如何将带有 GROUP BY 子句的查询移植到 PostgreSQL？
How do I port query with GROUP BY clause to PostgreSQL?
我正在将一个简单的费用数据库移植到 Postgres，并卡在一个使用 GROUP BY 和多个 JOIN 子句的视图上。我认为 Postgres 希望我使用 GROUP BY 子句中的所有表。
Table定义在最后。请注意，account_id、receiving_account_id 和 place 列可能是 NULL，而 operation 可以有 0 个标签。
原始CREATE声明
CREATE VIEW details AS SELECT
op.id,
op.name,
c.name,
CASE --amountsign
WHEN op.receiving_account_id IS NOT NULL THEN
CASE
WHEN op.account_id IS NULL THEN '+'
ELSE '='
END
ELSE '-'
END || ' ' || printf("%.2f", op.amount) || ' zł' AS amount,
CASE --account
WHEN op.receiving_account_id IS NOT NULL THEN
CASE
WHEN op.account_id IS NULL THEN ac2.name
ELSE ac.name || ' -> ' || ac2.name
END
ELSE ac.name
END AS account,
t.name AS type,
CASE --date
WHEN op.time IS NOT NULL THEN op.date || ' ' || op.time
ELSE op.date
END AS date,
p.name AS place,
GROUP_CONCAT(tag.name, ', ') AS tags
FROM operation op
LEFT JOIN category c ON op.category_id = c.id
LEFT JOIN type t ON op.type_id = t.id
LEFT JOIN account ac ON op.account_id = ac.id
LEFT JOIN account ac2 ON op.receiving_account_id = ac2.id
LEFT JOIN place p ON op.place_id = p.id
LEFT JOIN operation_tag ot ON op.id = ot.operation_id
LEFT JOIN tag ON ot.tag_id = tag.id
GROUP BY IFNULL (ot.operation_id, op.id)
ORDER BY date DESC
Postgres 中的当前查询
我做了一些更新，我现在的声明是：
BEGIN TRANSACTION;
CREATE VIEW details AS SELECT
op.id,
op.name,
c.name,
CASE --amountsign
WHEN op.receiving_account_id IS NOT NULL THEN
CASE
WHEN op.account_id IS NULL THEN '+'
ELSE '='
END
ELSE '-'
END || ' ' || op.amount || ' zł' AS amount,
CASE --account
WHEN op.receiving_account_id IS NOT NULL THEN
CASE
WHEN op.account_id IS NULL THEN ac2.name
ELSE ac.name || ' -> ' || ac2.name
END
ELSE ac.name
END AS account,
t.name AS type,
CASE --date
WHEN op.time IS NOT NULL THEN to_char(op.date, 'DD.MM.YY') || ' ' || op.time
ELSE to_char(op.date, 'DD.MM.YY')
END AS date,
p.name AS place,
STRING_AGG(tag.name, ', ') AS tags
FROM operation op
LEFT JOIN category c ON op.category_id = c.id
LEFT JOIN type t ON op.type_id = t.id
LEFT JOIN account ac ON op.account_id = ac.id
LEFT JOIN account ac2 ON op.receiving_account_id = ac2.id
LEFT JOIN place p ON op.place_id = p.id
LEFT JOIN operation_tag ot ON op.id = ot.operation_id
LEFT JOIN tag ON ot.tag_id = tag.id
GROUP BY COALESCE (ot.operation_id, op.id)
ORDER BY date DESC;
COMMIT;
这里我在添加列出的错误时得到 Column 'x' must appear in GROUP BY clause 个错误：
GROUP BY COALESCE(ot.operation_id, op.id), op.id, c.name, ac2.name, ac.name, t.name, p.name
当我添加 p.name 列时，我得到 Column 'p.name' is defined more than once error. 我该如何解决？
Table定义
CREATE TABLE operation (
id integer NOT NULL PRIMARY KEY,
name character varying(64) NOT NULL,
category_id integer NOT NULL,
type_id integer NOT NULL,
amount numeric(8,2) NOT NULL,
date date NOT NULL,
"time" time without time zone NOT NULL,
place_id integer,
account_id integer,
receiving_account_id integer,
CONSTRAINT categories_transactions FOREIGN KEY (category_id)
REFERENCES category (id) MATCH SIMPLE
ON UPDATE NO ACTION ON DELETE NO ACTION,
CONSTRAINT transactions_accounts FOREIGN KEY (account_id)
REFERENCES account (id) MATCH SIMPLE
ON UPDATE NO ACTION ON DELETE NO ACTION,
CONSTRAINT transactions_accounts_second FOREIGN KEY (receiving_account_id)
REFERENCES account (id) MATCH SIMPLE
ON UPDATE NO ACTION ON DELETE NO ACTION,
CONSTRAINT transactions_places FOREIGN KEY (place_id)
REFERENCES place (id) MATCH SIMPLE
ON UPDATE NO ACTION ON DELETE NO ACTION,
CONSTRAINT transactions_transaction_types FOREIGN KEY (type_id)
REFERENCES type (id) MATCH SIMPLE
ON UPDATE NO ACTION ON DELETE NO ACTION
);
大多数数据库要求您 group by 在 select 中出现的每一列都未聚合。未聚合意味着不包含在聚合中，如 min、max 或 string_agg。所以你需要分组：op.id, op.name, c.name, op.receiving_account_id, ...，等等
此要求的原因是数据库必须确定组的值。通过将列添加到 group by 子句，您可以确认组中的每一行都具有相同的值。对于其他组，您必须指定要与聚合一起使用的值。例外是 MySQL，如果您没有做出有意识的选择，它只会选择一个任意值。
如果您的 group by 只是创建一个标签列表，您可以将其移至子查询：
left join
(
select id
, string_agg(tag.name, ', ') tags
from tag
group by
id
) t
on ot.tag_id = t.id
并且您可以避免为外部查询使用非常长的分组依据。
与 类似：大多数 RDBMS 要求按未聚合的每一列进行分组 - 查询中的任何其他位置（包括 SELECT 列表，但也在 WHERE 子句等中.)
