I run following command : use db1; insert into db2.c select a.a, a.b, a,c, b.d, b,e ... from a left join b on (a.id=b.id);
size of table a and table b is: [EMAIL PROTECTED] ls -l ../db1/a.* -rw-rw---- 1 mysql mysql 9230 May 10 15:41 ../db1/a.frm -rw-rw---- 1 mysql mysql 880880528 Jul 17 01:15 ../db1/a.MYD -rw-rw---- 1 mysql mysql 383653888 Jul 17 01:15 ../db1/a.MYI [EMAIL PROTECTED] ls -l ../db1/b.* -rw-rw---- 1 mysql mysql 11277 May 10 15:47 ../db1/b.frm -rw-rw---- 1 mysql mysql 1494998192 Jul 17 01:15 ../db1/b.MYD -rw-rw---- 1 mysql mysql 619606016 Jul 17 01:15 ../db1/b.MYI It has took 16776 seconds , and now the command is still running. the size of table c is [EMAIL PROTECTED] ls -l c.* -rw-rw---- 1 mysql mysql 11623 Jul 27 05:37 c.frm -rw-rw---- 1 mysql mysql 4633395200 Jul 27 10:08 c.MYD -rw-rw---- 1 mysql mysql 165481472 Jul 27 10:08 c.MYI the number of records in a is : 3101692 my question is : why is size of c big? and increase of size of MYI is so slow than MYD? Will sql 'insert into .. select ' write temp data into table c ? It is too slow. Has anyone better methods to do that work? thanks. -- -- -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe: http://lists.mysql.com/[EMAIL PROTECTED]
