从ajax形式返回值

Question

I have a login form in php. 我在php中有一个登录表单。 I have used ajaxform so that the page does not get reloaded when the username or password is wrong. 我使用了ajaxform，以便在用户名或密码错误时不会重新加载页面。 I have another page checklogin.php which checks whether the username and pw is there in the database and it returns the count of the number of rows. 我还有另一个页面checklogin.php，它检查数据库中是否存在用户名和密码，并返回行数的计数。 I want to compare the count in the login.php and display the error message if the count=1 and redirect to another page if count=2. 我想比较login.php中的计数，如果count = 1则显示错误消息，如果count = 2则重定向到另一页。 I tried to display the count in an errordiv using target: 'errordiv' and checking its innerHTML but it failed to do so. 我试图使用target：'errordiv'并在errordiv中显示计数，并检查其innerHTML，但这样做失败。

My login.php 我的login.php

<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.8/jquery.min.js"></script>
  <script type="text/javascript" src="scripts/jquery.form.js"></script>
<!--Slider-in icons-->
<script type="text/javascript">


    function checklogin()
    {
     var request=$("#login-form").ajaxForm(
                    {

                  target:'#errordiv'

                    }).abort();
                request.submit();
        var divmsg=document.getElementById("errordiv");
        if (divmsg.childNodes[0].nodeValue == "1")
            {
        divmsg.childNodesp[0].nodeValue="Sorry";
        alert ("Invalid username or password");} //I didn't get the alert as well a the divmsg content didn't change


                    };
</script>

</head>
<body>

<!--WRAPPER-->
<div id="wrapper">

<!--LOGIN FORM-->
<form name="login-form" id="login-form" class="login-form" action="checklogin.php" method="post">


    <!--CONTENT-->
    <div class="content">
    <!--USERNAME--><input name="un" type="text" class="input username" placeholder="Username" onfocus="this.value=''" required="required" /><!--END USERNAME-->
    <!--PASSWORD--><input name="pw" type="password" class="input password" placeholder="Password" onfocus="this.value=''" required="required" /><!--END PASSWORD-->
    </div>
    <!--END CONTENT-->

    <!--FOOTER-->
    <div class="footer">
    <!--LOGIN BUTTON--><input type="submit" name="submit" value="Login" class="button" onclick="checklogin()" /><!--END LOGIN BUTTON-->

    </div>
    <!--END FOOTER-->

</form>
<div id="errordiv" class="errordiv"></div>
<!--END LOGIN FORM-->

checklogin.php checklogin.php

<?php

 include ('dbconn.php');
       $user = trim($_POST['un']);
        $pass = trim($_POST['pw']);
        $user = mysql_real_escape_string($user);
        $pass = mysql_real_escape_string($pass);
       $result=mysql_query("select Id from login where Username='$user' and Password='$pass'") or die (mysql_error());
      $result= mysql_fetch_assoc($result);
     $num= count ($result);
    echo $num;
?>

How can I get the value of $num in login.php without displaying it to a div and compare the value of $num and accordingly display the errormsg in errordiv or if $num==2 redirect to another page. 如何在login.php中获取$ num的值而不将其显示到div并比较$ num的值，并相应地在errordiv中显示errormsg，或者如果$ num == 2重定向到另一页。

Answer 1

Change your code to include the code which change the div and provide alert to be included in the callback. 更改您的代码以包括更改div并提供要包含在回调中的警报的代码。

     var request=$("#login-form").ajaxForm(
                 {
                  target:'#errordiv', 
                  success: function (html) {
                      if ($("#errordiv).html() == "1") {
                          $("#errordiv).html("Sorry");
                          alert ("Invalid username or password");  
                      }
                  }
                    }).abort();
                request.submit();

Answer 2

Maybe try to make the checking request asynchronously and process user feedback in AJAX callback: 也许尝试异步发出检查请求并在AJAX回调中处理用户反馈：

JS: JS：

    $(function(){
        $("#login-form").ajaxForm(function(response){
            if(response.count===1){
                $("#errordiv").html("msg you want to show");
            }else if(response.count===2){
                //redirect
            }
        });
    });

php: 的PHP：

<?php
    header("Content-Type: application/json");

    //your database check logic
    $user = trim($_POST['un']);
    $pass = trim($_POST['pw']);
    $user = mysql_real_escape_string($user);
    $pass = mysql_real_escape_string($pass);
    $result=mysql_query("select Id from login where Username='$user' and Password='$pass'") or die (mysql_error());
    $result= mysql_fetch_assoc($result);
    $num= count ($result);

    //and return response in JSON
    echo json_encode(array("count"=>$num));

?> ？>

Hope this is helpful for you. 希望这对您有帮助。

[EDIT] [编辑]

remove the form submit button inline JS function invoking: 删除表单提交按钮内联JS函数调用：

onclick="checklogin()"

and put checklogin function logic to document ready callback, initialize the form when the DOM is ready 并将checklogin函数逻辑记录到文档就绪回调中，在DOM准备就绪时初始化表单

从ajax形式返回值

问题描述

2 个解决方案

解决方案1
1 2013-07-11 02:47:35

解决方案2
0 2013-07-11 02:56:15

从ajax形式返回值

问题描述

2 个解决方案

解决方案1 1 2013-07-11 02:47:35

解决方案2 0 2013-07-11 02:56:15

解决方案1
1 2013-07-11 02:47:35

解决方案2
0 2013-07-11 02:56:15