Re: [SLUG] bash shell script question

From: Jan Mason (td376@mail.anonymizer.com)
Date: Sun Sep 15 2002 - 02:22:33 EDT


Using ~/ works the same as using just ~ in both of the test cases.

Norbert Cartagena wrote:
> Jan Mason wrote:
>
>> In test case 1, when a "~" is passed in as a parameter it is
>> substituted for $HOME and the script works.
>>
>> In test case 2, when no parameter is passed and the "~" is assigned to
>> the var $directory using the read command, $HOME is not substituted
>> for "~" and the script does not work.
>>
>> How do I get test case 2 to work?
>>
>> bash shell script named green:
>>
>> directory=$1
>> if [[ $directory = "" ]]
>> then
>> echo -n "prompt for dir: "
>> read directory
>> fi
>> echo "dir = "$directory
>> pwd
>> cd $directory
>> pwd
>>
>> Test case 1 output:
>>
>> $ green ~
>> dir = /home/jan
>> /home/jan/bin
>> /home/jan
>>
>> Test case 2 output:
>>
>> $ green
>> prompt for dir: ~
>> dir = ~
>> /home/jan/bin
>> /home/jan/bin/green: cd: ~: No such file or directory
>> /home/jan/bin
>>
>
> Perhaps I'm wrong, but shouldn't it be ~/ ?
>
> Gnorb
>
>

-- 
Jan Mason
td376@mail.anonymizer.com



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