PHP中的2个MySQL连接显示错误
我的网页有问题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, $web); 这是我的连接部分 if(isset($_
$server = mysql_connect($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($server)
和var\u dump($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);