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PHP中的2个MySQL连接显示错误_Php_Html_Mysql - Fatal编程技术网

PHP中的2个MySQL连接显示错误

PHP中的2个MySQL连接显示错误,php,html,mysql,Php,Html,Mysql,我的网页有问题 $server = mysql_connect($server_host, $server_user, $server_pass, true); mysql_select_db($server_db, $server); $web = mysql_connect($foorum_host, $foorum_user, $foorum_pass); mysql_select_db($foorum_db, $web); 这是我的连接部分 if(isset($_

我的网页有问题

$server = mysql_connect($server_host, $server_user, $server_pass, true); 
mysql_select_db($server_db, $server);

$web = mysql_connect($foorum_host, $foorum_user, $foorum_pass); 
mysql_select_db($foorum_db, $web);
这是我的连接部分

        if(isset($_POST['logisisse'])) {

        $user = mysql_real_escape_string($_POST['user']);
        $pass = mysql_real_escape_string($_POST['pass']);

        $osqlq = "SELECT * FROM mybb_users WHERE username='$user'";
        $asqlq = "SELECT username FROM mybb_users WHERE username='$user'";

        $osql = mysql_fetch_array(mysql_query($osqlq, $web));
        $asql = mysql_num_rows(mysql_query($asqlq, $web));

        $salt = $osql['salt'];

        if(empty($user) || empty($pass)) {
            echo '<div class="alert alert-danger">Täida kõik väljad.</div>';
        }
        else if($asql < 1) { // pole kasutajat
            echo'<div class="alert alert-danger">Sellist kasutajat ei eksisteeri.</div>';
        }
        else if($osql['pass'] != $hashedPassword) { // vale parool
            echo '<div class="alert alert-danger">Sisestasid vale parooli.</div>';
        }
    }
    echo '
                <aside class="widget widget_search">
                  <h3 class="widget-title">Logi sisse</h3>
                  <form role="login" method="post" class="validate-form" />
                    <div>
                      <label class="screen-reader-text" for="s">Logi sisse:</label>
                      <input type="text" value="" id="user" name="user" placeholder="Foorumi kasutajanimi" class="required" />
                      <input type="password" value="" id="pass" name="pass" placeholder="Parool" class="required" />
                      <br /><br />
                      <button id="logisisse" name="logisisse" value="Logi sisse" />
                    </div>
                  </form>
                </aside>
    ';
此外,我的MySQL连接工作正常


我一直在尝试从谷歌搜索问题,但没有帮助,也许你们可以帮助我。

添加如下错误处理:

<?php
$link = mysql_connect($server_host, $server_user, $server_pass, true); 
if (!$link) {
    die('Could not connect: ' . mysql_error());
}

?>

我发现了问题。我的网页登录脚本是一个函数,我将该函数复制到了登录脚本应该位于的位置,它的工作方式很有魅力。

请指出哪行是32行和33行。您确定已正确连接到这两个数据库吗?
var\u dump($server)
var\u dump($web)
输出是什么?我的MySQL连接工作正常,不会出现任何错误。好的,但是做错误处理总是很好的做法……在查询之前尝试var\u dump($web)和var\u dump($server)。。。
<?php
$link = mysql_connect($server_host, $server_user, $server_pass, true); 
if (!$link) {
    die('Could not connect: ' . mysql_error());
}

?>
$server = mysql_connect($server_host, $server_user, $server_pass); 
mysql_select_db($server_db, $server);

$web = mysql_connect($foorum_host, $foorum_user, $foorum_pass, true); 
mysql_select_db($foorum_db, $web);