- PGError: ERROR: aggregates not allowed in WHERE clause on a AR query of an object and its has_many objects
SQL 标准还定义了 GROUP BY 子句中的表达式也应涵盖功能相关的表达式。 Postgres 实现了 PK 列覆盖相同 table.
的所有列
- PostgreSQL - GROUP BY clause
所以 op.id 涵盖了整个 table 这应该适用于您当前的查询：
GROUP BY op.id, c.name, 5, t.name, p.name
5 是对 SELECT 列表的 位置引用 ，这在 Postgres 中也是允许的。它只是符号 shorthand 用于重复长表达式：
CASE
WHEN op.receiving_account_id IS NOT NULL THEN
CASE
WHEN op.account_id IS NULL THEN ac2.name
ELSE ac.name || ' -> ' || ac2.name
END
ELSE ac.name
END
- Concatenate multiple result rows of one column into one, group by another column
- Select first row in each GROUP BY group?
我从你的名字中得出你在 operation 和 tag 之间有一个 n:m 关系，用 operation_tag 实现。所有其他连接似乎都不会乘以行，因此单独聚合标签会更有效 - 就像@Andomar 暗示的那样，只要逻辑正确即可。
这应该有效：
SELECT op.id
, op.name
, c.name
, CASE -- amountsign
WHEN op.receiving_account_id IS NOT NULL THEN
CASE WHEN op.account_id IS NULL THEN '+' ELSE '=' END
ELSE '-'
END || ' ' || op.amount || ' zł' AS amount
, CASE -- account
WHEN op.receiving_account_id IS NOT NULL THEN
CASE
WHEN op.account_id IS NULL THEN ac2.name
ELSE ac.name || ' -> ' || ac2.name
END
ELSE ac.name
END AS account
, t.name AS type
, <b>to_char(op.date, 'DD.MM.YY') || ' ' || op.time AS date</b> -- see below
, p.name AS place
, ot.tags
FROM operation op
LEFT JOIN category c ON op.category_id = c.id
LEFT JOIN type t ON op.type_id = t.id
LEFT JOIN account ac ON op.account_id = ac.id
LEFT JOIN account ac2 ON op.receiving_account_id = ac2.id
LEFT JOIN place p ON op.place_id = p.id
<b>LEFT JOIN (
SELECT operation_id, string_agg(t.name, ', ') AS tags
FROM operation_tag ot
LEFT JOIN tag t ON t.id = ot.tag_id
GROUP BY 1
) ot ON op.id = ot.operation_id</b>
<b>ORDER BY op.date DESC, op.time DESC</b>;
旁白
您可以替换：
CASE --date
WHEN op.time IS NOT NULL THEN to_char(op.date, 'DD.MM.YY') || ' ' || op.time
ELSE to_char(op.date, 'DD.MM.YY')
END AS date
用这个较短的等价物：
concat_ws(' ', to_char(op.date, 'DD.MM.YY'), op.time) AS date
但由于两列都已定义 NOT NULL，您可以进一步简化为：
to_char(op.date, 'DD.MM.YY') || ' ' || op.time AS date
小心你的 ORDER BY 你至少有一个输入列也命名为 date。如果您使用非限定名称，它将引用 output 列 - 这就是您想要的（如评论中所述）。详情：
- PostgreSQL: How to return rows with respect to a found row (relative results)?
