PHP SQL注册表格
[英]PHP SQL registration form
问题描述
我正在尝试将数据保存到sql db中的注册表。 我现在还没有进行表单验证，因为最困扰我的是将这些东西放到sql上。 我将不胜感激任何建议！
我有一个form.php文件，应该进行艰苦的工作。 提交表单时，此时，我得到一个空白屏幕，并且没有任何内容加载到数据库中。
<?php
$fname = $_POST['fname'];
$lname = $_POST['lname'];
$password = $_POST['password'];
$email = $_POST['email'];
$cell = $_POST['cell'];
$experience = $_POST['experience'];
$ip = $_POST['ip'];
$password = md5($_POST['password']);
$connection = msql_connect(localhost, USERNAME, PASSWORD);
$db = mysql_select_db(registration,$connection);
mysql_query("INSERT INTO userTable (fname,lname,password,email,cell,experience,ip) VALUES ('$fname', '$lname', '$password', '$email', '$cell', '$experience', '$ip')")
or die (mysql_error());
echo "Thank you for your registration";
?>
我有一个包含以下内容的html文件：
<form method = "post" action = "form.php">
<h2>User Information</h2>
<div><label>First Name:</label>
<input type = "text" name = "fname"></div>
<div><label>Last Name:</label>
<input type = "text" name = "lname"></div>
<div><label>Password:</label>
<input type = "password" name = "password"></div>
<div><label>Email:</label>
<input type="text" name="email"></div>
<div><label>Cellphone:</label>
<input type="text" name="cell"></div>
<input type="hidden" name="ip" value='<?php echo $IP ?>'/>
<h2>What Is Your Experience Mountain Biking?</h2>
<p><input type="radio" name="experience" value="n00b"
checked>n00b
<input type="radio" name="experience" value="intermediate">Intermediate
<input type="radio" name="experience" value="extreme">Extreme
</p>
<p><input type="submit" name="submit" value="Register"></p>
</form>
最后，我有一个名为“ registration”的sql数据库（我正在本地运行xampp），我创建的表称为“ userTable”，它包含8个字段，包括ID（自动递增）和我包含的其他7个值顶上。 知道我做错了什么吗？
2 个解决方案
解决方案1
2 2012-08-03 01:45:54
问题是什么？
1）问题是，每次您加载该页面时，它都会进行INSERT查询-意味着每次它插入空值。 为什么？
仅仅因为没有条件检查是否所有字段都已过帐，所以代替：
<?php
$fname = $_POST['fname'];
$lname = $_POST['lname'];
$password = $_POST['password'];
$email = $_POST['email'];
$cell = $_POST['cell'];
$experience = $_POST['experience'];
$ip = $_POST['ip'];
您应该检查$_POST super-global是否具有某些密钥。 因此，在执行任何查询之前-首先检查$_POST是否为空
<?php
//This means that user did submit the form
if ( !empty($_POST) ){
//all your stuff goes here
}
?>
<html>
.....
</html>
2）确定要控制代码吗？ 显然不是。
您必须检查某个函数是否返回TRUE ，然后根据该函数执行以下操作。
例如，您确定mysql_query("your sql query")成功了吗？
3）启用error_reporting到E_ALL，因此只需将error_reporting(E_ALL)放在页面顶部，如下所示：
<?php
error_reporting(E_ALL);
这样您就可以随时“动态”调试脚本
4）您正在尽一切努力使此代码难以维护，为什么？ 看这个：
<?php
//Debug mode:
error_reporting(E_ALL);
//Sure you want to show some error if smth went wrong:
$errors = array();
/**
*
* @return TRUE if connection established
* FALSE on error
*/
function connect(){
$connection = mysql_connect(localhost, USERNAME, PASSWORD);
$db = mysql_select_db(registration,$connection);
if (!$connection || !$db ){
return false;
} else {
return true;
}
}
//So this code will run if user did submit the form:
if (!empty($_POST)){
//Connect sql server:
if ( !connect() ){
$errors[] = "Can't establish link to MySQL server";
}
$fname = $_POST['fname'];
$lname = $_POST['lname'];
$password = $_POST['password'];
$email = $_POST['email'];
$cell = $_POST['cell'];
$experience = $_POST['experience'];
//Why post ip? not smth like $_SERVER['REMOTE_ADDR']...
$ip = $_POST['ip'];
$password = md5($_POST['password']);
//No error at this point - means that it successfully connected to SQL server:
if ( empty($errors) ){
//let's prevent sql injection:
$fname = mysql_real_escape_string($fname);
//Please do this for all of them..
}
//Now we should try to INSERT the vals:
$query = "INSERT INTO `userTable` (`fname`,`lname`,`password`,`email`,`cell`,`experience`,`ip`) VALUES ('$fname', '$lname', '$password', '$email', '$cell', '$experience', '$ip')";
//So try it:
if ( !mysql_query($query) ){
//
//die (mysql_error());
$errors[] = "Can't insert the vals";
} else {
//Or on success:
print ("Thank you for your registration");
//or you can do redirect to some page, like this:
//header('location: /thanks.php');
}
}
?>
<form method="post">
<h2>User Information</h2>
<div><label>First Name:</label>
<input type = "text" name = "fname"></div>
<div><label>Last Name:</label>
<input type = "text" name = "lname"></div>
<div><label>Password:</label>
<input type = "password" name = "password"></div>
<div><label>Email:</label>
<input type="text" name="email"></div>
<div><label>Cellphone:</label>
<input type="text" name="cell"></div>
<input type="hidden" name="ip" value='<?php echo $IP ?>'/>
<h2>What Is Your Experience Mountain Biking?</h2>
<p><input type="radio" name="experience" value="n00b"
checked>n00b
<input type="radio" name="experience" value="intermediate">Intermediate
<input type="radio" name="experience" value="extreme">Extreme
</p>
<?php if ( !empty($errors) ) : ?>
<?php foreach($errors as $error): ?>
<p><b><?php echo $error; ?></b></p>
<?php endforeach; ?>
<?php endif; ?>
<p><input type="submit" name="submit" value="Register"></p>
</form>
解决方案2
-1 已采纳 2012-08-03 01:59:11
Solved my own problem
<?php
$fname = $_POST['fname'];
$lname = $_POST['lname'];
$password = $_POST['password'];
$email = $_POST['email'];
$cell = $_POST['cell'];
$experience = $_POST['experience'];
$ip = $_POST['ip'];
$fname = mysql_real_escape_string($fname);
$lname = mysql_real_escape_string($lname);
$email = mysql_real_escape_string($email);
$password = md5($_POST['password']);
$servername="localhost";
$username="user";
$conn= mysql_connect($servername,$username, password)or die(mysql_error());
mysql_select_db("registration",$conn);
$sql="insert into userTable (fname,lname,password,email,cell,experience,ip) VALUES ('$fname', '$lname', '$password', '$email', '$cell', '$experience', '$ip')";
$result=mysql_query($sql,$conn) or die(mysql_error());
print "<h1>you have registered sucessfully</h1>";
echo "Thank you for your registration to the ";
mysql_close($connection);
？>
PHP注册表格-SQL
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