[英]How to insert many rows into two relative tables in mysql?
我想在两个相对表中插入许多行。 我使用LAST_INSERTED_ID()将数据插入第二个表中,但是此ID不变
BEGIN;
insert into themes (tutorID, year, theme_name, degreeID)
select tutorID, year, work_name, degreeID from journal;
INSERT INTO assigned_students (studentID, tutorID, themeID, writing_language_id, work_typeID)
select studentID, tutorID, LAST_INSERT_ID(), 0, 4 from journal;
COMMIT
尝试一下,在此声明一个INT
,然后设置该值,请参见下文。
BEGIN;
declare lastid INT;
insert into table1 (tutorID, year, name, degreeID)
select sm.tutorID, year, namework, dt.degreeID from table3;
set lastid = (SELECT LAST_INSERT_ID());
INSERT INTO table2 (studentID, tutorID, table1ID, writing_language_id, work_typeID)
select distinct StudentID, tutorID, lastid , 0, 4 from table3;
COMMIT;
您也可以尝试一下。
BEGIN;
insert into table1 (tutorID, year, name, degreeID)
select sm.tutorID, year, namework, dt.degreeID from table3;
INSERT INTO table2 (studentID, tutorID, table1ID, writing_language_id, work_typeID)
select distinct StudentID, tutorID, (SELECT LAST_INSERT_ID()), 0, 4 from table3;
COMMIT;
还可以使插入的assigned_students引用themes
刚插入的行(这样您就可以获得其自动生成的ID):
BEGIN;
insert into themes (tutorID, year, theme_name, degreeID)
select themeID, year, work_name, degreeID from journal;
INSERT INTO assigned_students (studentID, tutorID, themeID, writing_language_id, work_typeID)
select j.studentID, j.tutorID, t.THEME_ID_COLUMN_NAME, 0, 4
from
journal j
inner join themes t on j.themeID = t.tutorid;
COMMIT
在这里,我们看到首先我们在主题中插入了一些内容,我想它会为主题的pk列自动生成一些ID。
因此,我们然后将日记加入到我们插入的那些行中,以便我们可以检索生成的ID
我不知道themes
的PK列叫什么,因此您必须将THEME_ID_COLUMN_NAME
替换为正确的值
请注意,您可能必须on j.themeID = t.tutorid
指定其他列,而不仅仅是tutorid,如果这不能唯一地标识一行
我还读到 ,对于自动增量列,可以确保ID是连续的,因此您可以获得LAST_INSERTED_ID()(即mos中间插入的行)以及ROW_COUNT,因此知道了插入ID的范围,并可以使用选择/加入日记数据的方法
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