doutao5499 2013-09-27 17:53
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如何通过$ _session []来搜索用户和密码? [重复]

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>
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2条回答 默认 最新

  • douyi1899 2013-09-27 18:23
    关注

    There is a lot wrong here.

    I would read the PHP Documentation before proceeding.

    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";
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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