第一部分 --36转10进制 create or replace function f_36to10 (str varchar) return int is returnValue int; str36 varchar(36); subWork varchar(1); workIndex int; len int; i int; begin returnValue:= 0; str36 := '123456789ABCDEFGHIJK
import java.util.HashMap; public class Ten2Thirty { private static final String X36 = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"; private static final String[] X36_ARRAY = "0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z
#include "stdio.h" int main() { int num=0;int a[100]; int i=0; int m=0;int yushu; char hex[16]={'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F'};printf("请输入一个十进制整数:"); scanf("%d",&num); while(num>0) { y
查询跟索引有关的数据字典时,可以用下面这条SQL语句: SQL>select * from dictionary where instr(comments,'index')>0; 如果我们想知道user_indexes表各字段名称的详细含义,可以用下面这条SQL语句: SQL>select column_name,comments from dict_columns where table_name='USER_INDEXES'; 依此类推,就可以轻松知道数据字典的详细名称和解释,不用查
0 order by asc/desc 默认升序 order by 列的名字|表达式|别名|序号 把空放在后边:order by desc nulls last 1分组函数--会自动滤空值 count(*|distinct|clumn) max min sum avg select sum(comm)/count(*) 一, sum(comm)/count(comm) 二,avg(comm) 三from emp 2 过滤解决,空值替换函数,NVL(comm,0) 1 select count(*