你对该字符串一个个的进行判断,遇到一个','是就认为得到一个数字就行了。
#include <stdio.h>void main(void)
{
char *oristr = "2,5,6,8,10,13,14,16,17,18";
char digit[20];
char *psrc, *pdst;
int num; psrc = oristr;
pdst = digit;
num = 0;
while(*psrc)
{
if(*psrc == ',') //遇到',',则输出得到的数字,然后重新进行
{
psrc++;
num++;
*pdst = '\0';
printf("str%d:=%s\n", num, digit);
pdst = digit;
}
*pdst++ = *psrc++;
}
num++; //输出最后一个
*pdst = '\0';
printf("str%d:=%s\n", num, digit);
}
#include <stdio.h>void main(void)
{
char *oristr = "2,5,6,8,10,13,14,16,17,18";
char digit[20];
char *psrc, *pdst;
int num; psrc = oristr;
pdst = digit;
num = 0;
while(*psrc)
{
if(*psrc == ',') //遇到',',则输出得到的数字,然后重新进行
{
psrc++;
num++;
*pdst = '\0';
printf("str%d:=%s\n", num, digit);
pdst = digit;
}
*pdst++ = *psrc++;
}
num++; //输出最后一个
*pdst = '\0';
printf("str%d:=%s\n", num, digit);
}
create or replace type mytabletype as table of number;
/create or replace function strtab(p_str in varchar2)
return mytabletype
as
lstr varchar2(1000) default p_str||',';
ln number;
ldata mytabletype:=mytabletype();
begin
loop
ln:=instr(lstr,',');
exit when (nvl(ln,0)=0);
ldata.extend;
ldata(ldata.count):=ltrim(rtrim(substr(lstr,1,ln-1)));
lstr:=substr(lstr,ln+1);
end loop;
return ldata;
end;
/SQL> select * from table(cast(strtab('11,12,13') as mytabletype));COLUMN_VALUE
------------
11
12
13SQL> create table bb(id varchar2(2),name varchar2(10));Table createdSQL> insert into bb values('11','张三');1 row insertedSQL> insert into bb values('12','李四');1 row insertedSQL> insert into bb values('13','王五');1 row insertedSQL> select * from bb where id in (select * from table(cast(strtab('11,12,13') as mytabletype)));ID NAME
-- ----------
11 张三
12 李四
13 王五