我的查询如下:
SELECT
cc.chain_desc as chain_desc
,cc.chain_id as chain_id
,COUNT(distinct t.trans_id) as TranCount
FROM TRANSACTION AS t
LEFT OUTER JOIN location AS l
ON t.location_id = l.location_id
LEFT OUTER JOIN trans_line AS tl
ON t.trans_id = tl.trans_id
LEFT OUTER JOIN contract as c
ON t.contract_id = c.contract_id
LEFT OUTER JOIN chain_desc as cc
ON l.chain_id = cc.chain_id
WHERE
t.loc_country = 'U'
AND c.issuer_id IN (156966,166203)
AND t.trans_date >= '2016-10-01 00:00'
and t.trans_date < '2016-10-31 00:00'
AND tl.cat NOT IN ('DEF','DEFD','DEFC')
GROUP BY cc.chain_desc, cc.chain_id
UNION
SELECT
'TOTAL'
,0
,COUNT(distinct t.trans_id)
FROM TRANSACTION AS t
LEFT OUTER JOIN location AS l
ON t.location_id = l.location_id
LEFT OUTER JOIN trans_line AS tl
ON t.trans_id = tl.trans_id
LEFT OUTER JOIN contract as c
ON t.contract_id = c.contract_id
LEFT OUTER JOIN chain_desc as cc
ON l.chain_id = cc.chain_id
WHERE
t.loc_country = 'U'
AND c.issuer_id IN (156966,166203)
AND t.trans_date >= '2016-10-01 00:00'
and t.trans_date < '2016-10-31 00:00'
AND tl.cat NOT IN ('DEF','DEFD','DEFC')
执行上述查询会重新生成以下结果:
我需要将结果显示如下:
专栏&#34; Chain_Id&#34;是&#34;整数&#34;类型,我该怎么做那个空白?
答案 0 :(得分:3)
你可以简单地选择null
.....
UNION
SELECT
'TOTAL'
, NULL::INTEGER
,COUNT(distinct t.trans_id)
FROM TRANSACTION AS t
LEFT OUTER JOIN location AS l
ON t.location_id = l.location_id
LEFT OUTER JOIN trans_line AS tl
ON t.trans_id = tl.trans_id
LEFT OUTER JOIN contract as c
ON t.contract_id = c.contract_id
LEFT OUTER JOIN chain_desc as cc
ON l.chain_id = cc.chain_id
WHERE
t.loc_country = 'U'
AND c.issuer_id IN (156966,166203)
AND t.trans_date >= '2016-10-01 00:00'
and t.trans_date < '2016-10-31 00:00'
AND tl.cat NOT IN ('DEF','DEFD','DEFC')
因为null不是您可以尝试在第一个查询上方添加的类型
DEFINE test INT;
LET test = NULL;
......
SELECT
'TOTAL'
, test
,COUNT(distinct t.trans_id)
.....
或者像@Jonathan Leffler所说的那样使用NULL::INTEGER
或CAST(NULL AS INTEGER)
答案 1 :(得分:2)
一种方法是转换为NULL
:
(case when cc.chain_id <> 0 then cc.chain_id end) as chain_id
另一种方法是将所有内容转换为字符串:
(case when cc.chain_id <> 0 then cast(cc.chain_id as varchar(255)) else '' end) as chain_id
答案 2 :(得分:2)
在Informix中,您可以在投影列表中使用NULL
,但该列必须具有类型。由于在Informix NULL
中没有类型,因此您需要使用CAST
。
SELECT NULL::INTEGER AS id FROM systables WHERE tabid = 1;
SELECT CAST(NULL AS INTEGER) AS id FROM systables WHERE tabid = 1;
您可以查看其他问题的答案(Informix: Select null problem)。
您可以查看IBM知识中心(NULL Keyword)。