Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
526 views
in Technique[技术] by (71.8m points)

mysql - How to select from dynamic table name

I have tables like

changes201101
changes201102
changes201103
...
changes201201

And table whichchanges which contain rows Year and MONTH

How I can select * from changes from whichchanges?

I type this query

SET @b := SELECT CONCAT('changes',year,month) FROM whichchanges;
((((@b should contain now multiple rows of changesYearMonth)))))
SET @x := SELECT * FROM @b;
Prepare stmt FROM @b;
Prepare stmt FROM @x;
Execute stmt;

#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'SELECT CONCAT('changes',year,month) FROM changes)' at line 1

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

You open 1 ( and close 2 ). Remove the last:

SELECT CONCAT('changes',year,month) FROM changes

Edit

the second statement should probably be

SET @x := SELECT * FROM (@b) as b;

That works, but not sure if that is what you want:

SET @b := 'SELECT CONCAT(''changes'',`year`,`month`) FROM whichchanges';
SET @x := 'SELECT * FROM (SELECT CONCAT(''changes'',`year`,`month`) FROM whichchanges) as b';
Prepare stmt FROM @b;
Prepare stmt FROM @x;
Execute stmt;

Edit2

If I understood you right you are looking for that single query:

select * from changes
where change_column in (select distinct concat(`year`, `month`) from whichchanges)

Edit3

select @b := group_concat(concat(' select * from changes', `year`, `month`, ' union ') separator ' ') as w from whichchanges;
set @b := left(@b, length(@b) - 6);

Prepare stmt FROM @b;
Execute stmt;

SQLFiddle example


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...