脚本优化

显示全部楼层 · 2008-1-2 17:35:53

以下脚本是对不同的 v_bmonth进行相应的计算,请问,对以下脚本在逻辑上(或类型转换)上如何优化呀
if (to_number(substr(v_bmonth, 1, 4)) = 2008) then
if (to_number(substr(v_bmonth, 5, 6)) > 2) then
if (v_flag = '1') then

v_KR := v_temp + 2000;
else

v_templ := round((v_temp - v_datedef - 2000), 2);

if (v_templ > 0) then

v_KR := 0;

else

v_KR := abs(v_templ);

end if;

end if;
else
if (v_flag = '1') then

v_KR := v_datedef * 1600;
else

v_templ := round((v_temp - v_datedef - 1600), 2);

if (v_templ > 0) then

v_KR := 0;

else

v_KR := abs(v_templ);

end if;
end if;
end if;
elsif (to_number(substr(v_bmonth, 1, 4)) > 2008) then
if (v_flag = '1') then
v_KR := v_datedef * 2000;
else
v_templ := round((v_temp - v_datedef - 2000), 2);
if (v_templ > 0) then

v_KR := 0;
else

v_KR := abs(v_templ);
end if;

end if;
else
if (v_flag = '1') then
v_KR := v_datedef * 1600;
else

v_templ := round((v_temp - v_datedef - 1600), 2);
if (v_templ > 0) then

v_KR := 0;
else

v_KR := abs(v_templ);
end if;
end if;
end if;

千问 · 2008-1-2 17:35:53

光从语法的运用上是不是很合理?有优化的空间吗

千问 · 2008-1-2 17:35:53

逻辑没细看，但是我觉着这里写个function处理看起来不就更简单了~~~

千问 · 2008-1-2 17:35:53

类似to_number(substr(v_bmonth, 1, 4)) = 2008
to_number(substr(v_bmonth, 1, 4)) > 2008
等语句能不能优化

千问 · 2008-1-2 17:35:53

本人是写成了function,上面的语句是function体中的实现,感觉写的不好所以想请教

千问 · 2008-1-2 17:35:53

请大家帮忙

千问 · 2008-1-2 17:35:53

要实现什么功能？

千问 · 2008-1-2 17:35:53

原帖由 sxlcom 于 2008-8-12 16:12 发表
类似to_number(substr(v_bmonth, 1, 4)) = 2008
to_number(substr(v_bmonth, 1, 4)) > 2008
等语句能不能优化

单独的好像不能优化了。
不过像
if (to_number(substr(v_bmonth, 1, 4)) = 2008) then
if (to_number(substr(v_bmonth, 5, 6)) > 2) then
为什么不写成一个if语句呢？

千问 · 2008-1-2 17:35:53

if (v_flag = '1') then
if (to_number(substr(v_bmonth, 1, 4)) > 2008) then
v_KR := v_datedef * 2000;
elseif (to_number(substr(v_bmonth, 1, 4)) = 2008)

and(to_number(substr(v_bmonth, 5, 6)) > 2) then

v_KR := v_temp + 2000;

else

v_KR := v_datedef * 1600;
end if;
else
if ((to_number(substr(v_bmonth, 1, 4)) = 2008) and
(to_number(substr(v_bmonth, 5, 6)) > 2)) or
(to_number(substr(v_bmonth, 1, 4)) > 2008) then
v_templ := round((v_temp - v_datedef - 2000), 2);
if (v_templ > 0) then

v_KR := 0;
else

v_KR := abs(v_templ);
end if;
else
v_templ := round((v_temp - v_datedef - 1600), 2);
if (v_templ > 0) then

v_KR := 0;
else

v_KR := abs(v_templ);
end if;
end if;

end if;

千问 · 2008-1-2 17:35:53

通过v_KR(标识)和v_bmonth(年月)来进行相应的计算