javascript - Codeigniter JavaScript 错误

Question

我有脚本，检查用户是否可以注册:

PHP:

<?php  

if(isset($_POST['username']))//If a username has been submitted
{
    $username = mysql_real_escape_string($_POST['username']);//Some clean up :)

    $check_for_username = mysql_query("SELECT * FROM users WHERE username='$username'");
    //Query to check if username is available or not

    if(mysql_num_rows($check_for_username))
    {
       echo '1';//If there is a  record match in the Database - Not Available
    }
    else
    {
        echo '0';//No Record Found - Username is available
    }
}
?>

JAVASCRIPT:

<script>
$(document).ready(function()//When the dom is ready
{
    $("#username").change(function()
    { //if theres a change in the username textbox

        var username = $("#username").val();//Get the value in the username textbox
        if(username.length > 3)//if the lenght greater than 3 characters
        {
            $("#availability_status").html('Checking availability...');
            //Add a loading image in the span id="availability_status"

            $.ajax({  //Make the Ajax Request
                type: "POST",
                url: "http://mywebsite.com/auth/sign_up",
                data: "username="+ username,  //data
                success: function(server_response){

                    $("#availability_status").ajaxComplete(function(event, request){

                        if(server_response == '0')//if ajax_check_username.php return value "0"
                        {
                            $("#availability_status").html('<font color="Green"> Available </font>  ');
                           //add this image to the span with id "availability_status"
                        }
                        else  if(server_response == '1')//if it returns "1"
                        {
                            $("#availability_status").html('<font color="red">Not Available </font>');
                        }

                    });
                }

            });

        }
        else
        {

            $("#availability_status").html('Username too short');
           //if in case the username is less than or equal 3 characters only
        }
        return false;
    });
});
</script>

但是当我写入用户名字段时，我遇到了 firebug 错误: 不允许您请求的操作。 [14:32:30.980] 发布 http://mywebsite.com/auth/sign_up [HTTP/1.1 500 内部服务器错误 63ms]

有什么建议吗？

抱歉我的英语不好:)

Answer 1

由于多种原因，这绝对不是这样做的方法

1. 不推荐使用 mysql_ 扩展使用 PDO
2. 如果您使用的是 codeigniter，则使用 Active Record 类进行数据库调用，使用输入类获取 POST 变量

但是我想演示可能的流程

php

$response = array();

if(isset($_POST['username'])) //If a username has been submitted
{
    //assume username is not available
    $status = false;
    $statusText = "Not Available";
    $username = mysql_real_escape_string($_POST['username']); //Some clean up :)

    //Query to check if username is available or not
    $isAvailable = mysql_query("SELECT * FROM users WHERE username='$username'");

    if(mysql_num_rows($isAvailable)) {
        $status = true;
        $statusText = "Available";
    }

    $response = array(
        "status" => $status,
        "statusText" => $statusText
    );
}

return json_encode($response);

js

//alex >> comments usually go on top of the line you are commenting plus
//you dont have to comment every single line only when something 'big' or 'strange' happens

//When the dom is ready
$(document).ready(function()
{
    //if there is a change in the username textbox
    $("#username").change(function()
    {
        //Get the value in the username textbox
        var username = $("#username").val();
        //alex >> caching the div so i dont call it every time which improves performance
        var statusDiv = $("#availability_status");

        //if the length greater than 3 characters
        if(username.length > 3)
        {
            //TODO Add a loading image in the span id="availability_status"
            statusDiv.html('Checking availability...');

            $.ajax({
                type : "POST",
                url : "http://mywebsite.com/auth/sign_up",
                data : "username=" + username,
                success : function(server_response)
                {
                    var color = "red";
                    var statusText = "an error occurred !";

                    server_response = JSON.parse(server_response);

                    if(server_response)
                    {
                        statusText = server_response.statusText;

                        if(server_response.status)
                            color = "green";
                    }
                    statusDiv.html("span class='" + color + "'" + statusText + "span");
                }
            });
        }
        else
        {
            //if in case the username is less than or equal 3 characters only
            statusDiv.html('Username too short');
        }

        return false;
    });
});

只需记住将 span 设为适当的标签，因为“智能”编辑器正在杀死整个标签

statusDiv.html("<span class='" + color + "'>" + statusText + "</span>");

这让我想起了!

始终检查是否为空:P