我要打开我的网站时,我发现我的MODS的人给我这个错误:致命错误:无法使用类型mysqli_result的对象
Fatal error: Cannot use object of type mysqli_result as array in /var/www/vbsubscribetouser.php on line 303
我就去查看303行,这是我发现:
//Check if requested username can be followed.
if (in_array($followingdata['usergroupid'], explode("|", $vbulletin->options['subscribetouser_usergroups_cannot']))){
这里开始在行303所有代码:
//Check if requested username can be followed.
if (in_array($followingdata['usergroupid'], explode("|", $vbulletin->options['subscribetouser_usergroups_cannot']))){
exit;
}
if ($followinginfo[subscribers] > 0){
$user_followers = $followinginfo[followers].$userinfo[userid].'|';
}
else{
$user_followers = '|'.$userinfo[userid].'|';
}
$vbulletin->db->query_write("
UPDATE " . TABLE_PREFIX . "user
SET subscribers = subscribers + 1, `followers` = '$user_followers'
WHERE userid = $followinginfo[userid]
");
我不是在PHP代码方面的专家,所以双在打开网站之前,帮助会很好。任何帮助/建议?
非常感谢!
Cannot use object of type mysqli_result as array
使用mysqli_fetch_assoc或mysqli_fetch_array抓取结果一行作为关联阵列。
$query = "SELECT 1";
$result = $mysqli->query($query);
$followingdata = $result->fetch_assoc()
或
$followingdata = $result->fetch_array(MYSQLI_ASSOC);
这解决了我的问题,非常感谢你!我应该开始学习一些PHP技术!