oracle - Oracle 复杂查询帮助要求

Question

嗨，我有一张表格，其中包含以下字段：godwon、产品名称、数量我想获得结果仓库和产品明智请检查图片

记住仓库不能修复它取决于用户可能 2 个仓库或 100 个仓库

Answer 1

那么解决方案取决于数据 - 您是否只有 6 个“货仓”值？如果是这样，您可以尝试以下解决方案：

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
;

检查 SQLFiddle：http ://sqlfiddle.com/#!4/5f8e8/1

Answer 2

您可以将 CASE 语句与聚合函数一起使用。

  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;

您也可以在 11g 中使用 PIVOT 运算符，但它不会为您提供最终的总计列。您必须单独计算。

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));