PHP SQL注冊表格
[英]PHP SQL registration form
問題描述
我正在嘗試將數據保存到sql db中的注冊表。 我現在還沒有進行表單驗證，因為最困擾我的是將這些東西放到sql上。 我將不勝感激任何建議！
我有一個form.php文件，應該進行艱苦的工作。 提交表單時，此時，我得到一個空白屏幕，並且沒有任何內容加載到數據庫中。
<?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";
?>
我有一個包含以下內容的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，我有一個名為“ registration”的sql數據庫（我正在本地運行xampp），我創建的表稱為“ userTable”，它包含8個字段，包括ID（自動遞增）和我包含的其他7個值頂上。 知道我做錯了什么嗎？
2 個解決方案
解決方案1
2 2012-08-03 01:45:54
問題是什么？
1）問題是，每次您加載該頁面時，它都會進行INSERT查詢-意味着每次它插入空值。 為什么？
僅僅因為沒有條件檢查是否所有字段都已過帳，所以代替：
<?php
$fname = $_POST['fname'];
$lname = $_POST['lname'];
$password = $_POST['password'];
$email = $_POST['email'];
$cell = $_POST['cell'];
$experience = $_POST['experience'];
$ip = $_POST['ip'];
您應該檢查$_POST super-global是否具有某些密鑰。 因此，在執行任何查詢之前-首先檢查$_POST是否為空
<?php
//This means that user did submit the form
if ( !empty($_POST) ){
//all your stuff goes here
}
?>
<html>
.....
</html>
2）確定要控制代碼嗎？ 顯然不是。
您必須檢查某個函數是否返回TRUE ，然后根據該函數執行以下操作。
例如，您確定mysql_query("your sql query")成功了嗎？
3）啟用error_reporting到E_ALL，因此只需將error_reporting(E_ALL)放在頁面頂部，如下所示：
<?php
error_reporting(E_ALL);
這樣您就可以隨時“動態”調試腳本
4）您正在盡一切努力使此代碼難以維護，為什么？ 看這個：
<?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>
解決方案2
-1 已采納 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注冊表格-SQL
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