mysql - 删除一个表中未被另一表引用的所有行

Question

我有两张桌子：

CREATE TABLE  "status" (
    "id" integer NOT NULL PRIMARY KEY,
    "created_at" datetime NOT NULL,    
    "updated_at" datetime NOT NULL);


CREATE TABLE "device" (
    "id" integer NOT NULL PRIMARY KEY,
    "created_at" datetime NOT NULL,
    "updated_at" datetime NOT NULL,
    "last_status_object_id" integer REFERENCES "status" ("id"));

在“设备”表中，last_status_object_id 引用 status.id。

我想删除“last_status_object_id”未引用的所有状态行。我似乎无法弄清楚如何做到这一点。谁能帮我吗？

Answer 1

DELETE
    status
FROM
    status
    LEFT JOIN device ON (status.id = last_status_object_id)
WHERE
    device.id IS NULL

Answer 2

delete from status
where id not in (select last_status_object_id
                            from device);

我们开始吧，在sqlfiddle中有演示

Answer 3

我想删除“last_status_object_id”未引用的所有状态行。

一种简单的表达方式是使用NOT EXISTS：

DELETE FROM 
    status
WHERE
    NOT EXISTS (SELECT * FROM device WHERE last_status_object_id = status.id)