Php 如何通过$\u会话[]使用用户和密码进行搜索？
我创建图像滑块。必须使用用户和密码才能运行。这是Php 如何通过$\u会话[]使用用户和密码进行搜索？,php,mysql,session,Php,Mysql,Session,我创建图像滑块。必须使用用户和密码才能运行。这是 <? include 'connect.php'; $_session[user_login] = "zyxel"; $_session[pass_login] = "12345"; ?> <html> <head> <link rel="stylesheet" href="themes/default/de
<?
include 'connect.php';
$_session[user_login] = "zyxel";
$_session[pass_login] = "12345";
?>
<html>
<head>
<link rel="stylesheet" href="themes/default/default.css" type="text/css" media="screen" />
<link rel="stylesheet" href="themes/light/light.css" type="text/css" media="screen" />
<link rel="stylesheet" href="themes/dark/dark.css" type="text/css" media="screen" />
<link rel="stylesheet" href="lib/nivo-slider.css" type="text/css" media="screen" />
<link rel="stylesheet" href="lib/style.css" type="text/css" media="screen" />
</head>
<body>
<div class="slider-wrapper theme-default">
<div id="slider" class="nivoSlider">
<?php
$sql1 = "SELECT gload1,gload2,gload3,gload4,gload5 FROM member WHERE $_session[user_login] AND $_session[pass_login]";
$result1 = mysql_query($sql1);
$arry1 = mysql_fetch_array($result1);
$a = 0;
$b = 1;
$c = 2;
$d = 3;
$e = 4;
$f = 5;
$g = 6;
$h = 7;
$i = 8;
$j = 9;
if(empty($arry1)){
exit();
}
else {
$index = array_search(max($arry1),$arry1);
if($index == $a){
echo "<img src=images/01/01.jpg>";
echo "<img src=images/01/02.jpg>";
echo "<img src=images/01/03.jpg>";
echo "<img src=images/01/04.jpg>";
}else if($index == $b){
echo "<img src=images/02/01.jpg>";
echo "<img src=images/02/02.jpg>";
echo "<img src=images/03/03.jpg>";
echo "<img src=images/04/04.jpg>";
}else if($index == $c){
echo "<img src=images/03/01.jpg>";
echo "<img src=images/03/02.jpg>";
echo "<img src=images/03/03.jpg>";
echo "<img src=images/03/04.jpg>";
}else if($index == $d){
echo "<img src=images/04/01.jpg>";
echo "<img src=images/04/02.jpg>";
echo "<img src=images/04/03.jpg>";
echo "<img src=images/04/04.jpg>";
}else {exit();
}
?>
</body>
</html>
感谢您的回答。选择查询中的问题。。您没有比较任何值。像
FROM member WHERE $_session[user_login] AND $_session[pass_login]
^^^ ^^^
$sql1 = "SELECT gload1,gload2,gload3,gload4,gload5 FROM member WHERE mid = '".$_SESSION['user_login']."' AND password = '".$_SESSION['pass_login']."'";
另外，会话变量的顶部声明错误，请按如下所示进行更改
$_SESSION['user_login'] = "zyxel";
$_SESSION['pass_login'] = "12345";
这里有很多错误 在继续之前，我会先看一下报告。 以下列出了几个突出问题： 从不以明文形式存储密码。这是一个关于代码其余部分的假设
$_SESSION; // a super global should always be capitalized.
$_SESSION['key'] = "value";
$foo['bar'] = "example";
echo "This is an " . $foo['bar'] . " of escaping quotes";
第33行[退出（）；]。我想插入其他条件。请用适当的缩进插入您的问题。为什么要在会话中输入密码？登录后你没有身份证吗？你的代码有很多问题。首先，
$\u session用户登录'user\u login'错误报告=E\u all