Remove first portion of string

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# 1  
Old 08-10-2011
Remove first portion of string

I have a script which currently uses a file containing a list of directories as an argument. The file is read in to an array, and then the array is iterated in a for loop.
What I would like to do is cut off the first few directories of the directory path (they won't exist on the server where the items are being copied).

Here is an example:
On local server:

On remote server:
/logs/serverlogdir <- note that the /usr/local is removed

Is there any easy way to do this?
# 2  
Old 08-10-2011
See if this works for you:
echo /usr/local/logs/serverlogdir | sed 's#.*\(/.*/\)#\1#'

This User Gave Thanks to Shell_Life For This Post:
# 3  
Old 08-10-2011
Originally Posted by Shell_Life
See if this works for you:
echo /usr/local/logs/serverlogdir | sed 's#.*\(/.*/\)#\1#'

Haha, you know, immediately after posting it I remembered that sed exists. Its been a many-cup-of-coffee type of morning. Thanks for answering! I know it'll work Smilie
# 4  
Old 08-10-2011
Or if your array is called A, you could try this:
for i in "${A[@]#/usr/local}"
  echo "$i"

# 5  
Old 08-10-2011
echo "/usr/local/logs/serverlogdir" | sed 's!.*\/local!!'
echo "/usr/local/logs/serverlogdir" | awk -F"/" '{ $1=$2=$3=""; }1' OFS="/"

Althoguh awk produces excess "/" infront of the output , it should not cause any issue to functionality.
# 6  
Old 08-10-2011
Here is another way to do this in ksh:

var="/usr/local/logs/serverlogdir"; echo ${var#/*[a-z]/*[a-z]/}

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