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php - 2 PHP 中的 MySQL 连接显示错误

转载 作者:行者123 更新时间:2023-11-30 22:46:59 25 4
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<分区>

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>
';

这就是问题所在,它给我这样的错误。

Warning: mysql_query() expects parameter 2 to be resource, null given in /data03/virt47609/domeenid/www.example.com/htdocs/ucp/inc/mysql.php on line 32

Warning: mysql_fetch_array() expects parameter 1 to be resource, null given in /data03/virt47609/domeenid/www.example.com/htdocs/ucp/inc/mysql.php on line 32

Warning: mysql_query() expects parameter 2 to be resource, null given in /data03/virt47609/domeenid/www.example.com/htdocs/ucp/inc/mysql.php on line 33

Warning: mysql_num_rows() expects parameter 1 to be resource, null given in /data03/virt47609/domeenid/www.example.com/htdocs/ucp/inc/mysql.php on line 33

“var_dump($server)”给了我“类型为(mysql 链接)的资源(2)”,但就在 $osql 之前,它给了我“NULL”

“var_dump($web)”给我“类型为(mysql 链接)的资源(3)”,这里也是如此。

第32行和第33行是

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

我的 MySQL 连接也运行良好。

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。

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