sql字串中截取字元
--SQL Server(SUBSTRING)
select SUBSTRING('0123456789',0,5) ;--0123
select SUBSTRING('0123456789',1,5) ;--01234
select SUBSTRING('0123456789',2,5) ;--12345
select SUBSTRING(m.xTIME,1,6),count(*) from table m
where m.xTIME > '202210'
group by SUBSTRING(m.xTIME,1,6) ;
--Oracle SQL (substr)
select substr('0123456789',0,5) from dual;--01234
select substr('0123456789',1,5) from dual;--01234
select substr('0123456789',2,5) from dual;--12345
select substr('0123456789',-9,1) from dual;--1
select substr('0123456789',-9,2) from dual;--12
select substr(g.xdate,1,6),count(*) from table g
where g.xdate>'202206'
group by substr(g.xdate,1,6);