我已创建2页PHP/MySQL的更新将无法工作
update.php
edit.php
我们开始edit.php所以这里是edit.php的脚本
<?php
$id = $_SESSION["id"];
$username = $_POST["username"];
$fname = $_POST["fname"];
$password = $_POST["password"];
$email = $_POST["email"];
mysql_connect('mysql13.000webhost.com', 'a2670376_Users', 'Password') or die(mysql_error());
echo "MySQL Connection Established! <br>";
mysql_select_db("a2670376_Pass") or die(mysql_error());
echo "Database Found! <br>";
$query = "UPDATE members SET username = '$username', fname = '$fname',
password = '$password' WHERE id = '$id'";
$res = mysql_query($query);
if ($res)
echo "<p>Record Updated<p>";
else
echo "Problem updating record. MySQL Error: " . mysql_error();
?>
<form action="update.php" method="post">
<input type="hidden" name="id" value="<?=$id;?>">
ScreenName:<br> <input type='text' name='username' id='username' maxlength='25' style='width:247px' name="username" value="<?=$username;?>"/><br>
FullName:<br> <input type='text' name='fname' id='fname' maxlength='20' style='width:248px' name="ud_img" value="<?=$fname;?>"/><br>
Email:<br> <input type='text' name='email' id='email' maxlength='50' style='width:250px' name="ud_img" value="<?=$email;?>"/><br>
Password:<br> <input type='text' name='password' id='password' maxlength='25' style='width:251px' value="<?=$password;?>"/><br>
<input type="Submit">
</form>
现在这里是update.php页在那里我遇到的主要问题
<?php
session_start();
mysql_connect('mysql13.000webhost.com', 'a2670376_Users', 'Password') or die(mysql_error());
mysql_select_db("a2670376_Pass") or die(mysql_error());
$id = (int)$_SESSION["id"];
$username = mysql_real_escape_string($_POST["username"]);
$fname = mysql_real_escape_string($_POST["fname"]);
$email = mysql_real_escape_string($_POST["email"]);
$password = mysql_real_escape_string($_POST["password"]);
$query="UPDATE members
SET username = '$username', fname = '$fname', email = '$email', password = '$password'
WHERE id='$id'";
mysql_query($query)or die(mysql_error());
if(mysql_affected_rows()>=1){
echo "<p>($id) Record Updated<p>";
}else{
echo "<p>($id) Not Updated<p>";
}
?>
现在edit.php我填写表格编辑帐户“测试”当我登录到它没有如果填写完表格,我点击| Submit!|按钮 它需要我update.php并返回该
(0) Not Updated
(0) <= id of user logged in
Not Updated <= MySql Error from
mysql_query($query)or die(mysql_error());
if(mysql_affected_rows()>=1){
我希望它更新用户登录,如果我不是在这个脚本错误的,它说
$id = (int)$_SESSION["id"];
巫更新与谁在
记录的人的ID的用户,但它不是更新它说,没有表已影响
如果它可以帮助继承人我们M ySql数据库图片 只需点击此处http://i50.tinypic.com/21juqfq.png
即使
session_start();
它不会工作之前
SQL注入攻击是可能的,我知道这病的修复,一旦我的Git这个工作 –
尝试呼应完整的SQL UPDATE语句,看看是什么它试图运行 –