number of "/"


 
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# 1  
Old 12-02-2004
number of "/"

Hi!

I have a file :
/usr
/var
/tmp
/base_DORA/rbs
/base_DORA/reorg
/base_DORA/tmp


And I want find only the first directory to find:
usr
var
tmp
base_DORA


Thinks for help! Smilie
# 2  
Old 12-02-2004
Tested on your sample data,
Code:
#!/bin/bash

while read line
do
  var1=${line#/*}
  var2=${var1%%/*}
  echo "${var2}"
done < dirlist > rootdirs

sort rootdirs | uniq

rm rootdirs

Gives the output
Code:
base_DORA
tmp
usr
var

Cheers
ZB
# 3  
Old 12-02-2004
Thanks... but I have find a better solution for me...

for i in `cat myfile`
do
echo $i |awk -F "/" '{print $2}' >> $df_rep
done
sort $df_rep | uniq


It is easier for me to understand
But thanks a lot!!!
# 4  
Old 12-02-2004
No problem Smilie

You might consider re-writing your script using a while loop - not only will this remove a uuoc, but will also have a performance boon... something like
Code:
while read i
do
  echo "$i" |awk -F "/" '{print $2}' >> $df_rep
done < myfile
sort $df_rep | uniq

Also; the script I posted will be faster as it uses built in mechanisms of the shell for processing the variables rather than using an external command such as awk which will be spawned for each iteration of the loop.

Just my 2p Smilie

Cheers
ZB
# 5  
Old 12-02-2004
Thank you zazzybob! Smilie

I would do as you say!
# 6  
Old 12-03-2004
Re: number of "/"

Quote:
Originally posted by Castelior
Hi!

I have a file :
/usr
/var
/tmp
/base_DORA/rbs
/base_DORA/reorg
/base_DORA/tmp


And I want find only the first directory to find:
usr
var
tmp
base_DORA


Thinks for help! Smilie



Hi there,

Another one line solution,

df -k | awk '{print $6}' | awk -F\/ '{print $2}' | sort

Regards
# 7  
Old 12-05-2004
$ cat file1 | sed 's/^.//g' | cut -d"/" -f 1 | sort -u
base_DORA
tmp
usr
var
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