我有,顯然,一個簡單的問題。作爲一個文檔,我可以檢查DataBase上是否存在TABLE或VIEWS。在我的應用程序,我需要檢查,如果兩個已經存在數據庫返回自定義錯誤。問題是我可以檢查是否有TABLE,但不是VIEWS。例如:PHP SQLITE PDO檢查是否存在視圖OR表
**Working case**
$checkTable = $this->db->query("SELECT * FROM sqlite_master WHERE name ='$table' and type='table'", PDO::FETCH_ASSOC);
**Not working case**
$checkTable = $this->db->query("SELECT * FROM sqlite_master WHERE name ='$table' and type='table' OR type='views'", PDO::FETCH_ASSOC);
**Not working case**
$checkTable = $this->db->query("SELECT * FROM sqlite_master WHERE name ='$table' and type='views'", PDO::FETCH_ASSOC);
奇怪的是,如果我運行的意見的查詢,這將返回結果:
**Working case**
$checkTable = $this->db->query("SELECT * FROM VIEWSpeople WHERE name ='$person'", PDO::FETCH_ASSOC);
通用查詢上SQLITE_MASTER找到的觀點:
**Working case**
$checkTable = $this->db->query("SELECT * FROM sqlite_master WHERE name ='$table'", PDO::FETCH_ASSOC);
問題:爲什麼查詢無法將視圖識別爲type ='views'? 感謝您的時間。
豈不是隻是'view'而不是'views'? – RamRaider
嗨RRAider,謝謝你,我也試過「查看」沒什麼好 – sundsx