当我使用以下代码时
foreach($admin_info as $key => $val)
{
echo $key." ".$val."\n";
}
我会得到真实的结果:
id 1 username Moein7tl password 0937afa17f4dc08f3c0e5dc908158370ce64df86 mail moein7tl@gmail.com added_time super_admin last_time last_ip see_user_per 1 change_user_per 1 see_people_per 1 change_people_per 1 add_people_per 1 remove_people_per 1 see_album_per 1 add_album_per 1 change_album_per 1 remove_album_per 1 see_music_per 1 add_music_per 1 change_music_per 1 remove_music_per 1 admin_per 1 yahoo_per 1 status_per 1 pm_per 1 ip_blocking_per 1
但是当我使用echo ($admin_info['id']);
或其他键时,它会出错并且不会显示任何结果。
哪里有问题?
- 编辑
var_dump($admin_info);
将返回
object(stdClass)#21 (27) { ["id"]=> string(1) "1" ["username"]=> string(8) "Moein7tl" ["password"]=> string(40) "0937afa17f4dc08f3c0e5dc908158370ce64df86" ["mail"]=> string(18) "moein7tl@gmail.com" ["added_time"]=> NULL ["super_admin"]=> string(0) "" ["last_time"]=> NULL ["last_ip"]=> NULL ["see_user_per"]=> string(1) "1" ["change_user_per"]=> string(1) "1" ["see_people_per"]=> string(1) "1" ["change_people_per"]=> string(1) "1" ["add_people_per"]=> string(1) "1" ["remove_people_per"]=> string(1) "1" ["see_album_per"]=> string(1) "1" ["add_album_per"]=> string(1) "1" ["change_album_per"]=> string(1) "1" ["remove_album_per"]=> string(1) "1" ["see_music_per"]=> string(1) "1" ["add_music_per"]=> string(1) "1" ["change_music_per"]=> string(1) "1" ["remove_music_per"]=> string(1) "1" ["admin_per"]=> string(1) "1" ["yahoo_per"]=> string(1) "1" ["status_per"]=> string(1) "1" ["pm_per"]=> string(1) "1" ["ip_blocking_per"]=> string(1) "1" }
答案 0 :(得分:0)
我发现问题,CodeIgniter会将数据库结果作为Object返回。
所以我应该像$admin_info->id
那样做,它解决了问题。
答案 1 :(得分:0)
尝试$admin_info->password
,这应该有效,因为$admin_info
似乎是stdClass
的对象。
答案 2 :(得分:-3)
您的逻辑是正确的,但您可以尝试使用echo $ admin_info ['id'];没有括号。