I have an old_table and a new_table:
CREATE TABLE `old_table` ( `id` bigint(20) NOT NULL AUTO_INCREMENT, `col1` varchar(15) DEFAULT NULL, `col2` int(15) DEFAULT NULL, ..., PREMARY_KEY (`id`) ); CREATE TABLE `new_table` LIKE `old_table`;
Then both tables are populated with some values. After that, I want to select some rows from old_table and insert into new_table:
INSERT INTO `old_table` SELECT * FROM `new_table` WHERE col2 > 100;
But this will cause errors due to duplicate key. I'm too lazy to specify the columns in the SELECT clause, because in the real system the tables have a lot of columns.
What would be the best way to get around the problem?
set @sql = (select concat('insert into new_table SELECT NULL,',
group_concat(column_name),' from ','old_table') from information_schema.columns
where table_name = 'old_table' and table_schema = '<database>' and column_name != 'id'
order by ordinal_position);
prepare stmt1 from @sql;
execute stmt1;
deallocate prepare stmt1;
where
<database> - your database
You could probably create a before-insert trigger that tests the existence of the id
value and replaces it with the correct new value (or null it out if that works - I don't know if MySQL would barf or give you the next auto-incremented value) if it's a duplicate.
Personally, I don't think I would want this kind of trigger to persist, but if you're just doing a one-time insert you could create the trigger, do the insert, then drop the trigger right after.
我遇到了同样的问题并解决了这个问题: - 修改new_table的id列(删除自动增量和主键,允许NULL) - 将new_table中的所有ID设置为NULL - 现在你可以做
INSERT INTO 'old_table' SELECT * FROM new_table
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