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zcat - clarification?

Posted on 2010-11-17
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I have 5 zipped files (.gz ) with header. All files header is same. I want to create new single file with limited columns from these 5 files without Header. I thought of using below command but I don’t know how to skip header from each file Please let me know how to do this task and below command is fine or i need to change anything.
zcat C:\filename1.gz| Cut -f 4,11-13,15,17-20 >> e:\test.txt
zcat C:\filename2.gz| Cut -f 4,11-13,15,17-20 >> e:\test.txt
zcat C:\filename3.gz| Cut -f 4,11-13,15,17-20 >> e:\test.txt
zcat C:\filename4.gz| Cut -f 4,11-13,15,17-20 >> e:\test.txt
zcat C:\filename5.gz| Cut -f 4,11-13,15,17-20 >> e:\test.txt
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Question by:PKTG
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Tintin earned 1000 total points
ID: 34160457
zcat file.gz | sed 1d | cut ......
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by:PKTG
ID: 34160536
I tried the below command and got the error. Please advise

zcat C:\filename1.gz | sed 1d | Cut -d "," -f 4,11-13,15,17-20 >> e:\test.txt

Error:
Cut: the delimiter must be a single character
Try `Cut --help' for more information.
sed: couldn't write 1212 items to {standard output}: Invalid argument
zcat: stdout: Broken pipe

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by:Tintin
ID: 34161689
Are you using cygwin or something else?
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