
把論壇的批處理精華下載按照自己的思路寫callpl的位置問題最后由 wokong 于 -1-4 19:46這個是論壇的原題: 要求通過批處理輸出10行每行30個由特殊字符隨機組成的字符串(見舉例)這個是標準答案
這個是正確結果<^=|>*)(~$=!$==###^!!&#*^~')^:**)>^&<%:^>^'|(^~!)&&|&%)'(^)$!|~'!%||##(%*)(!>)%:&^!'@~$>$*@~:'=='(=)|(^=|&%|%)'&*'&&&>@^~*^*||>$~|<!'<%=&(|~**'^@^|))#@%$|^%)|#&**%<~)|!(%:#%>>#$&<&~!)@~~~=>!'*|*'&<<$$!%**:$*'=!$&|)#(*|~#=@@)<=!%'(~):#<&%'*!^>&~:^@^^*>!~%$%$==$>|%$~@~&#*下面這個是我的代碼 僅僅換了下位置,為啥就截然不同了,我的2種寫法結果都不對。《1》nclick="copycode($('code0'));">復制代碼
- 代碼:
- :?^~!#$%^&*()='><|/@
- @echo off
- set/p str=<%~fs0
- setlocal enabledelayedexpansion
- for /l %%a in (1 1 10) do (
- for /l %%i in (1 1 30) do (
- set/a n=!random!%%22
- call :lp
- )
- echo !var!&set var=
- )
- pause>nul&exit/b
- :lp
- set var=!var!!str:~%n%,1!
《2》nclick="copycode($('code1'));">復制代碼
- :?^~!#$%^&*()='><|/@
- @echo off
- set/p str=<%~fs0
- setlocal enabledelayedexpansion
- for /l %%a in (1 1 10) do (
- for /l %%i in (1 1 30) do (
- set/a n=!random!%%22
- set var=!var!!str:~!n!,1!
- )
- echo !var!&set var=
- )
- pause>nul&exit/b
- :lp
nclick="copycode($('code2'));">復制代碼
- :?^~!#$%^&*()='><|/@
- @echo off
- set/p str=<%~fs0
- setlocal enabledelayedexpansion
- for /l %%a in (1 1 10) do (
- for /l %%i in (1 1 30) do (
- set/a n=!random!%%22
- set var=!var!!str:~%n%,1!
- )
- echo !var!&set var=
- )
- pause>nul&exit/b
- :lp

