Shell Programming and Scripting

the original file had "C 56" as the line but to test I also tried the other possibilities
"C 56" , "C 6", "C156"
Each iteration gave a following space in the outputted file and I just quoted the output from one of the other tests

It is only the first line I need to parse. The second and subsequent lines will be dealt with in another part of the script. I probably should not have included them as it unnecessarily confused the requirements.

The part of the input file I need to parse is

Code:

"C 56","Home athletics center","LAC"

The output I require is to be

Code:

D;56;Home athletics center;;M;;

As i mentioned, the code given in the first reply works, but adds a space after the numbers

Ken

So, with you new statement of what you want, try:

Code:

#!/bin/bash
#set initial variables
file="FILE.EXT"     # input filename

IFS='"' read -r x cn x centnam x < "$file"
cn=${cn##*[!1-9]} # strip off leading non-digits and zeros.
centnum=${cn:-0} # Add a zero if the input had no non-zero digits.
printf "D;%s;%s;;M;;\n" "$centnum" "$centnam" > MM_file.txt

PS Note that although your current input has 1, 2, or 3 digit numbers; the code above will work with numbers that are 1 or more digits long (as long as your total input line length is less than LINE_MAX bytes long for whatever value LINE_MAX has on your system).

Excellent Thanks that works.

Now all I need to do is review it in detail to understand how it works and hopefully remember it, but that is a task for me.

Re reading back thru the thread I realised the first reply also has a working Output if the Output line enclosed the Variable in a another set of brackets.

Code:

echo "D;$(($centnum));$centnam;;M;;" > MM_file.txt

I assume you knwo why, but can you explain why that format removes the leading space?

Thanks again
Ken

It's converting the variable into a numeric value, then operates on it. BTW, you don't need the "$" then:

Code:

echo ">"$((centnum))"<"
>56<

Cool Thank you all for your help.
I am quite amazed that so many replies on a sunday

Ken

The arithmetic expansion of $(($x)) when $x expands to the string " 56" is 56.
Similarly, the parameter expansion of ${x##*[!1-9]} (which removes everything in the string before the 1st non-zero digit) when $x expands to the string "c 56" is also 56.