parsing a string

parsing a string

am 10.08.2004 02:08:01 von Dave Pomeory

I'm new to bash and am trying to write a script that puts 3 substrings in 3
variables. I have a string with password,firstname,lastname. and I need it to
be in 3 variables. I bought a bash scripting book but I can't seem to make it
work. Thanks in advance.
Dave


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Re: parsing a string

am 10.08.2004 04:12:26 von John Aspinall

dave@dpomeroy.com wrote:
> I'm new to bash and am trying to write a script that puts 3 substrings in 3
> variables. I have a string with password,firstname,lastname. and I need it to
> be in 3 variables.

I'm just an apprentice here, but I think "cut" is a good choice of tool.

$ echo 'password,firstname,lastname' | cut -d',' -f1
password
$ echo 'password,firstname,lastname' | cut -d',' -f2
firstname
$ echo 'password,firstname,lastname' | cut -d',' -f3
lastname

-d is the field delimiter, -f says which fields you want

John

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Re: parsing a string

am 10.08.2004 10:48:15 von Jeff Woods

At 8/9/2004 10:12 PM -0400, John Aspinall wrote:
>dave@dpomeroy.com wrote:
>>I'm new to bash and am trying to write a script that puts 3 substrings in 3
>>variables. I have a string with password,firstname,lastname. and I need
>>it to
>>be in 3 variables.
>
>I'm just an apprentice here, but I think "cut" is a good choice of tool.
>
>$ echo 'password,firstname,lastname' | cut -d',' -f1
>password
>$ echo 'password,firstname,lastname' | cut -d',' -f2
>firstname
>$ echo 'password,firstname,lastname' | cut -d',' -f3
>lastname
>
>-d is the field delimiter, -f says which fields you want

I agree that "cut" is a wonderful tool on the command line. However, a
more purely bash solution is to use a powerful feature of variable
expansion in bash. For example:

pass_first_last='password,firstname,lastname'
pass=${pass_first_last%%,*}
first_last=${pass_first_last#*,}
first=${first_last%,*}
last=${first_last#*,}
or, alternatively,
last=${pass_first_last##*,}

The "#" operator in variable expansion takes a pattern argument and removes
as little as possible from the left side of the variable value that matches
the pattern.
The "##" operator in variable expansion takes a pattern argument and
removes as much as possible from the left side of the variable value that
matches the pattern.
The "%" operator in variable expansion takes a pattern argument and removes
as little as possible from the right side of the variable value that
matches the pattern.
The "%%" operator in variable expansion takes a pattern argument and
removes as much as possible from the right side of the variable value that
matches the pattern.

A common use for this is to extract the final filename from a pathname:

filename=${pathname##*/}

Another is to extract the directory path name from a file's path:

dirname=${pathname%/*}

P.S. I was trying to use cut to extract one field from a file using FS
(\0x1C) as the field delimiter. I discovered it was non-trivial to get the
FS into my cut command. The solution turned out to be a "string expansion"
in bash:

cut -d$'\x1C' -f2 filename

--
Jeff Woods


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