检查用户名是否可用AJAX [英] Checking if the username is available with AJAX

问题描述

我使用的AJAX基本登记表，但形式不能连接到数据库。我显然忽视的东西。

I'm using a basic registration form with AJAX, but the form is not connecting to the database. I'm obviously overlooking something.

因此​​，这里是我想要验证的字段。

So here's the field I want to validate.

Username:<input type="text" name="user" id="user" maxlength="30">
<span id="msgbox" style="display:none"/></input>

然后我使用jQuery，这里的code：

Then I use jQuery, here's the code:

$(document).ready(function() {
    $("#user").blur(function() {

        //remove all the class add the messagebox classes and start fading
        $("#msgbox").removeClass().addClass('messagebox').text('Checking...').fadeIn("slow");
        //check the username exists or not from ajax
        $.post("user_availability.php",{ user_name:$(this).val() },
            function(data) {
                if(data=='no') { //if username not avaiable
                    $("#msgbox").fadeTo(200,0.1,function() {//start fading the messagebox
                        //add message and change the class of the box and start fading
                        $(this).html('This User name Already exists').addClass('messageboxerror').fadeTo(900,1);
                    });       

                } else {
                    $("#msgbox").fadeTo(200,0.1,function() { //start fading the messagebox
                        //add message and change the class of the box and start fading
                        $(this).html('Username available to register').addClass('messageboxok').fadeTo(900,1);   
                    });
                } // else
             } // function

        ); // $.post
    }); // blur
}); // ready

和我有这个code， user_availability.php ：

And I have this code, user_availability.php:

mysql_connect(localhost,$user,$password);
    or die('There is error to connect to server:-'.mysqli_connect_error());

$db_selected = mysql_select_db($database);
$user_name=$_POST['user_name'];
$sql = "select * from members where username='$user_name'";
$result = mysql_query($sql);

while($row = mysql_fetch_assoc($result))   
{
    $existing_users[] = $row['username'];
}

if (in_array($user_name, $existing_users))
{
    echo "no"; //user name is not availble
}
else
{
    echo "yes"; //user name is available
}

我没有得到任何数据库错误。该形式将更加可观，但我不能把它与这个领域的工作。有什么建议？

I get no database errors. The form will be more substantial, but I can't get it to work with this field. Any suggestions?

推荐答案

SELECT语句只是鸡蛋里挑骨头。因为你只需要获得多少行所输入的用户名存在，你不需要做一个SELECT *为依据列在用户表，你可能会返回相当多的数据的数量。我将修改您的查询要像这样：

Just a nitpick on the select statement. Since you only need to get how many rows exist for the entered username you don't need to do a "select *" as depending on the number of columns in your users table you could be returning quite a bit of data. I would revise your query to be like so:

$sql = "select COUNT(username) from members where username=$user_name";

然后检查，以确保查询的结果等于零。如果是这样，则用户名是可用的。

Then you check to make sure that the result of the query equals zero. If it does then the username is available.

我知道上面是不是一个回答你的问题，但只是想我要指出来，因为这看起来是，是要被调用了很多依赖于流量的报名表和创意功能您的访客。

I know that the above wasn't an answer to your question but just thought I would point it out since this looks to be a function that is going to get called a lot depending on the traffic on your registration form and the creativity of your visitors.

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