Question

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

我认为这意味着还有其他人需要疯狂的查询才能让我们看到可能的事情，并确保我们的数据库层可以处理我们向他们投掷的任何内容。

那么，任何人都可以分享一些查询，这些查询甚至可以通过最核心的ORM-all-way方式进行循环吗？

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]]

Answer 1

在2009年11月的OpenSQLCamp期间查看此演讲。

标题：SQL For the Insane

另外 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。父级跟踪供应项属于哪个子类。例如。你有纸质父母有子类，如再生，非再生。当有人拿走耗材时，您需要返回完全限定的名称，例如纸 - >再循环 - > 20Lb

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

非常复杂（Postgre / My）SQL示例？

1 个答案: