功能内的DB2临时表

Question

我有一些简单的函数可以在DB2中工作，但是我无法使任何声明全局临时表的函数工作。 我知道DB2的版本和平台有很大的不同，但是我认为这应该可行，因为它适用于此发布者 。

CREATE FUNCTION FuncTest1 ()
    RETURNS TABLE
    (
        USE_NAME VARCHAR(48),
        USE_PARTNER_LINK FLOAT
    )
        LANGUAGE SQL
        MODIFIES SQL DATA
        DETERMINISTIC
        BEGIN
        DECLARE GLOBAL TEMPORARY TABLE USE_TRUNC
        (
           USE_NAME VARCHAR(48) NULL,
           USE_PARTNER_LINK FLOAT NULL
        );
        INSERT INTO SESSION.USE_TRUNC
            (USE_NAME,USE_PARTNER_LINK)
        SELECT USE_NAME,USE_PARTNER_LINK FROM F_USERS;
        RETURN
        SELECT USE_NAME,USE_PARTNER_LINK FROM F_USERS;
        END

当我尝试不同的操作时，错误的差异很大，但这是当前输出：

An unexpected token "USE_NAME" was found following " 
  RETURN 
  SELECT".  Expected tokens may include:  "(".. SQLCODE=-104, SQLSTATE=42601, DRIVER=4.9.78

添加了小提琴

Answer 1

CREATE FUNCTION（SQL标量，表或行）语句，v9.7 ：

-| (3)| '-MODIFIES SQL DATA-----' ... |--+-RETURN----------------------+------------------------------| | (5) | +-Compound SQL (compiled)-----+ '-Compound SQL (inlined)------'
...
3.如果RETURNS指定一个表（即TABLE column-list），则有效。 如果RETURNS指定标量结果并且SQL-function-body是复合SQL（已编译）语句，则该命令也有效。
...
5.仅在SQL标量函数定义中的SQL函数体支持复合SQL（编译）语句。 SQL表函数定义不支持此功能 。

CREATE FUNCTION（SQL标量，表或行）语句，v11.1 ：

-| (4) | '-MODIFIES SQL DATA-----'
...
4.仅对编译的标量函数定义和内联表函数定义有效 。

您不能将MODIFIED SQL DATA与Compound SQL (compiled) （ BEGIN ... END ）一起用于表函数。 Compound SQL (inlined)不支持DECLARE GTT语句（ BEGIN ATOMIC ... END ）。
因此，不要使用声明的GTT和INSERT重写表函数。 或尝试Created GTT ，但要事先创建它们（不在函数主体中）。

至于你的情况

您可以在Common Table Expression使用单个SELECT语句：

CREATE FUNCTION ...
...
RETURN
WITH 
SESSION_ALLREFERRALS AS 
(
SELECT
    REF.AL_NO AS AL_NO,        
    ENQ.E_KEY AS R_KEY,
    APP.A_2ND_NAME AS R_TRADING_NAME,
    CASE WHEN DOC1.DT_NAME is null THEN 'NONE' ELSE DOC1.DT_NAME END AS R_DROP1,
    REF.AL_DATE AS R_DATE,
    CASE WHEN DOC2.DT_NAME is null THEN 'Incomplete' ELSE DOC2.DT_NAME END AS R_DROP2,
    CAST(ENQ2.E_TOOLCOMM1 AS VARCHAR(48)) AS R_COMM1
FROM
    F_ENQUIRY ENQ INNER JOIN 
        F_APPLICANT_LINK REF ON REF.AL_KEY1 = ENQ.E_KEY INNER JOIN 
        F_APPLICANT APP ON APP.A_KEY = REF.AL_KEY2 AND REF.AL_TYPE1 = 2 AND REF.AL_TYPE2 = 1 LEFT JOIN 
        F_DOC_TYPES DOC1 ON DOC1.DT_NO = REF.AL_DROP1 LEFT JOIN 
        F_DOC_TYPES DOC2 ON DOC2.DT_NO = REF.AL_DROP2 INNER JOIN 
        F_ENQUIRY ENQ2 ON ENQ2.E_KEY = REF.AL_ENQ_LINK
WHERE
    ENQ.E_PRIORITY_LINK = 204 AND   
    ENQ.E_JOB_TYPE_LINK = 0
)
, 
SESSION_REFERRAL1 AS 
(
SELECT 
    MIN(AL_NO) AL_NO, 
    R_KEY
    FROM 
    SESSION_ALLREFERRALS 
    GROUP BY R_KEY
)
,
SESSION_REFERRAL2 AS 
(
SELECT 
    MIN(AL_NO) AL_NO, 
    R_KEY 
FROM 
    SESSION_ALLREFERRALS a2
WHERE
    NOT EXISTS (SELECT 1 FROM SESSION_REFERRAL1 R1 WHERE r1.AL_NO = a2.AL_NO fetch first 1 rows only)
GROUP BY 
    R_KEY
)
,
SESSION_REFERRAL3 AS 
(
SELECT 
    MIN(AL_NO) AL_NO, 
    R_KEY 
FROM 
    SESSION_ALLREFERRALS A3
WHERE
    NOT EXISTS (SELECT 1 FROM SESSION_REFERRAL1 R1 WHERE R1.AL_NO = A3.AL_NO fetch first 1 rows only) AND 
    NOT EXISTS (SELECT 1 FROM SESSION_REFERRAL2 R2 WHERE R2.AL_NO = A3.AL_NO fetch first 1 rows only)
GROUP BY 
    R_KEY
)
SELECT 
    *
FROM
    SESSION_REFERRAL1 R1 INNER JOIN
    SESSION_ALLREFERRALS A ON A.R_KEY = R1.R_KEY AND R1.AL_NO = A.AL_NO LEFT JOIN
    SESSION_REFERRAL2 R2 ON R2.R_KEY = R1.R_KEY LEFT JOIN
    SESSION_ALLREFERRALS A2 ON A2.R_KEY = R2.R_KEY AND R2.AL_NO = A2.AL_NO LEFT JOIN 
    SESSION_REFERRAL3 R3 ON R3.R_KEY = R1.R_KEY LEFT JOIN
    SESSION_ALLREFERRALS A3 ON A3.R_KEY = R3.R_KEY AND R3.AL_NO = A3.AL_NO
;

Answer 2

这将是编写该函数的更简单方法

CREATE FUNCTION FuncTest1 ()
RETURNS TABLE
(
    USE_NAME VARCHAR(48),
    USE_PARTNER_LINK FLOAT
)
    LANGUAGE SQL
    RETURN  SELECT USE_NAME,USE_PARTNER_LINK FROM F_USERS;