`我试图从注册表格输入数据到数据库,但它似乎不工作,我认为我做了正确的编码我不确定我在哪里做错了我阅读代码代码和我创建的所有数据库都遵循。我无法从表单到数据库输入信息
<form method='post' action='login.php'>
<table width='400' border='5' align='CENTER'>
<tr>
<td><h1>Registration</h1></td>
</tr>
<tr>
<td>User Name:</td>
<td><input type='text' name='name'/></td>
</tr>
<tr>
<td>Password:</td>
<td><input type='password' name='pass'/></td>
</tr>
<tr>
<td>Email:</td>
<td><input type='text' name='email'/></td>
</tr>
<tr>
<td><input type='submit' name='register' value='register'/></td>
</tr>
</table>
</form>
<?php
$connect=mysql_connect("localhost","root","");
$db_selected = mysql_select_db("users_db", $connect);
if(isset($_POST['submit'])){
$users_name = $_POST['name'];
$users_pass = $_POST['pass'];
$users_email = $_POST['email'];
if($users_name==''){
echo "<script>alert('Please enter your Username')</script>";
exit();
}
if($users_pass==''){
echo "<script>alert('Please enter your password')</script>";
exit();
}
if($users_email==''){
echo "<script>alert('Please enter your email')</script>";
exit();
}
$check_email="select*from
users where
users_email='$users_email'";
$run =
mysql_query($check_email) or
die(mysql_error());
if(mysql_num_rows($run)>0){
echo"<script>alert('Email $users_email
is already exist in our databse, please try another
one')</script>"; exit();
}
$query = "insert into users
(users_name,users_pass,users_email) values('
$users_name','$users_pass','$users_email')";
{
$result = mysql_query($query)
or die(mysql_error());
}
echo"<script>alert
window.open('','_self')</script>";
}
?>
**危险**:您正在使用[an **过时的**数据库API](http://stackoverflow.com/q/12859942/19068)并应使用[现代替换](http:// php.net/manual/en/mysqlinfo.api.choosing.php)。你也**易受[SQL注入攻击](http://bobby-tables.com/)**,现代的API会使[防御]更容易(http://stackoverflow.com/questions/60174/best-way-to-prevent-sql-injection-in-php)自己从。 – Quentin
您是否尝试过'echo'echo'$ query'并在phpMyAdmin或其他数据库工具中运行最终查询以确保它是正确的? –
我没有看到任何'INSERT'语句。所以没有数据会进入你的数据库 – Machavity