[SOLVED] Sed using a pattern variable at the start of the line doesn't work

zbyesk · 10-09-2014, 12:10 PM

I'm trying to make an script that replaces lines that starts with a pattern given by the lines of other file.

Example lines of the base file:

Code:

& NAME 1
UH   24   1        26.03               1
& NAME 2
UH   27   1        50.05               1

Example lines of the input file:

Code:

UH   24   1        27.68               1
UH   27   1        37.21               1

I just want the script to replace the values (i.ex: 26.03 for 27.68).

I've made the script below but the sed doesn't work.

Code:

while IFS= read -r line
do
    pattern=${line:0:15}
    sed "s/$pattern.*/$line/" BASE.TXT > OUTPUT.TXT
done < INPUT.TXT

Thanks in advance!

Keith Hedger · 10-09-2014, 01:46 PM

This should do it

Code:

#!/bin/bash
echo "Base file:"
cat BASE.TXT
cp BASE.TXT BASE.TXT.HOLD
while read
	do
		pattern="${REPLY:0:15}"
		sed -i "s/${pattern}.*/$REPLY/" BASE.TXT.HOLD
	done< <(cat INPUT.TXT)
echo
echo
echo "Converted base file:"
cat BASE.TXT.HOLD
echo

You don't need to set IFS, read without a value assigns to the variable REPLY, INPUT.TXT should end in a new line otherwise the last line won't get read in.

After the conversion has been done you can then copy BASE.TXT.HOLD to BASE.TXT if you need to.