非常复杂的（Postgre/My）SQL 示例？答案

【问题标题】：Very Complex (Postgre/My)SQL examples?非常复杂的（Postgre/My）SQL 示例？
【发布时间】：2009-12-18 02:15:07
【问题描述】：

在测试abstraction concepts 或仅在使用 PostgreSQL、MySQL 甚至 SQLite 时比较数据库样式和结构时，我经常发现自己需要非常复杂的 SQL 示例。

我认为这意味着还有其他人需要疯狂的查询来让我们看到可能发生的事情，并确保我们的 DB 层可以处理我们向他们抛出的任何东西。

那么，任何人都可以分享一些查询，即使是最顽固的 ORM-all-the-way 家伙也可以进行循环？

SELECT [ ALL | DISTINCT [ ON ( expression [, ...] ) ] ]
    * | expression [ [ AS ] output_name ] [, ...]
    [ FROM from_item [, ...] ]
    [ WHERE condition ]
    [ GROUP BY expression [, ...] ]
    [ HAVING condition [, ...] ]
    [ WINDOW window_name AS ( window_definition ) [, ...] ]
    [ { UNION | INTERSECT | EXCEPT } [ ALL ] select ]
    [ ORDER BY expression [ ASC | DESC | USING operator ] [ NULLS { FIRST | LAST } ] [, ...] ]
    [ LIMIT { count | ALL } ]
    [ OFFSET start [ ROW | ROWS ] ]
    [ FETCH { FIRST | NEXT } [ count ] { ROW | ROWS } ONLY ]
    [ FOR { UPDATE | SHARE } [ OF table_name [, ...] ] [ NOWAIT ] [...] ]

MySQL

SELECT
    [ALL | DISTINCT | DISTINCTROW ]
      [HIGH_PRIORITY]
      [STRAIGHT_JOIN]
      [SQL_SMALL_RESULT] [SQL_BIG_RESULT] [SQL_BUFFER_RESULT]
      [SQL_CACHE | SQL_NO_CACHE] [SQL_CALC_FOUND_ROWS]
    select_expr [, select_expr ...]
    [FROM table_references
    [WHERE where_condition]
    [GROUP BY {col_name | expr | position}
      [ASC | DESC], ... [WITH ROLLUP]]
    [HAVING where_condition]
    [ORDER BY {col_name | expr | position}
      [ASC | DESC], ...]
    [LIMIT {[offset,] row_count | row_count OFFSET offset}]
    [PROCEDURE procedure_name(argument_list)]
    [INTO OUTFILE 'file_name'
        [CHARACTER SET charset_name]
        export_options
      | INTO DUMPFILE 'file_name'
      | INTO var_name [, var_name]]
    [FOR UPDATE | LOCK IN SHARE MODE]]

【问题讨论】：

是否应该将其标记为社区 Wiki？

标签： sql mysql postgresql

【解决方案1】：

查看 2009 年 11 月 OpenSQLCamp 期间的这次演讲。

标题：SQL For the Insane

还有 Recursion with SQL

问题

假设您正在跟踪耗材，并且有一个名为 si_item 的字段和另一个名为 si_parentid 的字段。父级跟踪供应项目所属的子类。例如。您的纸质父母具有回收的、非回收的等子类。当有人拿走用品时，您想返回完全限定的名称，例如纸->回收->20磅

解决方案

CREATE TABLE supplyitem(si_id integer PRIMARY KEY, si_parentid integer, si_item varchar(100));

--load up the table (multirow constructor introduced in 8.2)
INSERT INTO supplyitem(si_id,si_parentid, si_item)
VALUES (1, NULL, 'Paper'),
(2,1, 'Recycled'),
(3,2, '20 lb'),
(4,2, '40 lb'),
(5,1, 'Non-Recycled'),
(6,5, '20 lb'),
(7,5, '40 lb'),
(8,5, 'Scraps');

--Recursive query (introduced in 8.4 returns fully qualified name)
WITH RECURSIVE supplytree AS
(SELECT si_id, si_item, si_parentid, CAST(si_item As varchar(1000)) As si_item_fullname
FROM supplyitem
WHERE si_parentid IS NULL
UNION ALL
SELECT si.si_id,si.si_item,
    si.si_parentid,
    CAST(sp.si_item_fullname || '->' || si.si_item As varchar(1000)) As si_item_fullname
FROM supplyitem As si
    INNER JOIN supplytree AS sp
    ON (si.si_parentid = sp.si_id)
)
SELECT si_id, si_item_fullname
FROM supplytree
ORDER BY si_item_fullname;

结果看起来像

si_id |      si_item_fullname
------+-----------------------------
 1    | Paper
 5    | Paper->Non-Recycled
 6    | Paper->Non-Recycled->20 lb
 7    | Paper->Non-Recycled->40 lb
 8    | Paper->Non-Recycled->Scraps
 2    | Paper->Recycled
 3    | Paper->Recycled->20 lb
 4    | Paper->Recycled->40 lb

【讨论】：