表格提交 - 成功

好吧，我有一个非常奇怪的表单提交问题，我浪费了几天，仍然无法弄清楚： 1）当我正确填写表格时（对于新用户）效果很好 2）if我尝试使用现有电子邮件注册用户...它只是刷新而不是抛出错误 3）如果我在尝试失败后提交表单，我没有收到任何成功标志，尽管数据已提交到数据库此处是我的代码：表格提交 - 成功

$(function() { 
$("#send").click(function() { 
    // validate and process form here 
//var email = $("#email").val(); 
//var lname = $("input#lname").val(); 
var dataString = $("#pForm").serialize(); 
//alert (dataString); 
     $.ajax({ 
      type: "POST", 
      url: "../php/insertUser.php", 
      data: dataString, 
      success:function(msg, status){ 
     //alert (dataString); 
     var reply = parseInt(msg); 
     alert(status+ "" + msg); 
     if(reply ==1){ 
      alert('Email address already exists in our members database.\nPlease try another address and then submit it again!'); 
     } 
     else if(reply ==2){ //do nothing 
     // alert('You have one or more empty fields!\nPlease provide all the information required and then submit it again!'); 
     } 
     else if(reply == 0){ 
     $('#pForm').hide('fast'); 
     $('#accForm').show('slow'); 
     } 
           } 
       });  

    }); 
});

和我的PHP是这样的：

<?php 
include ('databaseCon.php'); 
$err = 0; 
//echo $data= $_POST['dataString']; 
$accType = $_POST['accType']; 
$fname =$_POST['fname']; 
$lname = $_POST['lname']; 
$email = $_POST['email']; 

$country = $_POST['country']; 
$state = $_POST['state']; 
$city = $_POST['city']; 



// check for empty fields 
if (($accType == "") || ($fname == "") || ($lname == "") || ($email == "") || ($country == "") || ($city == "")){ 
$empty = true; 
} 

if ($email != ""){ 
    $em = mysql_query("Select Email from Users where Email = '$email';") or die(mysql_error()); 
    $row = mysql_num_rows($em); 

    } 
//else{echo "unknown error 2 " + $email + "\n";} 
    if ($empty == false){ 
     //echo ""+ $accType +"\n"+ $fname +"\n"+ $lname +"\n"+ $email +"\n"+ $country +"\n"+ $state +"\n"+ $city + "" ; 
     if($row == 0){ 
     echo $users = mysql_query("Insert into Users values('','$fname', '$lname', '$email', '$accType', '$country', '$state', '$city');") or die(mysql_error()); 
     } 
     else{ 
      $err = 1;} 
     } 
    else{ 
    $err = 2; 
    } 
echo $err; 

?>

任何想法和感谢您的帮助！

来源

2009-12-16 uran

检查您的HTML并查看提交按钮是否确实在

尝试使用类似Firebug的Firefox扩展，然后您可以看到您的发布数据和响应是否处于有效格式。它将帮助您识别问题 – 2009-12-16 06:20:04

您是否通过在ajax调用上实现了错误处理程序进行了测试？

在一个不相关的说明中，您确实需要对查询使用某种形式的数据转义/数据库保护，最好是使用prepare。

来源

2009-12-16 04:14:16 Aaron

if(reply ==1){ 
    alert('Email address already exists in our members database.\nPlease try another address and then submit it again!'); 
    return false; 
}

试试这个。

来源

2012-03-02 15:18:32 gokturk

表格提交 - 成功

回答

相关问题