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

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

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!

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.