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Posted by 皮皮熊 on 2008-06-09 05:00 下午
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各位达达
我想将tableA中的field1的值由'x'变成''，该如何做呢？
是否采用
update tableA set fileld1 = '' where ...
这样 还是？？
谢谢
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Posted by im0o海星 on 2008-06-10 02:32 下午
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请说具体 ...
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Posted by Apple1 on 2008-06-11 10:19 上午
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data: wa1 like tablea.
select * from tablea into wa1.
wa1-field1 = ''.
update tablea from wa1.
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Posted by 皮皮熊 on 2008-06-12 11:39 上午
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谢谢
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Posted by 皮皮熊 on 2008-06-12 11:40 上午
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不用循环wa1吗？
data: wa1 like tablea.
select * from tablea into wa1.
wa1-field1 = ''.
update tablea from wa1.
还是
data: wa1 like tablea.
select * from tablea into wa1.
loop at wa1.
wa1-field1 = ''.
modify wa1.
endloop.
update tablea from wa1.
上面两种哪种正确啊？
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Posted by im0o海星 on 2008-06-12 01:02 下午
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 皮皮熊 wrote: | 不用循环wa1吗？
data: wa1 like tablea.
select * from tablea into wa1.
wa1-field1 = ''.
update tablea from wa1.
还是
data: wa1 like tablea.
select * from tablea into wa1.
loop at wa1.
wa1-field1 = ''.
modify wa1.
endloop.
update tablea from wa1.
上面两种哪种正确啊？ |
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2种都有错
第1种: 你声明的是一个 wa1的工作区域 不是内表
所以不能 select all to the 工作区域 wa1.
data: wa1 like tablea.
select single * from tablea into wa1.
wa1-field1 = ''.
update tablea from wa1.
第2种: data: wa1 like table of tablea with header line..
select * from tablea into wa1.
loop at wa1.
wa1-field1 = ''.
modify wa1.
endloop.
update tablea from wa1.
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Posted by 浪客 on 2008-06-12 02:36 下午
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我用实际给你做了一下。目的是为了提高效率。代码如下：
DATA :
IT_WA TYPE STANDARD TABLE OF TABLEA,
STR_WA TYPE TABLEA。
SELECT * FROM TABLE INTO TABLE IT_WA .
STR_WA-field1 = ' '.
MODIFY IT_WA FROM STR_WA TRANSPORTING field1 WHERE field1 = 'x'.
关键是where条件必须追加。否则你就只能用LOOP AT 。。。。。。ENDLOOP了。
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Posted by thinklong on 2008-06-13 10:49 下午
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楼上的正解
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Posted by xiebinren on 2008-07-04 03:09 下午
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MODIFY 好像是里面没有记录，他还会添加上去吧
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Posted by im0o海星 on 2008-07-05 09:02 上午
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modify = insert + update
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