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How to trim the file names?

 
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# 1  
Old 02-28-2014
How to trim the file names?
How do i get the following output, irrespective of how many . 's i will have the file name.

Input:
FCBTexas_2013_12_31_16_V2.5_Masked.xml.filepart

Output:
FCBTexas_2013_12_31_16_V2.5_Masked.xml

Thank you.
# 2  
Old 02-28-2014
Code:
ls | grep "\.filepart$" | while read FILENAME
do
        echo mv "$FILENAME" "${FILENAME/.filepart/}"
done

Remove the 'echo' once you're sure it does what you want.
# 3  
Old 02-28-2014
Quote:
Originally Posted by Corona688
Code:
ls | grep "\.filepart$" | while read FILENAME
do
        echo mv "$FILENAME" "${FILENAME/.filepart/}"
done

Remove the 'echo' once you're sure it does what you want.
i cann't move the file, is there any other way i can do this using sed or awk command.

Thank you.
# 4  
Old 02-28-2014
Oh, you just want to remove it from a string, not change the filename. That's much easier.

Code:
STRING="asdf.filepart"
echo "${STRING/.filepart/}"

You can put it in the loop above to do many.

Or you can also do it this way in awk:

Code:
ls *.filepart | awk -F"." '{ NF=(NF-1) } 1'

This User Gave Thanks to Corona688 For This Post:
Ariean
# 5  
Old 02-28-2014
Code:
file="FCBTexas_2013_12_31_16_V2.5_Masked.xml.filepart"

echo "${file%.*}"

This User Gave Thanks to Yoda For This Post:
Ariean
# 6  
Old 02-28-2014
Quote:
Originally Posted by Yoda
Code:
file="FCBTexas_2013_12_31_16_V2.5_Masked.xml.filepart"

echo "${file%.*}"

how do i just get the extention only the "filepart", i am also trying to get it on my side, please help.
# 7  
Old 02-28-2014
Hello,

Following may help.

Code:
echo "FCBTexas_2013_12_31_16_V2.5_Masked.xml.filepart" | awk 'gsub(/xml\..*/,"xml") 1'

Output will be as follows.
Code:
FCBTexas_2013_12_31_16_V2.5_Masked.xml


Code:
echo "FCBTexas_2013_12_31_16_V2.5_Masked.xml.filepart" | sed 's/\(.*xml\)\(.*\)/\1/g'


Output will be as follows.
Code:
FCBTexas_2013_12_31_16_V2.5_Masked.xml


To get only filepart please use the following.

Code:
echo "FCBTexas_2013_12_31_16_V2.5_Masked.xml.filepart" | awk 'gsub(/.*\.+/,X) 1'

Thanks,
R. Singh
This User Gave Thanks to RavinderSingh13 For This Post:
Ariean
 
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