Read and Replace

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# 1  
Old 06-29-2011
Read and Replace

I think my last post was a little confusing, so I will try an post a sample this time.
I have a script which is supposed to output two columns of a file to a different file, then take the first column, search a directory for it, and replace anything it finds with the second column. This is what I have:

for file in /XXXX/XXXX/XXXX/I
     echo $VAR|awk '{printf "%d\n",$1}'|read NEWVAR

awk < XXXX_table.txt '{print $1 $3}' >> var1

for line in /XXXX/XXXX/XXXX/var1
    x=$(awk < var1 '{print $1}') 
    y=$(awk < var1 '{print $2}')
    sed "s/$x/$y/g"
    echo $x
    echo $y

The first loop is supposed to remove all of the leading zeroes in all the files in that directory as well, but I guess that is wrong too.

Having lots of difficulty on this one, but I feel as though the answer is very simple

Last edited by fpmurphy; 06-30-2011 at 10:01 AM.. Reason: code tags please!
# 2  
Old 06-30-2011
Perhaps you would be good enough to explain to us what difficulty you are having?
# 3  
Old 06-30-2011
Actually now it is just the command to remove leading zeroes from all filenames in a directory.

They look like this:

And I want to remove just the leading zeroes on the second string if they exist. I have been trying to use this with no success.
echo $VAR|awk '{printf "%d\n",$1}'|read NEWVAR

Last edited by Franklin52; 06-30-2011 at 12:09 PM.. Reason: Please use code tags, thank you
# 4  
Old 06-30-2011

Fire this below command in the directory in which you want to rename:

$ for i in *
mv $i `echo $i | sed 's/\.0*/./'`

# 5  
Old 06-30-2011
Excellent! that is perfect.
And one last thing! I am trying to replace one variable (string) with antoher variable using:
sed "s/$x/$y/g"

however, it seems as though this is incorrect, as it makes a new directory.

Last edited by Franklin52; 06-30-2011 at 12:10 PM.. Reason: Please use code tags, thank you
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