$mysql = mysql_query("SELECT pass, user, id, folder, http, spacelimit, language, theme, permbrowse, permupload, permcreate, permuser, permadmin, permdelete, permmove, permchmod, permget, permdeleteuser, permedituser, permmakeuser, permpass, permrename, permedit, permsub, formatperm, status, recycle, permprefs FROM ".$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);
错误出现在第三行。它说
Warning: mysql_fetch_row() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\index.php on line 83
我尝试查找它，但找不到解决方法。
我知道有些方法已被弃用，但我只是想修复这个问题，这样我就可以让它工作(暂时)。任何帮助将不胜感激。
最佳答案
您正在运行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
关于PHP 我在这段代码中遇到错误，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/11908370/