Informix 到 HiveQL

【问题标题】：Informix to HiveQLInformix 到 HiveQL
【发布时间】：2022-01-08 19:07:58
【问题描述】：

我有一个为 Informix 编写的 SQL 查询

SELECT cols
FROM table1 t1, outer(table2 t2, table3 t3)
WHERE t1.id = t2.id and t2.type = t3.type

我们在 Hive 上有相同的表和数据，我想将其转换为 HiveQL

【问题讨论】：

    标签： sql join hive hiveql informix


    【解决方案1】：

    使用 LEFT JOIN 代替 OUTER

    SELECT cols
FROM table1 t1
     LEFT JOIN 
         (select t2.id as join_key, other_cols 
            from table2 t2
                 inner join table3 t3 ON t2.type = t3.type
         ) s ON s.join_key=t1.id

    【讨论】：

    • 你写的很接近，但它对应于FROM table1 t1, OUTER(table2 t2, OUTER table3 t3)。
    • @JonathanLeffler 谢谢，现在看起来正确吗？
    • 是的，应该是这样的。
    【解决方案2】：

    Informix 样式的外连接非常有趣、独特且完全不标准。 Informix 对 ANSI (ISO) 标准 SQL 外连接的实现应该是完全无聊的，并且与其他 DBMS 相同。

    这是一个 SQL 脚本，它创建和填充 3 个表并针对这些表执行 5 个不同的查询。

    CREATE TABLE table1
(
    id      SERIAL NOT NULL PRIMARY KEY,
    data    VARCHAR(32) NOT NULL
);

CREATE TABLE table2
(
    id      INTEGER NOT NULL,
    type    INTEGER NOT NULL,
    info    VARCHAR(32) NOT NULL,
    PRIMARY KEY(id, type)
);

CREATE TABLE table3
(
    type    INTEGER NOT NULL PRIMARY KEY,
    name    VARCHAR(32) NOT NULL
);

INSERT INTO table1 VALUES(100, 'Table 1 - ID 100');
INSERT INTO table1 VALUES(101, 'Table 1 - ID 101');
INSERT INTO table1 VALUES(102, 'Table 1 - ID 102');
INSERT INTO table1 VALUES(103, 'Table 1 - ID 103');
INSERT INTO table1 VALUES(104, 'Table 1 - ID 104');

INSERT INTO table2 VALUES(100, 300, 'Table 2 - ID 100, Type 300');
INSERT INTO table2 VALUES(100, 301, 'Table 2 - ID 100, Type 301');
INSERT INTO table2 VALUES(100, 302, 'Table 2 - ID 100, Type 302');
INSERT INTO table2 VALUES(101, 301, 'Table 2 - ID 101, Type 301');
INSERT INTO table2 VALUES(101, 400, 'Table 2 - ID 101, Type 400');
INSERT INTO table2 VALUES(101, 302, 'Table 2 - ID 101, Type 302');
INSERT INTO table2 VALUES(103, 302, 'Table 2 - ID 103, Type 302');
INSERT INTO table2 VALUES(103, 303, 'Table 2 - ID 103, Type 303');
INSERT INTO table2 VALUES(103, 300, 'Table 2 - ID 103, Type 300');
INSERT INTO table2 VALUES(107, 300, 'Table 2 - ID 107, Type 300');
INSERT INTO table2 VALUES(107, 400, 'Table 2 - ID 107, Type 400');

INSERT INTO table3 VALUES(300, 'Table 3 - Type 300');
INSERT INTO table3 VALUES(301, 'Table 3 - Type 301');
INSERT INTO table3 VALUES(302, 'Table 3 - Type 302');
INSERT INTO table3 VALUES(303, 'Table 3 - Type 303');
INSERT INTO table3 VALUES(304, 'Table 3 - Type 304');
INSERT INTO table3 VALUES(305, 'Table 3 - Type 305');

    SELECT t1.id as t1_id, t1.data, t2.id as t2_id, t2.type as t2_type, t2.info, t3.type as t3_type, t3.name
  FROM table1 t1, OUTER(table2 t2, table3 t3)
 WHERE t1.id = t2.id AND t2.type = t3.type;

SELECT t1.id as t1_id, t1.data, t2.id as t2_id, t2.type as t2_type, t2.info, t3.type as t3_type, t3.name
  FROM table1 t1, OUTER(table2 t2, OUTER table3 t3)
 WHERE t1.id = t2.id AND t2.type = t3.type;

