<?php
$name = $_require['username'];
$pwd = $_require['password'];if($name ==""||$pwd=="")
{
echo "请输入用户名和密码";
}include("config.php")
$sql = "SELECT * FROM user WHERE uname = $name";$rs = mysql_query($sql);
$num = mysql_num_rows($rs);
$row = mysql_fetch_array($rs);
$user =$row[0];
$infor =$row[1];
if($user = $name)
{
echo ("登陆成功");
}
else
{
echo ("登陆失败");
}include("login.html")?>报错 错误就是 $sql 语句那
Parse error: syntax error, unexpected T_VARIABLE in F:\PHP\xampp\htdocs\logic.php on line 18我初学者 大家帮帮忙~
$name = $_require['username'];
$pwd = $_require['password'];if($name ==""||$pwd=="")
{
echo "请输入用户名和密码";
}include("config.php")
$sql = "SELECT * FROM user WHERE uname = $name";$rs = mysql_query($sql);
$num = mysql_num_rows($rs);
$row = mysql_fetch_array($rs);
$user =$row[0];
$infor =$row[1];
if($user = $name)
{
echo ("登陆成功");
}
else
{
echo ("登陆失败");
}include("login.html")?>报错 错误就是 $sql 语句那
Parse error: syntax error, unexpected T_VARIABLE in F:\PHP\xampp\htdocs\logic.php on line 18我初学者 大家帮帮忙~
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<?php
$name = $_require['username'];
$pwd = $_require['password'];if($name ==""||$pwd=="")
{
echo "请输入用户名和密码";
}include("config.php");$sql = "SELECT * FROM user WHERE uname = $name";$rs = mysql_query($sql);$row = mysql_fetch_array($rs, MYSQL_ASSOC)
$user =$row['user'];
$infor =$row['infor'];
if($user = $name)
{
echo ("登陆成功");
}
else
{
echo ("登陆失败");
}include("login.html");?>报错: 24行是 $user =$row['user'];
$infor =$row['infor'];
Parse error: syntax error, unexpected T_VARIABLE in F:\PHP\xampp\htdocs\logic.php on line 24
$sql = "SELECT * FROM user WHERE uname = '".$name."'";
$pwd = $_require['password']; 你页面传值的方式是什么?
method ="??"如果是GET的话 把
$name = $_GET['username'];
$pwd = $_GET['password']; 如果是POST的话 把
$name = $_POST['username'];
$pwd = $_POST['password'];
$name = $_REQUEST['username'];
$pwd = $_REQUEST['password']; if($name =="" ¦ ¦$pwd=="")
{
echo "请输入用户名和密码";
} include("config.php")
$sql = "SELECT * FROM user WHERE uname = '".$name."'"; $rs = mysql_query($sql);
$num = mysql_num_rows($rs);
$row = mysql_fetch_array($rs);
$user =$row[0];
$infor =$row[1];
if($user == $name)
{
echo ("登陆成功");
}
else
{
echo ("登陆失败");
} include("login.html") ?>
$name = $_request['username'];
$pwd = $_request['password'];
$_request是通杀的哦!
$name = $_request['username'];
$pwd = $_request['password'];if($name ==""||$pwd=="")
{
echo "请输入用户名和密码";
}include("config.php");
$sql = "SELECT * FROM user WHERE uname = '".$name."'";
$rs = mysql_query($sql);
$row = mysql_fetch_array($rs);
$user =$row[0];
$infor =$row[1];
if($user = $name)
{
echo ("登陆成功");
}
else
{
echo ("登陆失败");
}
include("login.html");?>
这里报错
大家帮下 我初学PHP
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in F:\PHP\xampp\htdocs\logic.php on line 15
你打印 echo $sql.肯定是sql有问题了。
$sql = "SELECT * FROM user WHERE 'uname' = '".$name."'";
第一次也和你一样的错误!
$rs = mysql_query($sql);
$num = mysql_num_rows($rs);
$row = mysql_fetch_array($rs); 红色的地方对么?
$name = $_REQUEST['username'];
$pwd = $_REQUEST['password'];
include("config.php");
$sql = "SELECT * FROM userinfor WHERE 'user' = '".$name."'";
$rs = mysql_query($sql);
$row = mysql_fetch_array($rs);
$user =$row[0];
$infor =$row[1];
if($user == $name||$infor == $pwd)
{
echo ("登陆成功");
}
else
{
echo ("登陆失败");
}
include("login.html");?>
这个是我 最后的代码了,显示方面没有问题,运行也没报错
但是 输入 数据库正确的 数据 他说登陆失败. 汗
而且页面运行,还没点按钮 他就显示 登陆成功.
大家多帮忙 我初学
//……$rs = mysql_query($sql);
$num = mysql_num_rows($rs);
$row = mysql_fetch_array($rs);
print_r($row);//你把这里的$row打印出来看看是否有值//……if($user = $name)//这是是赋值,所以永远是登录成功,你应该修改为==//……
看看你的Sql语句有什么问题
不是mysql_fetch_array有什么错
$name = $_REQUEST['username'];
$pwd = $_REQUEST['password'];
include("config.php");
$sql = "SELECT * FROM userinfor";
$rs = mysql_query($sql);
$row = mysql_fetch_array($rs);
$user =$row[0];
$infor =$row[1];
if($user == $name and $infor == $pwd)
{
echo ("登陆成功");
}
else
{
echo ("登陆失败");
}
include("login.html");
?>
现在我一调试 页面就显示 登陆失败.
我数据库了有 帐号,密码: 111,222. 333,444. 555,666.
结果是 我输入 111 222 就显示登陆成功
但是输入 333,444 就说失败寻求答案
if(!empty($row)){
$user =$row['user'];
$infor =$row['infor'];
if($user = $name)
{
echo ("登陆成功");
}
else
{
echo ("登陆失败");
}
}
while(($row=mysql_fetch_array($result)))
{
if($name == $row['user'])
{ echo "Success!";
$Notfind = false ;
} }
if($Notfind)
{
echo "No Success !!!!";
}
$name = $_REQUEST['username'];
$pwd = $_REQUEST['password']; if($name =="" ||$pwd=="")
{
echo "请输入用户名和密码";
exit;
} include("config.php"); $sql = "SELECT * FROM user WHERE uname='".$name."'";$rs = mysql_query($sql);
$num = mysql_num_rows($rs);
if($num > 0)
{
$row = mysql_fetch_array($rs);
$user =$row[0];
$infor =$row[1];
echo ("登录成功");
}
else
{
echo ("登陆失败");
}include("login.html"); ?>
$pwd = trim($_REQUEST['password']);
是不是在加入数据时有空格的问题?
后面加上结束符;试试
<?
session_start();
$admin_name=$_POST['admin_name'];
$admin_password=$_POST['admin_password'];
$conn=mysql_connect ("localhost", "root", "root"); mysql_select_db("guest_book");
$exec="select * from admin where admin_name = '".$admin_name."' ";
$result=mysql_query($exec);
if ($rs = mysql_fetch_object($result))
{
if ($rs->admin_password == $admin_password)
{
$_SESSION['admin']="OK";
header("location:admin_index.php");
}
else echo"密码不正确";
}
else echo"用户名不正确";
mysql_close();
?>
因为你已经限定了$user =$row[0];
$infor =$row[1];
所以只能是第一个数据有效。