小编Mar*_*a T的帖子

我想让这个注册脚本告诉用户他们输入的密码不匹配.

我使用这段代码:

 if ($_POST['pass' != 'pass2'])
 {
  echo
   ("Oops! Password did not match! Try again. ");
 }

请帮我纠正我的编码.:-( 非常感谢!

我在数据库中创建表有一个简单的问题.这是我的代码:

$query="CREATE TABLE users (
    userid int(5) not null AUTO_INCREMENT, 
    firstname varchar(20),
    lastname varchar(20),
    username varchar(30),
    password varchar(32),
    email varchar(50),
    age int(2),
    PRIMARY KEY (userid)
)";

我希望从一USERID AUTO INCREMENT开始SPECIFIC NUMBER.

例如,从99001开始......

我怎样才能做到这一点？

伙计们,我有这些代码来更新数据库中的记录.

请..检查这些是否正确？

<?php
session_start(); 
include "db.php"; 

$username = $_SESSION['username'];


$query="SELECT * FROM members where username='".mysql_real_escape_string($username)."'";

$result=mysql_query($query);
$num=mysql_numrows($result);

mysql_close();


$userid             =mysql_result($result,$i, 'userid');
$firstname          =mysql_result($result,$i,'firstname');
$lastname           =mysql_result($result,$i,'lastname');
$username           =mysql_result($result,$i,'username');
$email              =mysql_result($result,$i,'email');
$age                =mysql_result($result,$i,'age');
?>

<form action="update.php" method="post">
<input type="hidden" name="u_userid" value="<? echo "$userid" ?>">
<table>
<tr><td>ID:</td> <td><? echo "$userid"?></td></tr>
<tr><td>First Name:</td> <td> <input type="text" name="u_firstname" value="<? echo "$firstname"?>"></td></tr>
<tr><td>Last Name: </td> <td><input type="text" name="u_lastname" value="<? echo "$lastname"?>"></td></tr>
<tr><td>Username:</td> <td> <input type="text" name="u_username" value="<? echo "$username"?>"></td></tr>
<tr><td>Email:</td> <td> <input type="text" name="u_email" value="<? echo …