SELECT t1.id as t1_id, t1.data, t2.id as t2_id, t2.type as t2_type, t2.info, t3.type as t3_type, t3.name
  FROM table1 t1
  LEFT JOIN table2 t2 ON t1.id = t2.id
  LEFT JOIN table3 t3 ON t2.type = t3.type;

SELECT t1.id as t1_id, t1.data, s.join_key as t2_id, s.t2_type, s.info, s.t3_type, s.name
  FROM table1 t1
  LEFT JOIN 
       (SELECT t2.id AS join_key, t2.info, t3.name, t2.type as t2_type, t3.type as t3_type
          FROM table2 t2
          JOIN table3 t3 ON t2.type = t3.type
       ) s ON s.join_key = t1.id;

SELECT t1.id as t1_id, t1.data, t2.id as t2_id, t2.type as t2_type, t2.info, t3.type as t3_type, t3.name
  FROM table1 t1
  LEFT JOIN table2 t2 ON t1.id = t2.id
  JOIN table3 t3 ON t2.type = t3.type;

    第一个查询是来自问题的查询，并指定了选定的列名。每个查询中的列名列表都是相同的。

    第二个查询是一个变体，使用了一个额外的 OUTER 关键字。它对应于第三个查询，即leftjoin 创建的first answer。第三个查询是@leftjoin 生成的第一个查询。第四个查询是@leftjoin 创建的'current answer'（严格来说是第三个修订版）。第五个查询是第四个查询的变体——它产生不同的答案，所以它不等价。

    这些是查询的输出。

    查询 1

    SELECT t1.id as t1_id, t1.data, t2.id as t2_id, t2.type as t2_type, t2.info, t3.type as t3_type, t3.name
  FROM table1 t1, OUTER(table2 t2, table3 t3)
 WHERE t1.id = t2.id AND t2.type = t3.type;
    t1_id data t2_id t2_type info t3_type name
    100 Table 1 - ID 100 100 300 Table 2 - ID 100, Type 300 300 Table 3 - Type 300
    100 Table 1 - ID 100 100 301 Table 2 - ID 100, Type 301 301 Table 3 - Type 301
    100 Table 1 - ID 100 100 302 Table 2 - ID 100, Type 302 302 Table 3 - Type 302
    101 Table 1 - ID 101 101 301 Table 2 - ID 101, Type 301 301 Table 3 - Type 301
    101 Table 1 - ID 101 101 302 Table 2 - ID 101, Type 302 302 Table 3 - Type 302
    102 Table 1 - ID 102
    103 Table 1 - ID 103 103 300 Table 2 - ID 103, Type 300 300 Table 3 - Type 300
    103 Table 1 - ID 103 103 302 Table 2 - ID 103, Type 302 302 Table 3 - Type 302
    103 Table 1 - ID 103 103 303 Table 2 - ID 103, Type 303 303 Table 3 - Type 303
    104 Table 1 - ID 104

    查询 2

    SELECT t1.id as t1_id, t1.data, t2.id as t2_id, t2.type as t2_type, t2.info, t3.type as t3_type, t3.name
  FROM table1 t1, OUTER(table2 t2, OUTER table3 t3)
 WHERE t1.id = t2.id AND t2.type = t3.type;
    t1_id data t2_id t2_type info t3_type name
    100 Table 1 - ID 100 100 300 Table 2 - ID 100, Type 300 300 Table 3 - Type 300
    100 Table 1 - ID 100 100 301 Table 2 - ID 100, Type 301 301 Table 3 - Type 301
    100 Table 1 - ID 100 100 302 Table 2 - ID 100, Type 302 302 Table 3 - Type 302
    101 Table 1 - ID 101 101 301 Table 2 - ID 101, Type 301 301 Table 3 - Type 301
    101 Table 1 - ID 101 101 302 Table 2 - ID 101, Type 302 302 Table 3 - Type 302
    101 Table 1 - ID 101 101 400 Table 2 - ID 101, Type 400
    102 Table 1 - ID 102
    103 Table 1 - ID 103 103 300 Table 2 - ID 103, Type 300 300 Table 3 - Type 300
    103 Table 1 - ID 103 103 302 Table 2 - ID 103, Type 302 302 Table 3 - Type 302
    103 Table 1 - ID 103 103 303 Table 2 - ID 103, Type 303 303 Table 3 - Type 303
    104 Table 1 - ID 104

