PHP --我收到了这段代码的错误

Question

$mysql = mysql_query("SELECT pass, user, id, folder, http, spacelimit, language, theme, permbrowse, permupload, permcreate, permuser, permadmin, permdelete, permmove, permchmod, permget, permdeleteuser, permedituser, permmakeuser, permpass, permrename, permedit, permsub, formatperm, status, recycle, permprefs FROM ".$GLOBALS['config']['db']['pref']."users WHERE user='".mysql_real_escape_string($user)."'");
    list ($dbpass, $dbuser, $userid, $userdir, $http, $limit, $language, $theme, $permbrowse, $permupload, $permcreate, $permuser, $permadmin, $permdelete, $permmove, $permchmod, $permget, $permdeleteuser, $permedituser, $permmakeuser, $permpass, $permrename, $permedit, $permsub, $formatperm, $status, $recycle, $permprefs) = mysql_fetch_row($mysql);

错误在第三行。上面写着

警告: mysql_fetch_row()期望参数1是资源，在第83行的C:\xampp\htdocs\index.php中给出布尔值

我试着查了一下，我很难找到解决办法。

我知道有些方法已经被废弃了，但我只是试图修复它，这样我就可以让它工作了(目前来说)。任何帮助都将不胜感激。

Answer 1

你在运行mysql_query(mysql_query(...))。

$mysql = "SELECT pass, user, id, folder, http, spacelimit, language, theme, permbrowse, permupload, permcreate, permuser, permadmin, permdelete, permmove, permchmod, permget, permdeleteuser, permedituser, permmakeuser, permpass, permrename, permedit, permsub, formatperm, status, recycle, permprefs FROM ".$GLOBALS['config']['db']['pref']."users WHERE user='".mysql_real_escape_string($user)."'"; 

echo $mysql; // Get the output here and run it directly to see if it succeeds

$result = mysql_query($mysql) or die(mysql_error()); // add `or die(mysql_error())` to output an error if the query fails

Answer 2

读查询！如果查询因任何原因失败，它将返回FALSE。您可以使用错误来找出原因。