具有FIRST选项的Informix子​​查询

时间:2009-06-23 16:34:24

标签: sql-server informix

将以下Transact-SQL代码转录到Informix Dynamic Server(IDS)9.40的最佳方法是什么:

目标:我需要前50个订单及其各自的订单行

select *
    from (select top 50 * from orders) a inner join lines b
            on a.idOrder = b.idOrder

我的问题在于子选择,因为Informix不允许子选择中的FIRST选项。

有什么简单的想法吗?

2 个答案:

答案 0 :(得分:7)

官方的答案是“请从IDS 9.40升级,因为它不再受IBM支持”。也就是说,IDS 9.40不是当前版本 - 并且(理想情况下)应该不使用。

IDS 11.50的解决方案

使用IDS 11.50,我可以写:

SELECT *
    FROM (SELECT FIRST 10 * FROM elements) AS e
         INNER JOIN compound_component AS a
            ON e.symbol = a.element
         INNER JOIN compound AS c
            ON c.compound_id = a.compound_id
    ;

这或多或少等同于您的查询。因此,如果您使用当前版本的IDS,则可以使用与Transact-SQL中几乎相同的表示法编写查询(使用FIRST代替TOP)。

IDS 9.40的解决方案

你能在IDS 9.40做什么?请原谅......我必须运行我的IDS 9.40.xC7服务器(此修订包于2005年发布;原始版本可能在2003年末)...

第一个问题 - IDS 9.40不允许FROM子句中的子查询。

第二个问题 - IDS 9.40在这两种情况下都不允许“FIRST n”表示法:

SELECT FIRST 10 * FROM elements INTO TEMP e;
INSERT INTO e SELECT FIRST 10 * FROM elements;

第三个问题 - IDS 9.40没有简单的ROWNUM。

所以,要解决这些问题,我们可以编写(使用临时表 - 我们稍后会删除它):

SELECT e1.*
    FROM elements AS e1, elements AS e2
    WHERE e1.atomic_number >= e2.atomic_number
    GROUP BY e1.atomic_number, e1.symbol, e1.name, e1.atomic_weight, e1.stable
    HAVING COUNT(*) <= 10
    INTO TEMP e;

SELECT *
    FROM e INNER JOIN compound_component AS a
                   ON e.symbol = a.element
           INNER JOIN compound AS c
                   ON c.compound_id = a.compound_id;

这与IDS 11.50中的单个查询产生相同的答案。我们可以避开临时表吗?是的,但它更详细:

SELECT e1.*, a.*, c.*
    FROM  elements AS e1, elements AS e2, compound_component AS a,
           compound AS c
    WHERE e1.atomic_number >= e2.atomic_number
      AND e1.symbol = a.element
      AND c.compound_id = a.compound_id
    GROUP BY e1.atomic_number, e1.symbol, e1.name, e1.atomic_weight,
             e1.stable, a.compound_id, a.element, a.seq_num,
             a.multiplicity, c.compound_id, c.name
    HAVING   COUNT(*) <= 10;

将其应用于原始订单和订单行示例留给读者练习。


“元素表”的相关模式子集:

-- See: http://www.webelements.com/ for elements.
-- See: http://ie.lbl.gov/education/isotopes.htm for isotopes.

CREATE TABLE elements
(
    atomic_number   INTEGER NOT NULL UNIQUE CONSTRAINT c1_elements
                    CHECK (atomic_number > 0 AND atomic_number < 120),
    symbol          CHAR(3) NOT NULL UNIQUE CONSTRAINT c2_elements,
    name            CHAR(20) NOT NULL UNIQUE CONSTRAINT c3_elements,
    atomic_weight   DECIMAL(8,4) NOT NULL,
    stable          CHAR(1) DEFAULT 'Y' NOT NULL
                    CHECK (stable IN ('Y', 'N'))
);

CREATE TABLE compound
(
    compound_id     SERIAL NOT NULL PRIMARY KEY,
    name            VARCHAR(100) NOT NULL UNIQUE
);

-- The sequence number is used to order the components within a compound.

CREATE TABLE compound_component
(
    compound_id     INTEGER REFERENCES compound,
    element         CHAR(3) NOT NULL REFERENCES elements(symbol),
    seq_num         SMALLINT DEFAULT 1 NOT NULL
                    CHECK (seq_num > 0 AND seq_num < 20),
    multiplicity    INTEGER NOT NULL
                    CHECK (multiplicity > 0 AND multiplicity < 20),
    PRIMARY KEY(compound_id, seq_num)
);

输出(在我的示例数据库中):

 1 H   Hydrogen        1.0079 Y     1 H    1     2       1 water
 1 H   Hydrogen        1.0079 Y     3 H    2     4       3 methane
 1 H   Hydrogen        1.0079 Y     4 H    2     6       4 ethane
 1 H   Hydrogen        1.0079 Y     5 H    2     8       5 propane
 1 H   Hydrogen        1.0079 Y     6 H    2    10       6 butane
 1 H   Hydrogen        1.0079 Y    11 H    2     5      11 ethanol
 1 H   Hydrogen        1.0079 Y    11 H    4     1      11 ethanol
 6 C   Carbon         12.0110 Y     2 C    1     1       2 carbon dioxide
 6 C   Carbon         12.0110 Y     3 C    1     1       3 methane
 6 C   Carbon         12.0110 Y     4 C    1     2       4 ethane
 6 C   Carbon         12.0110 Y     5 C    1     3       5 propane
 6 C   Carbon         12.0110 Y     6 C    1     4       6 butane
 6 C   Carbon         12.0110 Y     7 C    1     1       7 carbon monoxide
 6 C   Carbon         12.0110 Y     9 C    2     1       9 magnesium carbonate
 6 C   Carbon         12.0110 Y    10 C    2     1      10 sodium bicarbonate
 6 C   Carbon         12.0110 Y    11 C    1     2      11 ethanol
 8 O   Oxygen         15.9990 Y     1 O    2     1       1 water
 8 O   Oxygen         15.9990 Y     2 O    2     2       2 carbon dioxide
 8 O   Oxygen         15.9990 Y     7 O    2     1       7 carbon monoxide
 8 O   Oxygen         15.9990 Y     9 O    3     3       9 magnesium carbonate
 8 O   Oxygen         15.9990 Y    10 O    3     3      10 sodium bicarbonate
 8 O   Oxygen         15.9990 Y    11 O    3     1      11 ethanol

答案 1 :(得分:0)

如果我理解你的问题,你就会遇到“TOP”问题。尝试使用TOP-N查询。

例如:

select  *
  from  (SELECT  *
           FROM  foo
          where  foo_id=[number]
       order by  foo_id desc)
 where  rownum <= 50

这将获得前50个结果(因为我在子查询中通过desc命令)