    查询 3

    SELECT t1.id as t1_id, t1.data, t2.id as t2_id, t2.type as t2_type, t2.info, t3.type as t3_type, t3.name
  FROM table1 t1
  LEFT JOIN table2 t2 ON t1.id = t2.id
  LEFT JOIN table3 t3 ON t2.type = t3.type;
    t1_id data t2_id t2_type info t3_type name
    100 Table 1 - ID 100 100 300 Table 2 - ID 100, Type 300 300 Table 3 - Type 300
    100 Table 1 - ID 100 100 301 Table 2 - ID 100, Type 301 301 Table 3 - Type 301
    100 Table 1 - ID 100 100 302 Table 2 - ID 100, Type 302 302 Table 3 - Type 302
    101 Table 1 - ID 101 101 301 Table 2 - ID 101, Type 301 301 Table 3 - Type 301
    101 Table 1 - ID 101 101 302 Table 2 - ID 101, Type 302 302 Table 3 - Type 302
    101 Table 1 - ID 101 101 400 Table 2 - ID 101, Type 400
    102 Table 1 - ID 102
    103 Table 1 - ID 103 103 300 Table 2 - ID 103, Type 300 300 Table 3 - Type 300
    103 Table 1 - ID 103 103 302 Table 2 - ID 103, Type 302 302 Table 3 - Type 302
    103 Table 1 - ID 103 103 303 Table 2 - ID 103, Type 303 303 Table 3 - Type 303
    104 Table 1 - ID 104

    查询 4

    SELECT t1.id as t1_id, t1.data, s.join_key as t2_id, s.t2_type, s.info, s.t3_type, s.name
  FROM table1 t1
  LEFT JOIN 
       (SELECT t2.id AS join_key, t2.info, t3.name, t2.type as t2_type, t3.type as t3_type
          FROM table2 t2
          JOIN table3 t3 ON t2.type = t3.type
       ) s ON s.join_key = t1.id;
    t1_id data t2_id t2_type info t3_type name
    100 Table 1 - ID 100 100 300 Table 2 - ID 100, Type 300 300 Table 3 - Type 300
    100 Table 1 - ID 100 100 301 Table 2 - ID 100, Type 301 301 Table 3 - Type 301
    100 Table 1 - ID 100 100 302 Table 2 - ID 100, Type 302 302 Table 3 - Type 302
    101 Table 1 - ID 101 101 301 Table 2 - ID 101, Type 301 301 Table 3 - Type 301
    101 Table 1 - ID 101 101 302 Table 2 - ID 101, Type 302 302 Table 3 - Type 302
    102 Table 1 - ID 102
    103 Table 1 - ID 103 103 300 Table 2 - ID 103, Type 300 300 Table 3 - Type 300
    103 Table 1 - ID 103 103 302 Table 2 - ID 103, Type 302 302 Table 3 - Type 302
    103 Table 1 - ID 103 103 303 Table 2 - ID 103, Type 303 303 Table 3 - Type 303
    104 Table 1 - ID 104

    查询 5

    SELECT t1.id as t1_id, t1.data, t2.id as t2_id, t2.type as t2_type, t2.info, t3.type as t3_type, t3.name
  FROM table1 t1
  LEFT JOIN table2 t2 ON t1.id = t2.id
  JOIN table3 t3 ON t2.type = t3.type;
    t1_id data t2_id t2_type info t3_type name
    100 Table 1 - ID 100 100 300 Table 2 - ID 100, Type 300 300 Table 3 - Type 300
    103 Table 1 - ID 103 103 300 Table 2 - ID 103, Type 300 300 Table 3 - Type 300
    100 Table 1 - ID 100 100 301 Table 2 - ID 100, Type 301 301 Table 3 - Type 301
    101 Table 1 - ID 101 101 301 Table 2 - ID 101, Type 301 301 Table 3 - Type 301
    100 Table 1 - ID 100 100 302 Table 2 - ID 100, Type 302 302 Table 3 - Type 302
    101 Table 1 - ID 101 101 302 Table 2 - ID 101, Type 302 302 Table 3 - Type 302
    103 Table 1 - ID 103 103 302 Table 2 - ID 103, Type 302 302 Table 3 - Type 302
    103 Table 1 - ID 103 103 303 Table 2 - ID 103, Type 303 303 Table 3 - Type 303

    除非我错过了什么，否则 Q1 和 Q4 的输出是相同的（所以@leftjoin 的答案是正确的）；其他查询的输出都彼此不同（以及来自 Q1 和 Q4）。

    【讨论】：

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