Question

您好我的任务中遇到了一个SQL查询，并希望得到一些帮助。

我需要获取最佳执行项目的项目ID - 项目（VERY_GOOD记录计数+ GOOD记录计数） - （VERY_BAD记录计数+ BAD记录计数）最大

我在数据库中的架构和测试记录（HSQLDB）

CREATE TABLE
    PROJECT
    (
        ID IDENTITY NOT NULL PRIMARY KEY,
        PROJECT_NAME VARCHAR(255) NOT NULL
    );

CREATE TABLE
    RECORD
    (
        ID IDENTITY NOT NULL PRIMARY KEY,
        RESULT VARCHAR(255) NOT NULL,
    );

CREATE TABLE
    RECORD_PROJECT
    (
        PROJECT_ID INTEGER NOT NULL,
        RECORD_ID INTEGER NOT NULL,
        PRIMARY KEY(PROJECT_ID, RECORD_ID),
        FOREIGN KEY (PROJECT_ID) REFERENCES PROJECT(ID) ON DELETE CASCADE,
        FOREIGN KEY (RECORD_ID) REFERENCES RECORD(ID)
    );

测试数据：

INSERT INTO PROJECT (PROJECT_NAME) VALUES ('Bake a cake');
INSERT INTO PROJECT (PROJECT_NAME) VALUES ('Clean the house');

INSERT INTO RECORD (RESULT) VALUES ('GOOD');
INSERT INTO RECORD (RESULT) VALUES ('VERY_GOOD');
INSERT INTO RECORD (RESULT) VALUES ('VERY_GOOD');
INSERT INTO RECORD (RESULT) VALUES ('BAD');
INSERT INTO RECORD (RESULT) VALUES ('VERY_BAD');

INSERT INTO RECORD_PROJECT (PROJECT_ID, RECORD_ID) VALUES (0,0);
INSERT INTO RECORD_PROJECT (PROJECT_ID, RECORD_ID) VALUES (1,1);
INSERT INTO RECORD_PROJECT (PROJECT_ID, RECORD_ID) VALUES (1,2);
INSERT INTO RECORD_PROJECT (PROJECT_ID, RECORD_ID) VALUES (0,3);
INSERT INTO RECORD_PROJECT (PROJECT_ID, RECORD_ID) VALUES (1,4);

（我从表中删除了不相关的字段）

所以有了这些数据我有3个好的记录和2个坏的，我需要得到具有最高'评级'的项目，根据现在这个将是清理房子其他项目3个好评为2个负面。

也许有人会想到这一点，谢谢！

Answer 1

那应该是MySQL-Dialect中的（不是testet）SQL：

SELECT rp.PROJECT_ID, p.PROJECT_NAME
   SUM(CASE WHEN rp.RECORD_ID < 3 THEN 1 ELSE 0 END) AS rating
FROM RECORD_PROJEKT AS rp
JOIN PROJECT AS p ON p.ID = rp.PROJECT_ID
GROUP BY rp.PROJECT_ID
ORDER BY rating DESC

复杂的SQL计数查询

1 个答案: