使用AJAX将查询发送到php文件时，变量保持为空 [英] Variable remains empty when using AJAX in sending query to php file

问题描述

我有一个表单，在提交表单之前，我会进行一些AJAX错误处理，只是为了改善用户体验。我的问题是变量 $ user_password 在整个过程中似乎都为空，因此错误处理无关紧要。

I have a form in which I do some AJAX error handling before the form is submitted, just to improve user experience. My problem is that the variable $user_password seems to remain empty throughout the process, thus the error handling is irrelevant.

在下面的代码中，第一个键入功能用于检查密码是否长于最小长度，第二个用于检查密码是否匹配：

In the following code, the first keyup function is to check if the password is longer than the minimum length and the second is to check if the passwords match:

$(document).ready(function() {
    $("input[name=user_password]").keyup(function(){    
        var user_password = $("input[name=user_password]").val();
        //data to server...
        $.post("../server/hub.php", {
            //POST name and variable... 
            check_password: user_password
            //places fetched information...
        }, function(data, status) {
            $("#user_password").html(data);
        });
    });
});


$(document).ready(function() {

    $("input[name=user_password_2]").keyup(function(){
        var user_password = $("input[name=user_password]").val();
        var user_password_2 = $("input[name=user_password_2]").val();
        //data to server...
        $.post("../server/hub.php", {
            //POST name and variable... 
            password_match: user_password,
            password_match_2: user_password_2
            //places fetched information...
        }, function(data, status) {
            $("#user_password_2").html(data);
        });
    });

});

变量被重定向到实际执行错误处理的php文件：

The variables are redirected to a php file where the error handling is actually conducted:

if (isset($_POST['check_password'])) {
    $user_password = $_POST['check_password'];

    echo $user_password;

    if ($user_password == "") {
        echo "";
    } elseif (strlen($user_password) < 6) {
        echo "Password must contain at least six characters!";
    } else {
        echo "You are good to go!";
    }
}

if (isset($_POST['password_match'])) {
    $user_password = $_POST['password_match'];
    $user_password_2 = $_POST['password_match_2'];

    if ($user_password_2 == "") {
        echo "";
    } elseif ($user_password !== $user_password_2) {
        echo "Sorry, passwords do not match!";
    } else {
        echo "You are good to go!";
    }
}

尽管返回到html文件的数据仍然为空，并且回显 $ user_password 不会产生结果。

Although the data returned to the html file remains empty and echoing the $user_password yields no result.

以下是html段：

<form action="../server/register.php" method="POST">
                        <div class="input_table">                           
                            <table>
                                <thead>
                                    <tr>
                                        <th><h1>Register to LIMS</h1></th>
                                    </tr>
                                </thead>
                                <tbody>

                                    <tr>
                                        <td><input type="password" name="user_password" placeholder="Select Password"><p id="user_password"></p></td>
                                    </tr>
                                    <tr>
                                        <td><input type="password" name="user_password_2" placeholder="Repeat Password"><p id="user_password_2"></p></td>
                                    </tr>
                                </tbody>
                            </table>
                        </div>
                        <button type="submit" name="user_register" class="button_1">Register</button>
                        <button type="button" class="button_3">Cancel</button>              
                    </form>

有人可以解释为什么会这样吗？

Can anyone please explain why this is occurring?

推荐答案

因此，问题的答案非常简单，但很容易遗漏。这再次表明为什么编写成功的网站必须要有良好的编码习惯。我似乎在这种情况下， name = user_password 在代码的其他位置重复了，因此在使用以下命令调用时：

So the answer to the problem is quite simple but very easy to miss. Again this shows why good coding practice is imperative for writing a successful website. I seems that in this case the name=user_password was repeated somewhere else in the code, thus when called using:

var user_password = $("input[name=user_password]").val();

它是指代码另一部分中的输入，而不是我尝试应用的输入编码。只需将html代码中的 name = user_password 更改为 name = user_password_1 ，然后更改ajax代码即可：

it refered to input in another part of the code and not the input that I was trying to apply the code to. Simply by changing name=user_password to name=user_password_1 in the html code and then changing the ajax code:

$(document).ready(function() {
    $("input[name=user_password_1]").keyup(function(){  
        var user_password_1 = $("input[name=user_password_1]").val();
        //data to server...
        $.post("../server/hub.php", {
            //POST name and variable...
            check_password: user_password_1
            //places fetched information...
        }, function(data, status) {
            $("#user_password_1").html(data);
        });
    });


    $("input[name=user_password_2]").keyup(function(){
        var user_password_1 = $("input[name=user_password_1]").val();
        var user_password_2 = $("input[name=user_password_2]").val();
        //data to server...
        $.post("../server/hub.php", {
            //POST name and variable... 
            password_match_1: user_password_1,
            password_match_2: user_password_2
            //places fetched information...
        }, function(data, status) {
            $("#user_password_2").html(data);
        });
    });

});

代码完美运行。尽管我个人认为这是一个愚蠢的错误，但我仍然相信将其作为答案发布很重要，这样其他人可以从仔细检查其代码中的此类愚蠢错误中受益。

The code works perfectly. Although this was, in my own opinion, an idiotic mistake, I still believe it to be important to post it as an answer so that other people may benefit from double checking their code for silly mistakes such as this.

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