Oracle插入之 insert all、insert first介绍
利用insert first/all使得INSERT语句可以同时插入多张表,还可以根据判断条件来决定每条记录插入到哪张或哪几张表中。
insert first:对于每一行数据,只插入到第一个when条件成立的表,不继续检查其他条件。
insert all:对于每一行数据,对每一个when条件都进行检查,如果满足条件就执行插入操作。
create table edw_int ( agmt_no varchar2(40 byte) not null, agmt_sub_no varchar2(4 byte) not null, need_repay_int number(22,2), curr_period number(4) not null ); create table edw_int_1 ( agmt_no varchar2(40 byte) not null, agmt_sub_no varchar2(4 byte) not null, need_repay_int number(22,2), curr_period number(4) not null ); create table edw_int_2 ( agmt_no varchar2(40 byte) not null, agmt_sub_no varchar2(4 byte) not null, need_repay_int number(22,2), curr_period number(4) not null ); -- 插入数据 insert into edw_int (agmt_no, agmt_sub_no, need_repay_int, curr_period) values ('20003874', '2104', 3126.5, 7); insert into edw_int (agmt_no, agmt_sub_no, need_repay_int, curr_period) values ('20003874', '2104', 3290.76, 6); insert into edw_int (agmt_no, agmt_sub_no, need_repay_int, curr_period) values ('20003874', '2104', 3454.06, 5); insert into edw_int (agmt_no, agmt_sub_no, need_repay_int, curr_period) values ('20003874', '2104', 3616.41, 4); insert into edw_int (agmt_no, agmt_sub_no, need_repay_int, curr_period) values ('20017143', '2104', 2350.86, 0); insert into edw_int (agmt_no, agmt_sub_no, need_repay_int, curr_period) values ('20017143', '2104', 3566.55, 0); insert into edw_int (agmt_no, agmt_sub_no, need_repay_int, curr_period) values ('20018273', '2104', 1639.46, 0); insert into edw_int (agmt_no, agmt_sub_no, need_repay_int, curr_period) values ('20018273', '2104', 2080.49, 0); COMMIT;
insert all示例
insert all into edw_int_1 (agmt_no, agmt_sub_no, need_repay_int, curr_period) values (agmt_no, agmt_sub_no, need_repay_int, curr_period) into edw_int_2 (agmt_no, agmt_sub_no, curr_period) values (agmt_no, '1234', curr_period) select agmt_no, agmt_sub_no, need_repay_int, curr_period from edw_int; commit;
删除完数据继续测试 加上条件when then else
truncate table edw_int_1;
truncate table edw_int_2;
insert all when curr_period = 0 then into edw_int_1 (agmt_no, agmt_sub_no, need_repay_int, curr_period) values (agmt_no, agmt_sub_no, need_repay_int, curr_period) else into edw_int_2 (agmt_no, agmt_sub_no, need_repay_int, curr_period) values (agmt_no, agmt_sub_no, need_repay_int, curr_period) select agmt_no, agmt_sub_no, need_repay_int, curr_period from edw_int; commit;
删除数据
测试insert first
insert first when curr_period = 0 then into edw_int_1 (agmt_no, agmt_sub_no, need_repay_int, curr_period) values (agmt_no, agmt_sub_no, need_repay_int, curr_period) when agmt_sub_no = '2104' then into edw_int_2 (agmt_no, agmt_sub_no, need_repay_int, curr_period) values (agmt_no, agmt_sub_no, need_repay_int, curr_period) select agmt_no, agmt_sub_no, need_repay_int, curr_period from edw_int; commit;