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Shell Programming and Scripting

Printing empty subdirs before delete

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Top Forums Shell Programming and Scripting Printing empty subdirs before delete
# 8  
Old 01-12-2012
Code:
 
count=$(ls -lR | grep -c "total 0")
echo $count
[ "$count" -eq "0" ] && echo "There are no empty dirs in this dir" || echo $count
#To Print the Empty directory
ls -lR | nawk '/^\./{f=$1}/^total 0/{print f}'

# 9  
Old 01-12-2012
Thanks itkamaraj
This is working fine at command prompt but when i write it inside script I am getting
syntax error at line 1: `count=$' unexpected error

Quote:
Originally Posted by itkamaraj
Code:
 
count=$(ls -lR | grep -c "total 0")
echo $count
[ "$count" -eq "0" ] && echo "There are no empty dirs in this dir" || echo $count
#To Print the Empty directory
ls -lR | nawk '/^\./{f=$1}/^total 0/{print f}'

# 10  
Old 01-13-2012
try with

Code:
count=`ls -lR | grep -c "total 0"`

# 11  
Old 01-13-2012
i) This is storing ls -lR | grep -c "total 0" in $count ...and not the number of empty dirs...
ii)
ls -lR | nawk '/^\./{f=$1}/^total 0/{print f}'
displays those dirs also which has empty files ..is there anyway to exclude those.

Quote:
Originally Posted by itkamaraj
try with

Code:
count=`ls -lR | grep -c "total 0"`

# 12  
Old 01-13-2012
what is mean by ?

This is storing ls -lR | grep -c "total 0" in $count ...and not the number of empty dirs...


---------- Post updated at 09:59 AM ---------- Previous update was at 09:50 AM ----------

Use back ticks, see the posting carefully Smilie

Code:
 
count=`ls -lR | grep -c "total 0"`

# 13  
Old 01-13-2012
Sorry for confusion it is working fine...only thing it counts those dirs also which has empty files ..is there anyway to exclude those dir from the total count..
Quote:
Originally Posted by itkamaraj
what is mean by ?

This is storing ls -lR | grep -c "total 0" in $count ...and not the number of empty dirs...
# 14  
Old 01-13-2012
try this..

Code:
 
find . -type d | while read d; do [ $(( $(ls -a1 $d | wc -l) >= 3 )) == 0 ] && echo $d; done | wc -l
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