我用Php使用Jquery验证。成功将数据插入数据库后，它会在html页面中显示成功消息并重定向到不同的页面。好吧，它工作正常，但如果有错误，它也会重定向到不同的页面。

我想要的是什么：
如果有错误，应该留在此页面上。如果没有错误，那么它应该重定向到另一个页面。我怎么能用jquery做到这一点？

<center><img id="loading-image" src="images/loading-image.gif" style="display:none; margin-bottom:10px !important;"/></center>
<div id="result"></div>

$('body').on('click', '#request_tutor', function (e) {
    e.preventDefault();

    var formData = new FormData($(this).parents('form')[0]);
    $("#loading-image").show();
    $.ajax({
        url: 'request_tutor_process.php',
        type: 'POST',
        xhr: function () {
            var myXhr = $.ajaxSettings.xhr();
            return myXhr;
        },
        success: function (data) {
            $("#loading-image").hide(); //hide loading          
            $("#result").html(data);

            setTimeout(function () {
                $('input[type=submit]').attr('disabled', false);
                window.location.href = "login/index.php";
            }, 2000);

        },

        complete: function () {
            $("#loading-image").hide(); //hide loading here
        },

        data: formData,
        cache: false,
        contentType: false,
        processData: false
    });
});

Php页面：

$err =  array();
// required eata validation checking...
if(isset($name) && isset($tutoring_location) && isset($pcode) && isset($mphone) && isset($email) && isset($tutor_level)){
    if(empty($name) && empty($tutoring_location) && empty($pcode) && empty($mphone) && empty($email) && empty($tutor_level)){
        $err[] = "Please fill up required field which is indicated by *";
    }else{
        if(empty($name))    
            $err[] = "Your name required";
        elseif(strlen($name) > 30)
            $err[] = "Your name is too long";

        if(empty($tutoring_location))   
            $err[] = "Write your tutoring location";
        elseif(strlen($tutoring_location) > 255)
            $err[] = "Tutoring location address is too long";

        if(empty($pcode))   
            $err[] = "Postal code required";
        elseif(strlen($pcode) > 6)
            $err[] = "Invalid postal code"; 
        elseif(!is_numeric($pcode))
            $err[] = "Invalid postal code";

        if(empty($mphone))  
            $err[] = "Mobile phone number required";
        elseif(strlen($mphone) > 8)
            $err[] = "Invalid mobile phone number";     
        elseif(!is_numeric($mphone))
            $err[] = "Invalid mobile phone number"; 

        if(empty($email))   
            $err[] = "Your email address required.";    
        elseif(@!eregi("^[_a-z0-9-]+(\.[_a-z0-9-]+)*@[a-z0-9-]+(\.[a-z0-9-]+)*(\.[a-z]{2,3})$", $email))
            $err[] = "Invalid email address";
        elseif($num_email == 1)
            $err[] = "Email address <strong>$email</strong> is already exits, please choose another";       

        if(empty($tutor_level)) 
            $err[] = "Select your tutor level";         
    }

    if(!empty($err)){
        echo "<div class='error'>";
        foreach($err as $er){
            echo "$er. <br/>";
        }
        echo "</div>";
    }else{
        ecoh "success";
    }
}

Answer 1

这可能是罪魁祸首

setTimeout(function () {
    $('input[type=submit]').attr('disabled', false);         
    window.location.href = "login/index.php"; 
},   2000);

代码成功后，您将设置一个全局超时，每两秒重定向一次页面。出于某种原因，由于您没有显示PHP代码，因此它总是成功的。基本上，只要响应返回有效（HTTP响应代码200），它就会一直成功。如果PHP端存在错误，则需要设置相应的错误标头，以便调用jQuery错误函数。

结论：您很可能不会向您的AJAX函数返回HTTP错误代码，导致始终调用成功函数。

Answer 2

请勿使用前端语言来验证表单输入。

我看到对'request_tutor_process.php'的AJAX调用使用POST，所以request_tutor_process.php应该有类似的东西：

if($_POST['action'] == 'formSubmit'){
    //Let's say you have this function called 'add_to_database'
    attemptToAdd = add_to_database();
    //Let's say something went wrong with the form validation, and you return 'false'
    if(attemptToAdd === false) echo "Failed to submit, please try again later";
    //Let's say it worked, but it was unsuccessful (the username already exists)
    if(attemptToAdd === "unsuccessful") header('Location: /failed.php');
    //Let's say it worked fully...
    if(attemptToAdd === "successful") echo "It worked, click *here* to continue;
}

Answer 3

你的问题是它总是会进入成功模块，对吗？

如果是这样，请关注@Zarathuztra answer并在您的PHP代码中将HTTP代码设置为200以外的任何值，并添加一个错误块来处理错误。请参阅此答案Return errors from PHP run via. AJAX?

或者您可以检查数据的值，看看是否有错误。

Answer 4

试试这个：

     success: function (data) {
          $("#loading-image").hide(); //hide loading          
          $("#result").html(data);

        if(data == "success") {
          setTimeout(function () {
            $('input[type=submit]').attr('disabled', false);
            window.location.href = "login/index.php";
          }, 2000);
        } else {
          alert('error');
          return false;
        }

顺便说一句，PHP有点乱，但这不是问题。基本上因为数据有一个值，它正在成功，你没有处理错误

使用php和jquery成功将数据插入数据库后页面重定向

4 个答案: