Question

    echo "<form method='post' action='regprocess.php' id='registerform'>";
        echo '<fieldset class="register">';
        echo"<h2>Register</h2>";
            echo "<ul>";
                    echo '<li><label for="FirstName">First Name: </label> <input type="text" name="FirstName" id="FirstName"></li>';
                    echo '<li><label for="LastName">Last Name: </label> <input type="text" name="LastName" id="LastName"></li>';
                    echo '<li><label for="Email">Email: </label><input type="email" name="Email" id="Email"></li>';
                    echo '<li><label for="Username">Username: </label><input type="text" name="Username" id="Username"></li>';
                    echo '<li><input type="button" id="check_username_availability" value="Check Availability"></li>';  
                    echo '<div id="username_availability_result"></div>'; 
                    echo '<li><label for="Password">Password: </label><input type="password" name="Password" id="Password"></li>';
                    echo '<li><input type="submit" value="Register"></li>';
                    echo "</ul>";
        echo "</fieldset>";
        echo "</form>";
        }
        $username = mysql_real_escape_string($_POST['Username']);  

//mysql query to select field username if it's equal to the username that we check '  
$usernameresult = 'Select Username from User where Username = "'. $username .'"'; 
$uresult = $conn->query($usernameresult); 

//if number of rows fields is bigger them 0 that means it's NOT available '  
if($uresult->num_rows==1) {  
    //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;  
}  
?>
<script src="jquery-1.8.1.min.js" type="text/javascript"></script>
<SCRIPT LANGUAGE="JAVASCRIPT" TYPE="TEXT/JAVASCRIPT">
$(document).ready(function() {  

        var checking_html = 'Checking...';  

        //when button is clicked  
        $('#check_username_availability').click(function(){  
            //run the character number check  
            if($('#username').val().length < min_chars){   
                $('#username_availability_result').html(checking_html);  
                check_availability();  
            }  
        });  

  });  

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

        //get the username  
        var username = $('#username').val();  

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

}
</script>

大家好，所以我正在尝试制作一个表单来验证我的数据库中是否有可用的用户名。我之前有点使用过javascript，但是我没有用Ajax和JQuery做过很多。我一直在控制台中收到此错误消息 ReferenceError：找不到变量：$ 它响应这一行脚本 $（document）.ready（function（）{
我不确定为什么，我只是想知道是否有人能给我一些如何解决它的见解？谢谢。

Answer 1

代码似乎没问题。只是问题我认为你对jquery库的引用是不对的。查看源代码并单击包含行的jquery库来测试路径是否正常。将jquery-1.8.1.min.js保留在这些php文件所在的文件夹中，或者使用任何CDN图像

<script src="//ajax.googleapis.com/ajax/libs/jquery/1.8.1/jquery.min.js"></script>

希望它有所帮助!!

试图将jquery / ajax实现到html页面中

1 个答案: