Array declaration in Shell script


 
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# 8  
Old 02-28-2011
Question

Ok..ill try once more....thanks...

---------- Post updated at 04:38 PM ---------- Previous update was at 12:09 PM ----------

this is my code :

Code:
       cert=`cat $usr_cert`

                                for i in $cert ; do
                                        USERCERT[$COUNT]=$i
                                        ((COUNT++))
                                done

this is the error I'm getting:

Code:
File name too long
+ COUNT++
./clone.sh: 1: COUNT++: not found
./clone.sh: 1: Bad substitution


please explain the error and suggest a resolution....Smilie
# 9  
Old 02-28-2011
Code:
COUNT=0

while read i; do
  USERCERT[$COUNT]="$i"
  COUNT=$(( $COUNT + 1 ))
done < $usr_cert

# 10  
Old 02-28-2011
Code:
while IFS= read -r i
do
  USERCERT+=( "$i" )
done < "$usr_cert"


Or:
Code:
IFS=$'\n'
USERCERT=( $( < "$usr_cert" ) )

Or (bash 4):
Code:
mapfile -t USERCERT < "$usr_cert"


Last edited by cfajohnson; 02-28-2011 at 11:56 AM..
# 11  
Old 02-28-2011
what does this error indicate:

Code:
./clone.sh: 1: Bad substitution

# 12  
Old 02-28-2011
Quote:
Originally Posted by xerox
what does this error indicate:
Code:
./clone.sh: 1: Bad substitution


Probably that you used an invalid variable name in a parameter expansion.
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