如何做到這一點(diǎn)：Users::whereNotIn('user_id', [$all_users])->update(['disabled' => 1]); 產(chǎn)生這個(gè)正確的查詢： 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')我無法擺脫斜線。字符串是這樣構(gòu)建的： $all_users=$all_users.$user_id->id."','";如果我打印 $all_users 字符串，它會(huì)正確打印，見下文 echo $all_users; 產(chǎn)生：'642','532','539','588','488','601'用戶::whereNotIn('user_id', [addslashes($all_users)])->update(['disabled' => 1]); 產(chǎn)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\'')用戶::whereNotIn('user_id', [$all_users])->update(['disabled' => 1]); 產(chǎn)生：更新users集disabled= 1, users. updated_at= '2020-01-03 14:11:09' 其中user_id不在('\'642\',\'532\',\'539\',\'588\',\'488\',\' 601\'')//我也試過這個(gè) //Users::whereNotIn('user_id', [$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]);謝謝 ！