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want to remove last word.

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# 1  
Old 09-21-2011
want to remove last word.
Hi,

I have a file which has the following
Code:
 /u12/data/oracle/abc.dbf
/u12/data/oracle/def.dbf
/u12/data/oracle/daf.dbf
/u12/data/oracledb/fgh.dbf
/u12/data/oracledb/fkh.dbf
/u12/data/oracledb/kdq.dbf

I want to do something like this
Code:
  /u12/data/oracle
/u12/data/oracle
/u12/data/oracle
/u12/data/oracledb
/u12/data/oracledb
/u12/data/oracledb

I need to ignore the last word *.dbf
and use uniq command to find the mount point for checking space usage
Code:
df -m /u12/data/oracle
df -m /u12/data/oracledb

Thnx
Kaleem
Last edited by Scott; 09-22-2011 at 05:07 AM.. Reason: Use code tags, please...
# 2  
Old 09-21-2011
Code:
while read line; do
   dirname "$line"
done <INPUTFILE | sort -u | xargs df -m

Last edited by yazu; 09-21-2011 at 10:01 PM..
# 3  
Old 09-21-2011
Using awk:
Code:
$ awk ' /\/[^/]*.dbf$/ { sub("/[^/]*.dbf$","") ; A[$0]++ } 
END {for(dir in A) print "df -m " dir }' infile
df -m /u12/data/oracle
df -m /u12/data/oracledb

If your happy with the output you can use system() to run the df commands instead of print.
# 4  
Old 09-22-2011
Code:
$ nawk -F/ '{$NF= ""; print}' infile | sed 's, ,/,g;s,^,df -m ,g' | sh

# 5  
Old 09-22-2011
Code:
sed -e 's!/[^/]*$!!' infile | uniq | xargs df -m

# 6  
Old 09-22-2011
Code:
 
awk -F/ 'BEGIN{OFS="/"}{$NF="";print}' test | sort -u | xargs df -m

# 7  
Old 09-22-2011
Code:
awk -F/ '{NF--;a[$0]}END{for(i in a) print i}' OFS=/  file | xargs df -m
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