谁能帮我弄清楚这段代码有什么问题?

Can anyone help me figure out what is wrong with this code?(谁能帮我弄清楚这段代码有什么问题?)

问题描述

<?php
$con = mysql_connect("localhost", "root", '');

if (!$con)
{
    die('Cannot make a connection');
}


mysql_select_db('yumbox_table', $con) or die('Cannot make a connection');



$username = $_POST['user_name'];
$password = $_POST['password'];
$type = $_POST['user_type'];

$data = mysql_query("SELECT * from users ") or die(mysql_error());

$info = mysql_fetch_array($data);

$count = mysql_numrows($info);

if ($count==1)
{
    echo ("Success!!");
}
else 
{
    echo ("BIG FRIGGIN FAILURE!!");
}

mysql_close($con);
?>

每当我尝试运行此代码时，都会收到这些漂亮的大错误消息:

Whenever i try to run this code, I get these big beautiful error messages:

( ! ) Notice: Undefined index: user_name in C:wampwwwlogin.php on line 14
Call Stack
#   Time    Memory  Function    Location
1   0.0008  370104  {main}( )   ..login.php:0

( ! ) Notice: Undefined index: password in C:wampwwwlogin.php on line 15
Call Stack
#   Time    Memory  Function    Location
1   0.0008  370104  {main}( )   ..login.php:0

( ! ) Notice: Undefined index: user_type in C:wampwwwlogin.php on line 16
Call Stack
#   Time    Memory  Function    Location
1   0.0008  370104  {main}( )   ..login.php:0

( ! ) Warning: mysql_numrows() expects parameter 1 to be resource, array given in C:wampwwwlogin.php on line 22
Call Stack
#   Time    Memory  Function    Location
1   0.0008  370104  {main}( )   ..login.php:0
2   0.0157  380104  mysql_numrows ( )   ..login.php:22

我一直在努力弄清楚这些背后的含义并修复它们，遗憾的是没有看到任何解决方案.有人可以帮忙吗?

I have been pulling my hair out trying to figure out the meaning behind these and fix them, and sadly have seen no solutions. Can anyone out there help?

推荐答案

要么 POST 不包含相关变量，要么您一开始就没有执行 POST.

Either the POST doesn't contain the variables in question, or you aren't performing a POST in the first place.

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