关于字串的问题

[HeavenHe]

begin
--grade and grade_flag
begin
if(cc.grade='Z' and cc.grade_flag='N') then
B:='_Z';
elsif(cc.grade='P' and cc.grade_flag='N') then
B:='_P';
elsif(cc.grade='N' and cc.grade_flag='N') then
B:='_N';
elsif(cc.grade='V' and cc.grade_flag='N') then
B:='_V';
elsif(cc.grade='P' and cc.grade_flag='Y') then
B:='_Z'||'_P';
elsif(cc.grade='N' and cc.grade_flag='Y') then
B:='_Z'||'_P'||'_N';
elsif(cc.grade='V' and cc.grade_flag='Y') then
B:='_Z'||'_P'||'_N'||'_V';
end if;

end;
在这段程序中当cc.grade='V' and cc.grade_flag='Y'时B是不是('_Z','_P','_N','_V')啊？

[HeavenHe]

我们老板是这么解决的
and substr(a.subinventory_code,5,1) in
(select grade from
(
select 'Z' grade,1 seq from dual union
select 'P'grade ,2 seq from dual union
select 'N' grade,3 seq from dual union
select 'V' grade,4 seq from dual
order by seq
) grade_list
where seq <=
(select seq from
(
select 'Z' grade,1 seq from dual union
select 'P'grade ,2 seq from dual union
select 'N' grade,3 seq from dual union
select 'V' grade,4 seq from dual
)
where grade=cc.grade
)
)

[hrb_qiuyb]

B是_Z_P_N_V'

[HeavenHe]

那如何能让他是('_Z','_P','_N','_V')呢？

[bzszp]

B:='_Z'||'_P'||'_N'||'_V';
改成
B:=chr(39)||'_Z'||chr(39)||','||chr(39)||'_P'||chr(39)||','||chr(39)||'_N'||chr(39)||','||chr(39)||'_V'||chr(39);

[HeavenHe]

chr(39)是什么意思啊？

[bzszp]

代表一个'号

[HeavenHe]

哦
可是还是不好用啊。
我用in('_Z','_P','_V','_N')就能查到资料但是用in (B)就查不到，为什么啊？难道是我还有其他的地方写错了吗？

[HeavenHe]

if(cc.grade='Z' and cc.grade_flag='N') then
B:='_Z';

此时，in (B)是什么样的啊？

[bzszp]

这样
STR := 'SELECT * FROM TBNAME WHERE COLNAME IN ('||CHR(39)||B||CHR(39)||')';
EXECUTE IMMEDIATE STR;