Please explain this batch file of yours, zw19750516 - China DOS Union Forum

Sheesh, why does everyone want an explanation.
What comes before :lp is using suitable delimiters and a for+findstr statement to grab the wanted strings from 1.txt, so I won't explain that in detail.
I believe newbies may have a bit of trouble understanding the shift statement after the :lp label. Actually, if you just type shift /? in cmd, you can get a basic idea that what shift does is shift the parameters after call. And here shift is doing the same thing: first, four parameters are passed over from the for statement, namely %%i %%j %%k %%l, that is, %1 %2 %3 %4. Each time lp loops once, shift moves the parameters one place back. The first time, the parameter is %1=%%i, but the second time %1 moves to the second position, so now %1=%%j, and this continues until %1 has been shifted to the position of the last parameter.

[ Last edited by zw19750516 on 2008-6-20 at 06:34 PM ]

Floor1　hackerscans	Posted 2008-06-20 18:10
初级用户　Posts 51　Credits 73
@echo off for /f "tokens=3,5,7,8 delims==/> " %%i in ('findstr "FACTORY DUT" 1.txt^\|findstr /v "FIXTURE"') do call :lp %%i %%j %%k %%l pause>nul&goto :eof :lp set str=%1 if defined str echo %str:~1,-1% if defined str shift&goto lp Thanks, and experts are also welcome to chime in.

Floor2　bat-zw	Posted 2008-06-20 18:31
金牌会员　Posts 1,276　Credits 3,105
Sheesh, why does everyone want an explanation. What comes before :lp is using suitable delimiters and a for+findstr statement to grab the wanted strings from 1.txt, so I won't explain that in detail. I believe newbies may have a bit of trouble understanding the shift statement after the :lp label. Actually, if you just type shift /? in cmd, you can get a basic idea that what shift does is shift the parameters after call. And here shift is doing the same thing: first, four parameters are passed over from the for statement, namely %%i %%j %%k %%l, that is, %1 %2 %3 %4. Each time lp loops once, shift moves the parameters one place back. The first time, the parameter is %1=%%i, but the second time %1 moves to the second position, so now %1=%%j, and this continues until %1 has been shifted to the position of the last parameter. [ Last edited by zw19750516 on 2008-6-20 at 06:34 PM ]