sql - 计数所有状态和特定状态与分区的数量

Question

准备sql查询问题。

CREATE TABLE right_data(RID NUMBER(9,0), STATUS NUMBER(2,0));
CREATE TABLE left_data(LID NUMBER(9,0), URL VARCHAR(70 BYTE));
CREATE TABLE left_to_right(LID NUMBER(9,0), RID NUMBER(9,0));

INSERT INTO right_data (RID, STATUS) VALUES (1,0);
INSERT INTO right_data (RID, STATUS) VALUES (2,1);
INSERT INTO right_data (RID, STATUS) VALUES (3,0);
INSERT INTO right_data (RID, STATUS) VALUES (4,0);
INSERT INTO right_data (RID, STATUS) VALUES (5,1);

INSERT INTO left_data (LID, URL) VALUES (1,'data_1');
INSERT INTO left_data (LID, URL) VALUES (2,'data_2');
INSERT INTO left_data (LID, URL) VALUES (3,'data_3');
INSERT INTO left_data (LID, URL) VALUES (4,'data_4');
INSERT INTO left_data (LID, URL) VALUES (5,'data_5');

INSERT INTO left_to_right (LID, RID) VALUES (1,1);
INSERT INTO left_to_right (LID, RID) VALUES (1,2);
INSERT INTO left_to_right (LID, RID) VALUES (2,1);
INSERT INTO left_to_right (LID, RID) VALUES (1,4);
INSERT INTO left_to_right (LID, RID) VALUES (4,3);
INSERT INTO left_to_right (LID, RID) VALUES (5,1);
INSERT INTO left_to_right (LID, RID) VALUES (3,5);
INSERT INTO left_to_right (LID, RID) VALUES (4,2);

一个错误的查询。

SELECT
  DISTINCT left_to_right.LID
  , COUNT(left_to_right.RID) OVER (PARTITION BY left_to_right.LID) AS NUM_RID
  , COUNT(left_to_right.RID) OVER (PARTITION BY
                                      left_to_right.LID
                                  AND
                                      right_data.STATUS = 1
                                  ) AS NUM_RID_S1
FROM
    left_to_right
  JOIN
    right_data
  ON
    right_data.RID = left_to_right.RID
ORDER BY left_to_right.LID;

我如何在同一个查询中计算 LID/(所有 right_data.status 的数量)和 LID/(right_data.status 的数量为 1)？

Answer 1

如果我理解正确的话，应该这样做:

SELECT left_to_right.LID,
       COUNT(DISTINCT left_to_right.RID) all_right_data_statuses,
       COUNT(DISTINCT CASE WHEN right_data.STATUS = 1 THEN left_to_right.RID 
                      ELSE NULL END) right_data_status_1
FROM
    left_to_right
  JOIN
    right_data
  ON
    right_data.RID = left_to_right.RID
GROUP BY left_to_right.LID

Here is a demo供你尝试。