PHP更新表中的行
[英]PHP update rows in table
问题描述
您好，我在这里遇到了一些问题，我创建了一个脚本来更新用户帐户详细信息，但是当填写表格并单击提交按钮时，不会出现任何错误，但是与此同时，表中没有任何更改
这只是一个虚拟应用程序，因此一切都保持基本
<?php
session_start();
include('connect_mysql.php');
if(isset($_POST['update']))
{
$usernameNew = stripslashes(mysql_real_escape_string($_POST["username"]));
$passwordNew = stripslashes(mysql_real_escape_string($_POST["password"]));
$first_nameNew = stripslashes(mysql_real_escape_string($_POST["first_name"]));
$last_nameNew = stripslashes(mysql_real_escape_string($_POST["last_name"]));
$emailNew = stripslashes(mysql_real_escape_string($_POST["email"]));
$user_id = $_SESSION['user_id'];
$editQuery = mysql_query("UPDATE users SET username='$usernameNew', password='$passwordNew', first_name='$first_nameNew', last_name='$last_nameNew' , email='$emailNew' WHERE user_id='$user_id'");
if(!$editQuery)
{
echo mysql_error($editQuery);
die($editQuery);
}
}
?>
<html>
<head>
<title>Edit Account</title>
<meta http-equiv="content-type" content="text/html; charset=iso-8859-1" />
<link href="style.css" rel="stylesheet" type="text/css" />
</head>
<body>
<div id="wrapper">
<header><h1>E-Shop</h1></header>
<article>
<h1>Welcome</h1>
<h1>Edit Account</h1>
<div id="login">
<ul id="login">
<form method="post" name="editAccount" action="userEditAccount.php" >
<fieldset>
<legend>Fill in the form</legend>
<label>Select Username : <input type="text" name="username" /></label>
<label>Password : <input type="password" name="password" /></label>
<label>Enter First Name : <input type="text" name="first_name" /></label>
<label>Enter Last Name : <input type="text" name="last_name" /></label>
<label>Enter E-mail Address: <input type="text" name="email" /></label>
</fieldset>
<br />
<input type="submit" value="Edit Account" class="button">
<input type="hidden" name="update" value="update">
</form>
</div>
<form action="userhome.php" method="post">
<div id="login">
<ul id="login">
<li>
<input type="submit" value="back" onclick="index.php" class="button">
</li>
</ul>
</div>
</article>
<aside>
</aside>
<div id="footer">Text</div>
</div>
</body>
</html>
很抱歉，我忘了复制这部分面孔
login.php：
<?php
session_start();
require('connect_mysql.php');
if($_SERVER['REQUEST_METHOD'] == 'POST')
{
$username = $_POST["username"];
$password = $_POST["password"];
$username = stripslashes($username);
$password = stripslashes($password);
$username = mysql_real_escape_string($username);
$password = mysql_real_escape_string($password);
$query = mysql_query("SELECT * FROM users WHERE Username='$username' AND Password='$password'");
$numrow = mysql_num_rows($query);
if($username && $password){
$query = mysql_query("SELECT * FROM users WHERE username='$username'");
$numrow = mysql_num_rows($query);
if($numrow !=0){
while($row = mysql_fetch_assoc($query)){
$dbusername = $row['username'];
$dbpassword = $row['password'];
}
if($username == $dbusername && $password == $dbpassword ){
$_SESSION['user_id'] = $user_id;
header("Location: userhome.php");
}
else{
echo "Incorect password";
}
}
else{
die("This user dosent exists");
}
}
else{
$reg = die("Please enter username and password");
}
}
?>
2 个解决方案
解决方案1
1 已采纳 2013-03-02 06:13:43
您没有在文件的开头调用session_start() ，因此$username将是一个空字符串，并且update命令将仅更新用户名是空字符串的行。
编辑：实际上，该代码甚至不会运行，因为您尚未调用session_start() ， isset($_SESSION['update'])评估结果为false。
您是要写$_SESSION['update']吗？ 那不是$_POST['update']吗？
最后但并非最不重要的一点，我个人将替换为：
<input name="update" type="submit" submit="submit" value="Edit Account" class="button">
有了这个：
<input type="submit" value="Edit Account" class="button">
<input type="hidden" name="update" value="update">
至少为了清楚起见。 我不知道是否仍然如此，但是随着时间的流逝，并不是所有的浏览器都提交了提交按钮的名称/值。
解决方案2
0 2013-03-02 08:25:45
先生，从上面给出的代码中，我认为您的login.php错误
$_SESSION['user_id'] = $user_id;
您没有为$ user_id分配值，这就是为什么要将空白值设置为$ _SESSION ['user_id']。
<?php
session_start();
require('connect_mysql.php');
if($_SERVER['REQUEST_METHOD'] == 'POST')
{
$username = $_POST["username"];
$password = $_POST["password"];
$username = stripslashes($username);
$password = stripslashes($password);
$username = mysql_real_escape_string($username);
$password = mysql_real_escape_string($password);
$query = mysql_query("SELECT * FROM users WHERE Username='$username' AND Password='$password'");
$numrow = mysql_num_rows($query);
if($username && $password){
$query = mysql_query("SELECT * FROM users WHERE username='$username'");
$numrow = mysql_num_rows($query);
if($numrow !=0){
$user_id = 0;
while($row = mysql_fetch_assoc($query)){
$dbusername = $row['username'];
$dbpassword = $row['password'];
$user_id = $row['user_id'];
}
if($username == $dbusername && $password == $dbpassword ){
$_SESSION['user_id'] = $user_id;
header("Location: userhome.php");
}
else{
echo "Incorect password";
}
}
else{
die("This user dosent exists");
}
}
else{
$reg = die("Please enter username and password");
}
}
?>
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