I have a list that is 3 words to each line.
I know how to read the file.
I need only the first 12 character of the 3 word. This word is not
guaranteed to be 12 character long. This is where I'm stuck.

Is there an equivalent to the old basic 'left$'? Where do I find it?

What do I search Google for?

TIA Dave

Dave Kelly wrote:
> I have a list that is 3 words to each line.
> I know how to read the file.
> I need only the first 12 character of the 3 word. This word is not
> guaranteed to be 12 character long. This is where I'm stuck.
>
> Is there an equivalent to the old basic 'left$'? Where do I find it?
>

cut -c-12

--
Xicheng

On 2006-08-22, Dave Kelly wrote:
> I have a list that is 3 words to each line.
> I know how to read the file.
> I need only the first 12 character of the 3 word. This word is not
> guaranteed to be 12 character long. This is where I'm stuck.
>
> Is there an equivalent to the old basic 'left$'? Where do I find it?

To get the first 12 characters of the third word of every line:

awk '{ printf "%s\n", substr($3,1,12) }'

If you only want lines where the thrid word contains at least 12
letters:

awk 'length($3) >=12 { printf "%s\n", substr($3,1,12) }'

If you want to process it in the shell:

word=qwertyuiopasdfghjklzxcvbnm
mask=????????????
junk=${word#$mask}
printf "%s\n" "${word%"$junk"}"

In bash or ksh93:

printf "%s\n" "${word:0:12}"

For other POSIX shells, there is a substr() function in my book
that can do it. (I may previously have posted it here. All the
scripts from the book are available on line.)

--
Chris F.A. Johnson, author <http://cfaj.freeshell.org>
Shell Scripting Recipes: A Problem-Solution Approach (2005, Apress)
===== My code in this post, if any, assumes the POSIX locale
===== and is released under the GNU General Public Licence

Chris F.A. Johnson wrote:
> On 2006-08-22, Dave Kelly wrote:
>
>>I have a list that is 3 words to each line.
>>I know how to read the file.
>>I need only the first 12 character of the 3 word. This word is not
>>guaranteed to be 12 character long. This is where I'm stuck.
>>
>>Is there an equivalent to the old basic 'left$'? Where do I find it?
>
>
> To get the first 12 characters of the third word of every line:
>
> awk '{ printf "%s\n", substr($3,1,12) }'
>

Actually, that'll print the first 12 characters of the 3rd string of
non-blanks. For the OP - what's a "word"? For example, in this context:

Bob says "Hi!".

is <"Hi!".> a word, or <"Hi!">, or <Hi!> or <Hi> or something else?

Ed

On 2006-08-22, Ed Morton wrote:
> Chris F.A. Johnson wrote:
>> On 2006-08-22, Dave Kelly wrote:
>>
>>>I have a list that is 3 words to each line.
>>>I know how to read the file.
>>>I need only the first 12 character of the 3 word. This word is not
>>>guaranteed to be 12 character long. This is where I'm stuck.
>>>
>>>Is there an equivalent to the old basic 'left$'? Where do I find it?
>>
>>
>> To get the first 12 characters of the third word of every line:
>>
>> awk '{ printf "%s\n", substr($3,1,12) }'
>>
>
> Actually, that'll print the first 12 characters of the 3rd string of
> non-blanks. For the OP - what's a "word"? For example, in this context:
>
> Bob says "Hi!".
>
> is <"Hi!".> a word, or <"Hi!">, or <Hi!> or <Hi> or something else?

Very true, but the OP did say there were 3 words per line. whatever
a word is.

--
Chris F.A. Johnson, author <http://cfaj.freeshell.org>
Shell Scripting Recipes: A Problem-Solution Approach (2005, Apress)
===== My code in this post, if any, assumes the POSIX locale
===== and is released under the GNU General Public Licence

Chris F.A. Johnson wrote:
> On 2006-08-22, Ed Morton wrote:
>
>>Chris F.A. Johnson wrote:
>>
>>>On 2006-08-22, Dave Kelly wrote:
>>>
>>>
>>>>I have a list that is 3 words to each line.
>>>>I know how to read the file.
>>>>I need only the first 12 character of the 3 word. This word is not
>>>>guaranteed to be 12 character long. This is where I'm stuck.
>>>>
>>>>Is there an equivalent to the old basic 'left$'? Where do I find it?
>>>
>>>
>>> To get the first 12 characters of the third word of every line:
>>>
>>>awk '{ printf "%s\n", substr($3,1,12) }'
>>>
>>
>>Actually, that'll print the first 12 characters of the 3rd string of
>>non-blanks. For the OP - what's a "word"? For example, in this context:
>>
>> Bob says "Hi!".
>>
>>is <"Hi!".> a word, or <"Hi!">, or <Hi!> or <Hi> or something else?
>
>
> Very true, but the OP did say there were 3 words per line. whatever
> a word is.
>

Right, but he didn't say if "words" were strictly space-separated (e.g.
is "space-separated" one word or 2 to the OP?).

Ed.

Ed Morton wrote:
> Chris F.A. Johnson wrote:
> > On 2006-08-22, Dave Kelly wrote:
> >
> >>I have a list that is 3 words to each line.
> >>I know how to read the file.
> >>I need only the first 12 character of the 3 word. This word is not
> >>guaranteed to be 12 character long. This is where I'm stuck.
> >>
> >>Is there an equivalent to the old basic 'left$'? Where do I find it?
> >
> >
> > To get the first 12 characters of the third word of every line:
> >
> > awk '{ printf "%s\n", substr($3,1,12) }'
> >
>
> Actually, that'll print the first 12 characters of the 3rd string of
> non-blanks. For the OP - what's a "word"? For example, in this context:
>
> Bob says "Hi!".
>
> is <"Hi!".> a word, or <"Hi!">, or <Hi!> or <Hi> or something else?

No big deal, just trim all non-word with one more pipeline or y|tr
commands.

--
XC

Dave Kelly wrote:
> I have a list that is 3 words to each line.
> I know how to read the file.
> I need only the first 12 character of the 3 word. This word is not
> guaranteed to be 12 character long. This is where I'm stuck.
>
> Is there an equivalent to the old basic 'left$'? Where do I find it?
>
> What do I search Google for?
>
> TIA Dave

I did not provide enough information. My apologies.

I am developing a protected directory on my web server. This directory
will contain a membership list. If you are on the membership list you
can download the membership list.

I want the user name to be the first name and last name with no spaces
in between the 2 words. The password will be the format of the 1st 12
characters of the email address.

The membership is put together with excel, sent to me as a cvs file and
takes the form:

Kelly,Dave,10371 Renfaire
Drive,Plantersville,TX,77363,(936)xxx-1xx0,,,1/30/2001
0:00,daveekelly@xxxxxxxxx.net,R,6/30/2005 0:00,2005-2006,,

code I had started:

#!/bin/bash


main()
{
IFS=, ; while read "LastName" "FirstName" "Address" "City" "State" \
"ZipCode" "HomePh" "BusinessPh" "tmp" "DateJoined" "email" \
"MembershipType" "RenewalDate" "MembershipYear" "Paid" "Notes";

do make_list "$LastName" "$FirstName" "$email" ; done
}
make_list()
{
Email=cut -c12 "$email";
echo >> "$firstname""$lastname","$Email"
}
main

Im currently chasing this error "./pdlist.sh: line 14: -c12: command not
found"