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Shell Programming and Scripting

Report printing

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# 1  
Old 01-27-2013
Report printing
Hi. i am trying to print the contents of my text file with delimiter : separating them.

Code:
6:15:ABC ABD:ABB ABE:
2:20:AEE ABH:ACC AAA:

i have completed my header for the report but i am having problem with my contents. i am trying to read in the data with the code below.

Code:
{ while read 'num1:num2:alpha1:alpha2:'; do
print "%-12s%12d%12d%12d\n" "$num1" "$num2 "$alpha1" "$alpha2"
done } < $file

it tells me my
Code:
'num1:num2:alpha1:alpha2:'

not a valid identifier
Last edited by Scrutinizer; 01-27-2013 at 10:49 AM.. Reason: quote tags -> code tags for data sample
# 2  
Old 01-27-2013
Perhaps you need something more like:
Code:
while IFS=: read num1 num2 alpha1 alpha2 junk; do
  printf "%-12s%12d%12d%12d\n" "$num1" "$num2" "$alpha1" "$alpha2"
done < $file

Notes:
  • I missed where "$file" was set, but I presume it is
  • Changed "print" to "printf"
  • Added " after $num2
  • It's also useful to add "junk" after the last "while read" field, in case there are additional fields you don't care about, otherwise they become part of the last field read.
# 3  
Old 01-27-2013
Thanks for your reply. Sorry for the mistake as i typed it out instead of copying. $file was set already in my program. i get an error syntax error for my while line when i tried your code. it says read: 6:15:ABC ABD:ABB ABE: syntax error in expression ( error token is "6")
# 4  
Old 01-27-2013
Please tell me which shell you are using (please don't tell me it's C-Shell), and paste your exact script and input file (if possible!)

edit: seems to be Bash:
Code:
$ strings $(which bash) | grep -i token
%s%s%s: %s (error token is "%s")

edit 2: You need to change the formatting from "%d" to "%s" - these are strings, not numbers.
# 5  
Old 01-27-2013
Slight change to Scott's solution:
Code:
printf "%-12d%12d%12s%12s\n" "$num1" "$num2" "$alpha1" "$alpha2"

This is assuming that your actual data is as per the sample shown.
It's always better to use %s instead of %d when you are unsure about the data.
# 6  
Old 01-27-2013
Hi. its bash that im using. the error is still the same after changing my code. the error seems to happen when i start reading the 1st line.

edit: Hi scott. May i know what do i do with the code that u have provided?
Code:
$ strings $(which bash) | grep -i token
%s%s%s: %s (error token is "%s")

Last edited by One_2_three; 01-27-2013 at 11:21 AM..
# 7  
Old 01-27-2013
We're not going to second guess what you've done. Please post the code.

edit: You need do nothing with that code. It was only to find in which shell the string "error token" could be found.
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