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#1
04-23-2008
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While finding compare the result with given string
Hello,
There are directories named by "yyyyMMdd" format. The one directory name is given and I should find & print that directory & its previous days directory names if exist. Code:
For example: 20080401 20080402 20080403 20080404 ... The given date: 20080403 The result: 20080403 20080401 20080402 
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#2
04-24-2008
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Sounds like homework. Read the TOS.
If it's not homework, care to elaborate?
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#3
04-24-2008
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Hi,
If you have all the dates in a file called dates1 , then awk ' $1 <= 20080403 ' dates1 20080401 20080402 20080403
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#4
04-24-2008
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Thanks penchal_boddu,
I could do it by writing few lines of code, but I wanted to do it in one line. I wrote by the following way, it works, except for new line & space character. How can I remove new line & space characters? Thanks a lot Code:
[testuser@localhost proc]$ find -type d . ./20080401 ./20080405 ./20080404 ./20080414 ./20080423 ./20080413 ./20080415 ./20080420 ./20080406 ./20080417 ./20080418 ./20080419 ./20080410 ./20080407 ./20080412 ./20080403 ./20080416 ./20080411 ./20080408 ./20080402 ./20080409 [testuser@localhost proc]$ find -type d | sed "s#^.# #" | sed "s#/# #" | sort -k1 -u | awk '$1<=20080403' 20080401 20080402 20080403
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#5
04-24-2008
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Thanks all,
I found out it. Code:
[testuser@localhost proc]$ find -type d | sed "s#^.##" | sed "s#/##" | sed '/^$/d' | sort -k1 -u | awk '$1<=20080403' 20080401 20080402 20080403
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