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Re: [PROGRAMMERING] En 'for' lykke i konsolen



Kim Schulz wrote:

> On Thu, 25 May 2006 13:47:49 +0200
> Kim Schulz <sslug@sslug> wrote:
> 
>> On Thu, 25 May 2006 13:37:59 +0200
>> Jan Trøjgaard <sslug@sslug> wrote:
>> 
>> > Jeg var til foredrag den 18 og det var ganske fortræffeligt. Der
>> > blev også vist en 'for' lykke, hvor beep var det program der blev
>> > kaldt, men nu sidder jeg herhjemme og kan ikke få lykken til at
>> > virke, jeg spurgte også lige ude på CBC i går, onsdag, o gfik lidt
>> > svar, uden at jeg har haft held med mig her hjemme.
>> > 
>> > Lykken(erne) jeg skriver ser sådan ud:
>> > sslug@sslug:~$ for freq in $((seq 200 20 2000)): beep -f $freq: done
>> 
>> erstat : (kolon) med ; (semi kolon) og smid do efter det først ;
>> 
>> altså
>> 
>> for freq in ...; do beep -f $freq; done
>> 
>> jeg er dog ikke sikker på din seq ting. den virker umiddelbart ikke i
>> bash her.
> 
> men det virker sådan her:
> for freq in `seq 200 20 200`; do beep -f $freq; done

Den giver kune ét beep på frekvensen 200 hertz! Men den siger noget.
Jeg har også lige prøvet med et par andre variationer:
sslug@sslug:~$ for freq in $((`seq 200 20 200`)); do beep -f $freq; done
sslug@sslug:~$ for freq in $((seq 200 20 200)); do beep -f $freq; done
bash: seq 200 20 200: syntax error in expression (error token is "200 20
200")
sslug@sslug:~$ for freq in $((`seq 200 20 200`)); do beep -f $freq; done
sslug@sslug:~$ for freq i in $((`seq 200 20 200`)); do beep -f $freq; done
bash: syntax error near unexpected token `i'
sslug@sslug:~$ for i freq in $((`seq 200 20 200`)); do beep -f $freq; done
bash: syntax error near unexpected token `freq'

Men den vil stadig ikke udføre sekvensen i løkken!

Med venlig hilsen
Jan Trøjgaard





 
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