但是，按文本表示排序不会根据您的时间轴正确排序。按照我上面查询中的建议，按原始值排序。
我正在将一个简单的费用数据库移植到 Postgres，并卡在一个使用 GROUP BY 和多个 JOIN 子句的视图上。我认为 Postgres 希望我使用 GROUP BY 子句中的所有表。
Table定义在最后。请注意，account_id、receiving_account_id 和 place 列可能是 NULL，而 operation 可以有 0 个标签。
原始CREATE声明
CREATE VIEW details AS SELECT
op.id,
op.name,
c.name,
CASE --amountsign
WHEN op.receiving_account_id IS NOT NULL THEN
CASE
WHEN op.account_id IS NULL THEN '+'
ELSE '='
END
ELSE '-'
END || ' ' || printf("%.2f", op.amount) || ' zł' AS amount,
CASE --account
WHEN op.receiving_account_id IS NOT NULL THEN
CASE
WHEN op.account_id IS NULL THEN ac2.name
ELSE ac.name || ' -> ' || ac2.name
END
ELSE ac.name
END AS account,
t.name AS type,
CASE --date
WHEN op.time IS NOT NULL THEN op.date || ' ' || op.time
ELSE op.date
END AS date,
p.name AS place,
GROUP_CONCAT(tag.name, ', ') AS tags
FROM operation op
LEFT JOIN category c ON op.category_id = c.id
LEFT JOIN type t ON op.type_id = t.id
LEFT JOIN account ac ON op.account_id = ac.id
LEFT JOIN account ac2 ON op.receiving_account_id = ac2.id
LEFT JOIN place p ON op.place_id = p.id
LEFT JOIN operation_tag ot ON op.id = ot.operation_id
LEFT JOIN tag ON ot.tag_id = tag.id
GROUP BY IFNULL (ot.operation_id, op.id)
ORDER BY date DESC
Postgres 中的当前查询
我做了一些更新，我现在的声明是：
BEGIN TRANSACTION;
CREATE VIEW details AS SELECT
op.id,
op.name,
c.name,
CASE --amountsign
WHEN op.receiving_account_id IS NOT NULL THEN
CASE
WHEN op.account_id IS NULL THEN '+'
ELSE '='
END
ELSE '-'
END || ' ' || op.amount || ' zł' AS amount,
CASE --account
WHEN op.receiving_account_id IS NOT NULL THEN
CASE
WHEN op.account_id IS NULL THEN ac2.name
ELSE ac.name || ' -> ' || ac2.name
END
ELSE ac.name
END AS account,
t.name AS type,
CASE --date
WHEN op.time IS NOT NULL THEN to_char(op.date, 'DD.MM.YY') || ' ' || op.time
ELSE to_char(op.date, 'DD.MM.YY')
END AS date,
p.name AS place,
STRING_AGG(tag.name, ', ') AS tags
FROM operation op
LEFT JOIN category c ON op.category_id = c.id
LEFT JOIN type t ON op.type_id = t.id
LEFT JOIN account ac ON op.account_id = ac.id
LEFT JOIN account ac2 ON op.receiving_account_id = ac2.id
LEFT JOIN place p ON op.place_id = p.id
LEFT JOIN operation_tag ot ON op.id = ot.operation_id
LEFT JOIN tag ON ot.tag_id = tag.id
GROUP BY COALESCE (ot.operation_id, op.id)
ORDER BY date DESC;
COMMIT;
这里我在添加列出的错误时得到 Column 'x' must appear in GROUP BY clause 个错误：
GROUP BY COALESCE(ot.operation_id, op.id), op.id, c.name, ac2.name, ac.name, t.name, p.name
当我添加 p.name 列时，我得到 Column 'p.name' is defined more than once error. 我该如何解决？
Table定义
CREATE TABLE operation (
id integer NOT NULL PRIMARY KEY,
name character varying(64) NOT NULL,
category_id integer NOT NULL,
type_id integer NOT NULL,
amount numeric(8,2) NOT NULL,
date date NOT NULL,
"time" time without time zone NOT NULL,
place_id integer,
account_id integer,
receiving_account_id integer,
CONSTRAINT categories_transactions FOREIGN KEY (category_id)
REFERENCES category (id) MATCH SIMPLE
ON UPDATE NO ACTION ON DELETE NO ACTION,
CONSTRAINT transactions_accounts FOREIGN KEY (account_id)
REFERENCES account (id) MATCH SIMPLE
ON UPDATE NO ACTION ON DELETE NO ACTION,
CONSTRAINT transactions_accounts_second FOREIGN KEY (receiving_account_id)
REFERENCES account (id) MATCH SIMPLE
ON UPDATE NO ACTION ON DELETE NO ACTION,
CONSTRAINT transactions_places FOREIGN KEY (place_id)
REFERENCES place (id) MATCH SIMPLE
ON UPDATE NO ACTION ON DELETE NO ACTION,
CONSTRAINT transactions_transaction_types FOREIGN KEY (type_id)
REFERENCES type (id) MATCH SIMPLE
ON UPDATE NO ACTION ON DELETE NO ACTION
);
大多数数据库要求您 group by 在 select 中出现的每一列都未聚合。未聚合意味着不包含在聚合中，如 min、max 或 string_agg。所以你需要分组：op.id, op.name, c.name, op.receiving_account_id, ...，等等
此要求的原因是数据库必须确定组的值。通过将列添加到 group by 子句，您可以确认组中的每一行都具有相同的值。对于其他组，您必须指定要与聚合一起使用的值。例外是 MySQL，如果您没有做出有意识的选择，它只会选择一个任意值。
如果您的 group by 只是创建一个标签列表，您可以将其移至子查询：
left join
(
select id
, string_agg(tag.name, ', ') tags
from tag
group by
id
) t
on ot.tag_id = t.id
并且您可以避免为外部查询使用非常长的分组依据。
与 SELECT 列表，但也在 WHERE 子句等中.)
