This question already has an answer here:
I create images slider. It's must use user & password to run. Here is.
<?
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>
and error is : Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in line 21 [ $arry1 = mysql_fetch_array($result1);]
I known problem is in SQL !! Before this, i use [SELECT .... FROM member WHERE mid = '001';] and ,it's just show up. How can i fix that.
Thank for any answer.
</div>
There is a lot wrong here.
A few glaring issues are listed below:
Never store passwords in plain text. This is an assumption looking at the rest of your code.
$_SESSION; // a super global should always be capitalized.
Array keys:
$_SESSION['key'] = "value";
Escaping quotes
$foo['bar'] = "example";
echo "This is an " . $foo['bar'] . " of escaping quotes";
Problem in your SELECT query.. you didn't compare any values. like..
FROM member WHERE $_session[user_login] AND $_session[pass_login]
^^^ ^^^
Here you missing column name, also session variable must be in uppercase latters... try below query.
$sql1 = "SELECT gload1,gload2,gload3,gload4,gload5 FROM member WHERE mid = '".$_SESSION['user_login']."' AND password = '".$_SESSION['pass_login']."'";
Also at the top declaration of session variable is wrong, change it like
$_SESSION['user_login'] = "zyxel";
$_SESSION['pass_login'] = "12345";