甲骨文複雜的查詢求助requireds

Question

您好，我有一個表下面的字段godwon，產品名稱，數量我想導致 倉庫和產品明智 請檢查圖像

記住倉庫都沒有解決其依賴於用戶也許2個倉庫或100個倉庫

Answer 1

那麼解決方案取決於數據 - 你只有6個「倉庫」值？在SQLFiddle

CREATE TABLE my_table (
    godown VARCHAR2(20), 
    product_name VARCHAR2(20), 
    quantity NUMBER 
); 

INSERT INTO my_table VALUES ('Godown1', 'Rice', 5); 
INSERT INTO my_table VALUES ('Godown1', 'Tea', 6); 
INSERT INTO my_table VALUES ('Godown1', 'Milk', 4); 
INSERT INTO my_table VALUES ('Godown2', 'Rice', 6); 
INSERT INTO my_table VALUES ('Godown2', 'Tea', 7); 
INSERT INTO my_table VALUES ('Godown2', 'Milk', 5); 
INSERT INTO my_table VALUES ('Godown3', 'Rice', 8); 
INSERT INTO my_table VALUES ('Godown3', 'Tea', 3); 
INSERT INTO my_table VALUES ('Godown3', 'Milk', 5); 
INSERT INTO my_table VALUES ('Godown4', 'Rice', 3); 
INSERT INTO my_table VALUES ('Godown4', 'Tea', 2); 
INSERT INTO my_table VALUES ('Godown4', 'Milk', 1); 
INSERT INTO my_table VALUES ('Godown5', 'Rice', 4); 
INSERT INTO my_table VALUES ('Godown5', 'Tea', 6); 
INSERT INTO my_table VALUES ('Godown5', 'Milk', 7); 
INSERT INTO my_table VALUES ('Godown6', 'Rice', 9); 
INSERT INTO my_table VALUES ('Godown6', 'Tea', 24); 
INSERT INTO my_table VALUES ('Godown6', 'Milk', 12); 

COMMIT; 

SELECT 
    product_name, 
    godown1_sum_quantity AS Godown1, 
    godown2_sum_quantity AS Godown2, 
    godown3_sum_quantity AS Godown3, 
    godown4_sum_quantity AS Godown4, 
    godown5_sum_quantity AS Godown5, 
    godown6_sum_quantity AS Godown6, 
    (SELECT SUM(quantity) FROM my_table WHERE product_name = data.product_name) AS Total 
    FROM (
    SELECT 
     product_name, 
     godown, 
     quantity 
     FROM my_table 
) 
    PIVOT (SUM(quantity) AS sum_quantity 
    FOR (godown) IN (
    'Godown1' AS Godown1, 
    'Godown2' AS Godown2, 
    'Godown3' AS Godown3, 
    'Godown4' AS Godown4, 
    'Godown5' AS Godown5, 
    'Godown6' AS Godown6 
    ) 
) data 
; 

檢查：如果是這樣，你可以試試下面的解決方案http://sqlfiddle.com/#!4/5f8e8/1

Answer 2

可以使用CASE語句和聚合函數。

SELECT product_name, 
     MIN (CASE WHEN godown = 'godown1' THEN quantity END) godown1, 
     MIN (CASE WHEN godown = 'godown2' THEN quantity END) godown2, 
     MIN (CASE WHEN godown = 'godown3' THEN quantity END) godown3, 
     MIN (CASE WHEN godown = 'godown4' THEN quantity END) godown4, 
     MIN (CASE WHEN godown = 'godown5' THEN quantity END) godown5, 
     MIN (CASE WHEN godown = 'godown6' THEN quantity END) godown6, 
     SUM (quantity) 
    FROM products 
GROUP BY product_name; 

你也可以在11g中使用PIVOT運算符，但它不會給你最後的總列。你必須分開計算它。

SELECT * 
    FROM (SELECT godown, product_name, quantity FROM products) 
     PIVOT (MIN (
        quantity) 
     FOR (godown) 
     IN ('godown1' AS godown1, 
      'godown2' AS godown2, 
      'godown3' AS godown3, 
      'godown4' AS godown4, 
      'godown5' AS godown5, 
      'godown6' AS godown6));