由于字符串中的斜杠，PHP Mysql Laravel 雄辩错误，无法删除斜杠

Question

如何做到这一点：Users::whereNotIn('user_id', [$all_users])->update(['disabled' => 1]); 产生这个正确的查询： update usersset disabled= 1, users. updated_at= '2020-01-03 14:11:09' 其中user_id不在 ('642','532','539','588','488','601')我无法摆脱斜线。字符串是这样构建的： $all_users=$all_users.$user_id->id."','";如果我打印 $all_users 字符串，它会正确打印，见下文 echo $all_users; 产生：'642','532','539','588','488','601'用户::whereNotIn('user_id', [addslashes($all_users)])->update(['disabled' => 1]); 产生：更新users集disabled= 1, users. updated_at= '2020-01-03 13:53:02' 其中user_id不在('\'642\',\'532\',\'539\',\'588\',\'488\',\' 601\'')用户::whereNotIn('user_id', [$all_users])->update(['disabled' => 1]); 产生：更新users集disabled= 1, users. updated_at= '2020-01-03 14:11:09' 其中user_id不在('\'642\',\'532\',\'539\',\'588\',\'488\',\' 601\'')//我也试过这个 //Users::whereNotIn('user_id', [$all_users])->update(['disabled' => 1]); //Users::whereNotIn('user_id', [removeslashes($all_users)])->update(['disabled' => 1]); //Users::whereNotIn('user_id', [addslashes($all_users)])->update(['disabled' => 1]); //Users::whereNotIn('user_id', [str_replace("","\", $all_users)])->update(['disabled' => 1]);谢谢 ！

Answer 1

传递数组而不是字符串

$all_users[]=$user_id->id;

Users::whereNotIn('user_id', $all_users)->update(['disabled' => 1]);