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06.20.2008 at 09:24AM PDT, ID: 23502810
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Sed in a while read loop wont replace pattern stored in a variable

Asked by Jelcin in Bourne-Again Shell (bash)

Tags: , , ,

I am using sed to match patterns in csv File. The csv file is as follows:

 "VALUE";"VALUE";"VALUE";"VALUE";.......

to replace a value i use:

sed '2,4621s/[^;]*;/"'"$VALUE"'";/11' < file.csv

this works pretty fine if there is only one value given lets say:

VALUE=TEST

sed '2,4621s/[^;]*;/"'"$VALUE"'";/11' < file.csv

this replaces the 11th VALUE in every line from 2 to 4621 with TEST

but now i have a file2  where other values are stored. It's stored in this format

VALUE
VALUE
VALUE
VALUE


Now i want to read the values from the file above and replace it in the csv file

i tried this:

while read VALUE
do


        V=`sed '2,4621s/[^;]*;/"'"$VALUE"'";/11' < file.csv`

        echo $V > new_file.csv


done<file2<<EOF

but somehow this outputs only empty space on the place where the values of file2 should be inserted


 "VALUE";"VALUE";"VALUE";"EMPTY SPACE";"VALUE";.......

I tried many things and played around but without luck

i am using ubuntu hardy heron....

thx alot


 

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[+][-]06.20.2008 at 09:37AM PDT, ID: 21832811

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[+][-]06.20.2008 at 10:31PM PDT, ID: 21836467

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[+][-]06.23.2008 at 06:43PM PDT, ID: 21851946

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About this solution

Zone: Bourne-Again Shell (bash)
Tags: Open Source, Linux, latest, bash scripting
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Solution Provided By: Jelcin
Participating Experts: 2
Solution Grade: A
 
 
 
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