$query="select password from master where name=$name ";
这一句中没有指明要查询出name,但是后面,也就是第七行$arr['name']<>$name却在使用name,所以出错。修改为$query="select name,password from master where name=$name ";
这一句中没有指明要查询出name,但是后面,也就是第七行$arr['name']<>$name却在使用name,所以出错。修改为$query="select name,password from master where name=$name ";
这样试试
在你的数据库函数前加上@,如@mysql_query()
改为这样~~~
<?php
$query="select password from master where name=$name ";
mysql_connect('218.6.35.162','root','');
mysql_select_db(test);
$res=mysql_query($query);//注意这里!!!
$arr = mysql_fetch_array($res);{
if ($arr['name']<>$name) {
echo "对不起,请输入正确的用户名";
}
else if ($arr['password']<>$password){
echo "对不起,请输入正确的密码";
}
else {
include "./depot2.php";
}
}
?>
$query="select name,password from master where name='$name'";
@mysql_select_db("test");
isset($_POST['name'])?$name=$_POST['name']:$name=NULL;
isset($_POST['password'])?$password=$_POST['password']:$password=NULL;
$sql = "select * from master where name='$name' && password='$password'";
$res = @mysql_query($query,$ptconn);
$num = @mysql_num_rows($res)
$num>0?include "./depot2.php":echo "对不起,没有匹配的用户名和密码!";
$query="select password,name from master where name=$name ";
$ptconn=@mysql_connect('218.6.35.162','root','');
@mysql_select_db(test);
$res = @mysql_query($query,$ptconn);
$arr = @mysql_fetch_array($res);
if ($name!=NULL&&$arr['name']<>$name)
{
echo "对不起,请输入正确的用户名";
}
elseif ($password!=NULL&&$arr['password']<>$password)
{
echo "对不起,请输入正确的密码";
}
else
{
include "./depot2.php";
}
//试试先