exclude lines in a loop

Shell Programming and Scripting
1

I use while do - done loop in my shell script. It is working as per my expectations.
But I do not want to process all the lines. I am finding it difficult to exclude certain lines.

1) I do not want to process blank lines as well as lines those start with a space " "
2) I do not want to process the headings. The headlines start with the word AGENT or PRODUCT or TOTAL

Any help will be appreciated.

2

You can try something like this (and maybe you can do the whole processing with awk):

awk '!/^ / && !/^$/ && !/^AGENT/ && !/^PRODUCT/ && !/^TOTAL/' file | 
while read line
do
  echo "$line"
  ....
done

Regards

3

Hello there,

I think the following KornShell script can do the job

#!/bin/ksh

IFS='
'

while read LINE
do
    
    for TOKEN in $LINE
    do
        UPPER=$(print $TOKEN | tr '[a-z]' '[A-Z]')
        if [[ ($UPPER = +([     ])*) ||
              ($UPPER = *([     ])+(TOTAL|AGENT|PRODUCT)*) ]]
        then
            continue
        else
            #     And here you put all
            #     instructions to process the input
            #     don't forget the following break
            break
        fi
    done
    
done < $1

Actually, because here you ignore also the space character, you will need to modify the default IFS. In fact, the space characters (and I suppose the tabs) must be omitted. The IFS containing only the new line character is defined in the following way

IFS='
'

If you just write something like $'\n' or "\n" it doesn't work. The same is true about tabs. If you want to define a pattern including one or several space/tab characters you have to write [ ], that is, after [ you put a space character and then you push the tab key before closing ] (if instead of doing that you write \t it will not work)

I hope this can help

Regards,
:slight_smile: