标签:总结 function 1.4 ade print for overflow 2.7 app
给定一个字符串,为多个单字符串连接而成,单个字符串格式为
以以下详细字符串演示:
‘20151202142400000002(2,0)20151202142200000001(1,4)’
要求:
取出20151202142200000001(1,4)括号内的1和4.
直接上干货
var
i,j,k,m:integer;
str:=‘20151202142400000002(2,0)20151202142200000001(1,4)‘;
mstr:=‘20151202142200000001‘;
for i:=Pos(mstr,str)+length(mstr) to length(str) do
begin
j:=0;
k:=0;
m:=0;
if str[i]=‘(‘ then
begin
j=i;
end
else if str[i]=‘,‘ then
begin
k=i;
end
else if str[i]=‘)‘ then
begin
m=i;
end;
if (right_bracket>comma) and (comma>left_bracket) then
break;
end;
pro_num:=midstr(str,j+1,k-j-1);
pro_price:=midstr(str,k+1,m-k-1);
以上为个人总结。如有不妥之处,请在评论中指出。
标签:总结 function 1.4 ade print for overflow 2.7 app
原文地址:http://www.cnblogs.com/liguangsunls/p/7286076.html