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Substitute File name

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# 8  
Old 03-13-2007
Quote:
Originally Posted by Deal_NoDeal
When you use:
zcat $var2* | grep "^000$var1" >> my_output

It is actually translated to something like this:

zcat 01* | grep ....

So, there is the issue.

If you know that your $var2 contains the exact file name, then remove the "*". Use

zcat $var2 | grep "^000$var1" >> my_output

You can also use the file extension to be safe i.e "zcat $var2.zip".

Or if $var2 doesn't contain the exact file name then try using double quotes around $var*. Like

zcat "$var2*" | grep "^000$var1" >> my_output

Hope this helps.


Dear friend
Already tried all these. But all in vain. Folder contains the files starting with 01.xxxx, 02.xxxxxfg etc some I have to use *. But all attempts fails. Any other Way to do.
# 9  
Old 03-13-2007
Quote:
Originally Posted by ranj@chn
First simplify your code removing unnecessary commands -
Code:
while read var1 var2 
do
zcat ${var2}* | grep "^000$var1" >> my_output
done <myfile

Check if it works. Didnt have time to simulate it. Smilie. Make sure you are in the same directory as the zip files. Otherwise, your 'myfile' should contain the path or you should hardcode the path.

Thanks for trying,
But it does not worked and gave the same error. I have already checked the file names and path etc. No mistake in it. But why it is not replacing zcat $var2* with zcat 01* is not understood. Some other suggestions pl.
Thanks
# 10  
Old 03-13-2007
what shell are you using. I am using bash and this syntax works
Code:
# var=01
# zcat ${var}*

zcat: 01test.gz: not in gzip format

As you can see, it is able to detect files starting with 01. Also, zcat will complain if your extension is not .gz ..
# 11  
Old 03-13-2007
Quote:
Originally Posted by ghostdog74
what shell are you using. I am using bash and this syntax works
Code:
# var=01
# zcat ${var}*

zcat: 01test.gz: not in gzip format

As you can see, it is able to detect files starting with 01. Also, zcat will complain if your extension is not .gz ..

dear friend
is is still not working
Unix Gurus please help me for this script.....
# 12  
Old 03-13-2007
show your code, error outputs and whatever info that's will help us to help you.
# 13  
Old 03-13-2007
File location
I simulated your error and the only time this error occurs is when there are no files of type .Z in the directory where I am searching for the *.Z files. The script that I ran was a part of your code.
Code:
/home/nayakr/ranj/unix_forum> cat chkfile.sh
while read var1 var2
do
zcat tmp/$var2* | grep 'junk' >>myoutfile
done <myfile

There was 'tmp' directory with no *.Z files.
Code:
/home/nayakr/ranj/unix_forum/tmp> ls -l
total 0
/home/nayakr/ranj/unix_forum/tmp>

The content of myfile was
Code:
/home/nayakr/ranj/unix_forum> cat myfile
pattern testfile

/home/nayakr/ranj/unix_forum>

Note that, the file contained a newline at the end. On running the script,
Code:
/home/nayakr/ranj/unix_forum> chkfile.sh
tmp/testfile*.Z: A file or directory in the path name does not exist.
tmp/*.Z: A file or directory in the path name does not exist.

Hope that answers your question.
# 14  
Old 03-13-2007
Quote:
Originally Posted by ghostdog74
show your code, error outputs and whatever info that's will help us to help you.

The code I am trying is :
cat myfile | awk '{print $1, $2}' | while read var1 var2
do
zcat $var2* | grep "^000$var1" >> my_output
done
where myfile contains data like :
1231231 01
1451515 02
I just want to replace the var2 i.e. 01 with the zip files (.Z files) starting with 01.abcdfef
01.hjjjkjhjh and so on and then grep the var1 from these files.
Error I am getting is :
*.Z : No such file or directory.
*.Z : No such file or directory.
*.Z : No such file or directory.
*.Z : No such file or directory.
.
.
.
.
I am not getting how to substitute the file names read as var2 of myfile. Please help...
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