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Shell Programming and Scripting

Cut a string for last 8 characters

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# 8  
Old 08-26-2013
May this also help you..

Code:
Last_eight_char=$(echo ${filename#*.})
echo "$Last_eight_char"

the code will take out the all the extensions instead of last 8 characters. i think this is your requirement. i suggest this because extensions can be more than 3 characters also.

---------- Post updated at 02:20 PM ---------- Previous update was at 01:53 PM ----------

hi, you can use the below code also.. i think file extensions can be more tha 3 characters also.
Code:
TPT_PATH='/home/ctl/tpt'
CP_PATH='/home/ctl/copy'

for line in filelist.txt ;
do
  if [ -f $TPT_PATH/$line ]
  then
        line=$(echo ${line%*.*.*}).copy.vsql
        echo "###################################################################################################" >> $CP_PATH/$line
  fi
done

# 9  
Old 08-26-2013
@little:
Code:
$(echo ${line%*.*.*}).copy.vsql

could be this:
Code:
${line%.*.*}.copy.vsql

Also note that the first asterisk should not be there. It only works because you are using lazy matching (single percent sign) so the first asterisk is matching zero characters .
# 10  
Old 08-26-2013
Assuming the file is linear, a longhand version.
Inside a _variable_ as you originally stated:-

Code:
Last login: Mon Aug 26 09:37:04 on ttys000
AMIGA:barrywalker~> printf "abc.tpt.ctl\nbcd.tpt.ctl\ncdw.tpt.ctl\n" > /tmp/string.file
AMIGA:barrywalker~> text=$(cat < /tmp/string.file)
AMIGA:barrywalker~> printf "${text:0:3}\n"
abc
AMIGA:barrywalker~> printf "${text:12:3}\n"
bcd
AMIGA:barrywalker~> printf "${text:24:3}\n"
cdw
AMIGA:barrywalker~> _
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