Question

我遇到了代码不断显示用户名可用的问题，即使它存在于数据库中。任何人都可以解决我遇到的问题吗？下面是我的代码。

PHP

CoUninitialize

的jQuery

    // Define $username
    $username=$_POST["emailAddress"];

    // Create Connection
    $db = new mysqli($server, $user, $password, $database);

    // Check Connection
    if ($db->connect_error) {
        die("Connection failed: " . $db->connect_error);
    }

    // SQL query to fetch registerd users and finds user match.
    $query = "SELECT email_address FROM users WHERE email_address='$username'";

    $result = mysqli_query($db,$query);    
    $rows = mysqli_num_rows($result);

    //if number of rows fields is bigger them 0 that means it's NOT available '  
    if($rows>0){  
    //and we send 0 to the ajax request  
        echo 0;  
    }else{  
    //else if it's not bigger then 0, then it's available '  
    //and we send 1 to the ajax request  
        echo 1;  
    }  
    mysqli_close($db); // Closing Connection

}

HTML

$(document).ready(function () {

//when button is clicked  
$('[name="checkAvail"]').on("click", function () {
    check_availability();
});

});

//function to check username availability  
function check_availability() {

//get the username  
var username = $('[name="emailAddress"]').val();

//use ajax to run the check  
$.post("checkusers.php",
    function (result) {
        //if the result is 1  
        if (result == 1) {
            //show that the username is available  
            $('.existingUser').html(username + ' is Available');
        } else {
            //show that the username is NOT available  
            $('.existingUser').html(username + ' is NOT Available');
        }
    });

Answer 1

你的ajax没有发送任何数据......所以你的php中$_POST["emailAddress"];将是空的

您需要添加数据对象作为$.post

的第二个参数

$.post("checkusers.php",{emailAddress : username },function (result) {
     // success handing code left out for clarity
});

此外，您应该检查输入是否为空，然后才允许用户进行检查。向服务器发送无效的空值没有意义。

您可以使用许多简单的JavaScript验证插件。

同样在服务器上应该验证用户输入。这是最重要的验证点，因为javascript可以解决以允许发送任何内容

Answer 2

您没有向帖子页面发送任何数据。

$.post( "checkusers.php", { emailAddress: username})
  .done(function( data ) {
    if (data == "1") {
        //show that the username is available  
        $('.existingUser').html(username + ' is Available');
    } else {
        //show that the username is NOT available  
        $('.existingUser').html(username + ' is NOT Available');
    }
  });

检查jQueryDocumentation

Answer 3

根据你的php脚本尝试这个if（result == 0）

 function (result) {
    //if the result is 0  
    if (result == 0) {
        //show that the username is available  
        $('.existingUser').html(username + ' is Available');
    } else {
        //show that the username is NOT available  
        $('.existingUser').html(username + ' is NOT Available');
    }

使用Ajax和PHP检查数据库中的现有用户

3 个答案: