[英]How to Update table column in SQL SERVER DB based on condition
I have a table with the following information: 我有一张包含以下信息的表:
ID Value_N1 Date_N1 Value_N4 Date_N4
1 NULL 2017-05-31 0.236 2017-02-28
2 NULL 2017-05-31 0.589 2017-02-28
3 NULL 2017-08-30 0.898 2017-08-30
4 NULL 2017-11-30 0.789 2017-11-30
I want to update the Value_N1 column with values from Value_N4 where Date_N1 is equal to Date_N4 我想用Value_N4中的值更新Value_N1列,其中Date_N1等于Date_N4
I have tried to use the following query but am going nowhere with it: 我尝试使用以下查询,但是却无济于事:
Update TableName
set Value_N1 = (select Value_N4 from TbleName where Date_N1 = Date_N4)
Of-cause this query doesn't work because it returns more than one value. 由于该查询返回的值不止一个,因此该查询不起作用。 How can this be achieved?
如何做到这一点?
You can also use a CASE
expression. 您也可以使用
CASE
表达式。
Query 询问
update TableName
set Value_N1 = (
case when Date_N1 = Date_N4
then Value_N4
else Value_N1 end
);
更新TableName设置Value_N1 = Value_N4,其中Date_N1 = Date_N4
Your logic for setting the new value to Value_N1
is on the right track, but the restrictions for which records to update should appear in a WHERE
clause. 将新值设置为
Value_N1
逻辑是正确的,但是要更新记录的限制应出现在WHERE
子句中。
UPDATE TableName
SET Value_N1 = Value_N4
WHERE Date_N1 = Date_N4
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