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how to extract a certain part of a line

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# 1  
Old 06-28-2011
how to extract a certain part of a line
Hi friends,

I want to select and use the certain part of a line. For example I have the following line

home/1245/hgdf/acsdf/myhome/afolder/H2O/endfile


how can I extract the part " /myhome/afolder/H2O/endfile "



thanks
# 2  
Old 06-28-2011
Code:
echo "home/1245/hgdf/acsdf/myhome/afolder/H2O/endfile" | sed 's/^.\+\(\/myhome\)/\1/i'

This User Gave Thanks to Skrynesaver For This Post:
rpf
# 3  
Old 06-28-2011
it works well. thank you
# 4  
Old 06-28-2011
Code:
$ echo "home/1245/hgdf/acsdf/myhome/afolder/H2O/endfile" | awk -F"/" '{for(i=5;i<=NF;i++) {printf("/%s",$i); if(i==NF)print "\n"}}'
/myhome/afolder/H2O/endfile

This User Gave Thanks to itkamaraj For This Post:
rpf
# 5  
Old 06-28-2011
Is it possible to define a function that extracts the certain part of subsequent line when I copy the whole line after it. For example, when I type

rpf home/1245/hgdf/acsdf/myhome/afolder/H2O/endfile

output will be /myhome/afolder/H2O/endfile

can I define a function rpf like this ?
# 6  
Old 06-28-2011
Type in your .bashrc -or other, depending on your shell-
Code:
rpf() {
   #Get one of the two suggestions from above, replacing the string by "$1" (with double quotes)
}

and you're done Smilie
This User Gave Thanks to tukuyomi For This Post:
rpf
# 7  
Old 06-28-2011
thank you... thumbs up Smilie
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