Question

这是Informix更新的正确语法吗？

update table1
set table1.code = 100
from table1 a, table2 b, table3 c
where a.key = c.key
a.no = b.no
a.key = c.key
a.code = 10
b.tor = 'THE'
a.group = 4183
a.no in ('1111','1331','1345')

我得到通用的-201'发生语法错误'消息，但我看不出有什么问题。

Answer 1

您的语法错误是table1.code

set table1.code = 100

将此更改为

set a.code = 100

完整代码

update table1
set a.code = 100
from table1 a, table2 b, table3 c
where a.key = c.key
and a.no = b.no
and a.key = c.key
and a.code = 10
and b.tor = 'THE'
and a.group = 4183
and a.no in ('1111','1331','1345')

Answer 2

不幸的是，接受的答案会导致Informix Dynamic Server Version 11.50中出现语法错误。

这是避免语法错误的唯一方法：

update table1
set code = (
  select 100
  from table2 b, table3 c
  where table1.key = c.key
  and table1.no = b.no
  and table1.key = c.key
  and table1.code = 10
  and b.tor = 'THE'
  and table1.group = 4183
  and table1.no in ('1111','1331','1345')
)

BTW，到get Informix version，运行以下SQL：

select first 1 dbinfo("version", "full") from systables;

Answer 3

问题中的原始SQL是：

update table1
set table1.code = 100
from table1 a, table2 b, table3 c
where a.key = c.key
a.no = b.no
a.key = c.key
a.code = 10
b.tor = 'THE'
a.group = 4183
a.no in ('1111','1331','1345')

这无条件地缺少一系列AND关键字。接受的解决方案还使用table1而不是别名a来识别SET子句中的问题。那可能很重要;我无法测试它（见下面的讨论）。因此，假设完全接受了连接UPDATE，更正后的SQL应为：

UPDATE table1
   SET a.code = 100
  FROM table1 a, table2 b, table3 c
 WHERE a.key = c.key
   AND a.no = b.no
   AND a.key = c.key
   AND a.code = 10
   AND b.tor = 'THE'
   AND a.group = 4183
   AND a.no IN ('1111','1331','1345')

这与（语法修正的）接受的答案相同。但是，我很想知道您正在使用哪个版本的Informix接受FROM语法（可能是XPS？）。我在Mac OS X 10.7.4上使用的是IDS 11.70.FC2（当前11.70.FC5版本背后的3个修复包），我无法使用FROM语法获得UPDATE。此外，IBM Informix 11.70 Information Center更新UPDATE的手册没有提到它。如果您使用ODBC或JDBC，我不确定它是否会有任何区别;它不应该，但我正在使用ESQL / C，它将SQL不变地发送到服务器。

我尝试的符号是（+是提示符）：

+ BEGIN;
+ CREATE TABLE a(a INTEGER NOT NULL, x CHAR(10) NOT NULL, y DATE NOT NULL);
+ INSERT INTO a(a, x, y) VALUES(1, 'obsoletely', '2012-04-01');
+ INSERT INTO a(a, x, y) VALUES(2, 'absolutely', '2012-06-01');
+ CREATE TABLE b(b INTEGER NOT NULL, p CHAR(10) NOT NULL, q DATE NOT NULL);
+ INSERT INTO b(b, p, q) VALUES(3, 'daemonic', '2012-07-01');
+ SELECT * FROM a;
1|obsoletely|2012-04-01
2|absolutely|2012-06-01
+ SELECT * FROM b;
3|daemonic|2012-07-01
+ SELECT *
  FROM a, b
  WHERE a.a < b.b
    AND b.p MATCHES '*a*e*';
1|obsoletely|2012-04-01|3|daemonic|2012-07-01
2|absolutely|2012-06-01|3|daemonic|2012-07-01
+ UPDATE a
  SET x = 'crumpet'
  FROM a, b
  WHERE a.a < b.b
    AND b.p MATCHES '*a*e*';
SQL -201: A syntax error has occurred.
SQLSTATE: 42000 at <<temp>>:23
+ SELECT * FROM a;
1|obsoletely|2012-04-01
2|absolutely|2012-06-01
+ ROLLBACK;

Answer 4

对于Informix SE 7.25 ......

UPDATE ... FROM ...语法不存在
您还＆＃34;无法修改子查询中使用的表或视图＆＃34; 使用Rockallite's answer

另一个解决方案是将其分解为两个查询：

首先，获取所需记录的ROWID（在多个表上过滤）：

SELECT a.ROWID
  FROM table1 a, table2 b, table3 c
 WHERE a.key = c.key
   AND a.no = b.no
   AND a.key = c.key
   AND a.code = 10
   AND b.tor = 'THE'
   AND a.group = 4183
   AND a.no IN ('1111','1331','1345')

将结果放入逗号分隔的字符串中。

然后，仅更新在第一个查询中找到ROWID的主表的那些记录：

UPDATE table1 a
   SET a.code = 100
WHERE a.ROWID in ([comma separated ROWIDs found above])

Answer 5

这取决于您使用的版本。如果您使用至少11.50，最佳解决方案将是：

MERGE INTO table1 as t1
USING table2 as t2
   ON t1.ID = t2.ID
WHEN MATCHED THEN UPDATE set (t1.col1, t1.col2) = (t2.col1, t2.col2);

UPDATE - SET - FROM - 在大于11.50的版本中删除了语法。

如果您使用的是早期版本，则可以使用

UPDATE t SET a = t2.a FROM t, t2 WHERE t.b = t2.b;

使用连接更新Informix表

5 个答案: