值未使用php插入数据库
[英]Values not inserting in database with php
问题描述
I have created a user register form with php jquery and sql and i am trying to enter the details in database via ajax request, code is executing perfectly but values are not entering in the database, and i have checked my query too by running it in the sql editor query also working fine, 
我用php jquery和sql创建了一个用户注册表单，我试图通过ajax请求在数据库中输入详细信息，代码执行得很好，但是值没有输入数据库，我也通过在其中运行查询来检查我的查询sql编辑器查询也可以正常工作，
can you tell where is the error ? 你能知道错误在哪里吗？
<!DOCTYPE html>
<html>
<head>
<title>Login Register Test</title>
<script type="text/javascript" src="http://code.jquery.com/jquery-1.8.0.min.js"></script>
/**
* Created by pratyush on 8/28/14.
*/
$(function(){
$("input[name='btn_submit_reg']").click(function(){
registerUser();
});
$("input[name='btn_submit_login']").click(function(){
loginUser();
});
});
function registerUser(){
if(IsValidFormReg()){
var frm = $(".register").serialize();
$.ajax({
url : 'modal/registerdao.php',
type : 'POST',
data : frm,
success : function(result) {
if (result.indexOf("correct") > -1) {
alert(frm);
window.location.replace("registrationconfirm.php");
}
}
});
}
}
function IsValidFormReg()
{
var valid= true;
var username = $("input[name='username_reg']").val();
var userpass = $("input[name='userpass_reg']").val();
var email = $("input[name='useremail_reg']").val();
if(username.length==0){
valid = false;
$("input[name='username_reg']").addClass("formerror");
}
if(userpass.length==0){
valid = false;
$("input[name='userpass_reg']").addClass("formerror");
}
if(email.length==0){
valid = false;
$("input[name='useremail_reg']").addClass("formerror");
}
else{
if(checkemail(email)==false){
valid = false;
$("input[name='useremail_reg']").addClass("formerror");
alert("please enter valid email");
}
}
if(!valid)
$(".formentrieserror").html(" Please fill correct form entries...");
else
$(".formentrieserror").html(" ");
return valid;
}
function checkemail(email){
var filter = /^([a-zA-Z0-9_\.\-])+\@(([a-zA-Z0-9\-])+\.)+([a-zA-Z0-9]{2,4})+$/;
if(email.length>0){
if (!filter.test(email))
return false;
else
return true;
}
else
return false;
}
function loginUser(){}
<style>
.formerror{border: solid 2px red;}
</style>
</head>
<body>
<h2>Login Form</h2> <br><br>
<form class="login">
<input type="text" name="username_login" placeholder="user name"> <br> <br>
<input type="password" name="userpass_login" placeholder="password"><br> <br>
<input type="button" name="btn_submit_login" value="Login">
</form>
<br><br>
<h2>Registration Form</h2>
<br><br>
<form class="register">
<input type="text" name="username_reg" placeholder="user name"> <br> <br>
<input type="password" name="userpass_reg" placeholder="password"><br> <br>
<input type="email" name="useremail_reg" placeholder="email"><br> <br>
<input type="button" name="btn_submit_reg" value="Register">
</form>
<div class="formentrieserror"></div>
</body>
</html>
//registerDao.php..................................//
<?php
class RegisterUserInfo{
public $userName;
public $userPassword;
public $userEmail;
}
class userRegisterDao {
function RegisterUser($registration_info) {
include_once ("database.php");
$qry = "insert into userdetails(
userName,
userPassword,
userEmail)
values('".$registration_info->userName."','"
.$registration_info->userPassword."','"
.$registration_info->userEmail."')";
return Database::executeQuery($qry); // return true or false
}
}
$userName = mysql_escape_string($_REQUEST ['userName']);
$userPassword = mysql_escape_string($_REQUEST ['userPassword'] );
$userEmail = mysql_escape_string($_REQUEST ['userEmail'] );
$registration_info = new RegisterUserInfo();
$registration_info->userName=$userName;
$registration_info->userPassword=$userPassword;
$registration_info->userEmail=$userEmail;
$dao = new userRegisterDao();
$insert = $dao->RegisterUser($registration_info);
if($insert===true){
echo "correct";
}
else
echo "invalid";
?>
2 个解决方案
解决方案1
2 已采纳 2014-08-29 05:58:45
Change this line 更改此行
$userName = mysql_escape_string($_REQUEST ['userName']);
$userPassword = mysql_escape_string($_REQUEST ['userPassword'] );
$userEmail = mysql_escape_string($_REQUEST ['userEmail'] );
to this 对此
$userName = mysql_escape_string($_REQUEST ['username_reg']);
/* changed ^^ */
$userPassword = mysql_escape_string($_REQUEST ['userpass_reg'] );
/* changed ^^ */
$userEmail = mysql_escape_string($_REQUEST ['useremail_reg'] );
/* changed ^^ */
解决方案2
0 2014-08-29 06:11:59
I see this alot... Ajax with no error handler, just success handlers. 我看到了很多...没有错误处理程序的Ajax，只有成功处理程序。 Why not have php return errors to ajax and output to console IE console.log( some_error ) . 为什么不让php向ajax返回错误并输出到控制台IE console.log（some_error）。 It will make debugging a lot easier in the future. 将来，它将使调试变得更加容易。
I also see that there is no sql exception handling. 我也看到没有sql异常处理。 That would have shown you that no column exists by that name during attempted insertions. 那将向您显示在尝试插入期间该名称不存在任何列。
just a few debugging tips going forward, good luck. 只是一些调试技巧，祝您好运。
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