- PGError: ERROR: aggregates not allowed in WHERE clause on a AR query of an object and its has_many objects
SQL 标准还定义了 GROUP BY 子句中的表达式也应涵盖功能相关的表达式。 Postgres 实现了 PK 列覆盖相同 table.
- PostgreSQL - GROUP BY clause
所以 op.id 涵盖了整个 table 这应该适用于您当前的查询：
GROUP BY op.id, c.name, 5, t.name, p.name
5 是对 SELECT 列表的 位置引用 ，这在 Postgres 中也是允许的。它只是符号 shorthand 用于重复长表达式：
CASE
WHEN op.receiving_account_id IS NOT NULL THEN
CASE
WHEN op.account_id IS NULL THEN ac2.name
ELSE ac.name || ' -> ' || ac2.name
END
ELSE ac.name
END
- Concatenate multiple result rows of one column into one, group by another column
- Select first row in each GROUP BY group?
我从你的名字中得出你在 operation 和 tag 之间有一个 n:m 关系，用 operation_tag 实现。所有其他连接似乎都不会乘以行，因此单独聚合标签会更有效 - 就像@Andomar 暗示的那样，只要逻辑正确即可。
这应该有效：
SELECT op.id
, op.name
, c.name
, CASE -- amountsign
WHEN op.receiving_account_id IS NOT NULL THEN
CASE WHEN op.account_id IS NULL THEN '+' ELSE '=' END
ELSE '-'
END || ' ' || op.amount || ' zł' AS amount
, CASE -- account
WHEN op.receiving_account_id IS NOT NULL THEN
CASE
WHEN op.account_id IS NULL THEN ac2.name
ELSE ac.name || ' -> ' || ac2.name
END
ELSE ac.name
END AS account
, t.name AS type
, <b>to_char(op.date, 'DD.MM.YY') || ' ' || op.time AS date</b> -- see below
, p.name AS place
, ot.tags
FROM operation op
LEFT JOIN category c ON op.category_id = c.id
LEFT JOIN type t ON op.type_id = t.id
LEFT JOIN account ac ON op.account_id = ac.id
LEFT JOIN account ac2 ON op.receiving_account_id = ac2.id
LEFT JOIN place p ON op.place_id = p.id
<b>LEFT JOIN (
SELECT operation_id, string_agg(t.name, ', ') AS tags
FROM operation_tag ot
LEFT JOIN tag t ON t.id = ot.tag_id
GROUP BY 1
) ot ON op.id = ot.operation_id</b>
<b>ORDER BY op.date DESC, op.time DESC</b>;
旁白
您可以替换：
CASE --date
WHEN op.time IS NOT NULL THEN to_char(op.date, 'DD.MM.YY') || ' ' || op.time
ELSE to_char(op.date, 'DD.MM.YY')
END AS date
用这个较短的等价物：
concat_ws(' ', to_char(op.date, 'DD.MM.YY'), op.time) AS date
但由于两列都已定义 NOT NULL，您可以进一步简化为：
to_char(op.date, 'DD.MM.YY') || ' ' || op.time AS date
小心你的 ORDER BY 你至少有一个输入列也命名为 date。如果您使用非限定名称，它将引用 output 列 - 这就是您想要的（如评论中所述）。详情：
- PostgreSQL: How to return rows with respect to a found row (relative results)?
但是，按文本表示排序不会根据您的时间轴正确排序。按照我上面查询中的建议，按原始值排序。