SQL通過從同一個表中選擇來插入多個值和父ID

Question

如何通過從同一個表中選擇來插入多個值和父 ID

我在下面嘗試並收到錯誤

INSERT IGNORE INTO ctg_tbl (`ctg_name`,`ctg_img`,`ctg_parent_id`)
VALUES 
('Gravy', 'GravyImg', (SELECT id FROM ctg_tbl WHERE ctg_name='Foods')),
('Curry', 'CurryImg', (SELECT id FROM ctg_tbl WHERE ctg_name='Foods')),
('Ball', 'BallImg', (SELECT id FROM ctg_tbl WHERE ctg_name='Games'))

MySQL 說：文檔 #1093 - 表 'ctg_tbl' 被指定兩次，既作為 'INSERT' 的目標，也作為單獨的數據源

Answer 1

您的查詢可以通過簡單地將其轉換為INSERT INTO ... SELECT來工作：

INSERT IGNORE INTO ctg_tbl (ctg_name, ctg_img, ctg_parent_id)
SELECT 'Gravy', 'GravyImg', id FROM ctg_tbl WHERE ctg_name = 'Foods'
UNION ALL
SELECT 'Curry', 'CurryImg', id FROM ctg_tbl WHERE ctg_name = 'Foods'
UNION ALL
SELECT 'Ball', 'BallImg', id FROM ctg_tbl WHERE ctg_name = 'Games';

Answer 2

試試這個

    DECLARE @FoodID INT = (SELECT TOP (1) id FROM ctg_tbl WHERE ctg_name = 'Foods')
    DECLARE @GameID INT = (SELECT TOP (1) id FROM ctg_tbl WHERE ctg_name = 'Games')

    INSERT IGNORE INTO ctg_tbl (`ctg_name`,`ctg_img`,`ctg_parent_id`)
    VALUES 
    ('Gravy', 'GravyImg', @FoodID),
    ('Curry', 'CurryImg', @FoodID),
    ('Ball', 'BallImg', @GameID)

Answer 3

通過使用ROW 構造函數創建一個包含每行 3 個值的 CTE，您可以在INSERT語句中將其連接到表：

INSERT IGNORE INTO ctg_tbl (ctg_name , ctg_img , ctg_parent_id ) 
WITH cte(ctg_name , ctg_img , ctg_parent_name) AS (
  VALUES 
  ROW('Gravy', 'GravyImg', 'Foods'),
  ROW('Curry', 'CurryImg', 'Foods'),
  ROW('Ball', 'BallImg', 'Games')
)
SELECT c.ctg_name, c.ctg_img, t.id
FROM cte c INNER JOIN ctg_tbl t
ON t.ctg_name = ctg_parent_name;

請參閱演示。