linux - need bash script to strip version in a binary file and compare to db version -


trying setup bash script version in binary file versions included thinking have strip out audit_trail_#_##_a-z,1-9

this have far suggestions

#!/bin/bash echo searching fmx files audit_trail form version file in `/bin/ls *.fmx`   current_release=`strings $file |sed "s/\"audit_trail/notneeded/" |grep -i "audit_trail_8" |sed "s/audit\_trail\_/audit\-trail /" |sed "s/\_/\./g" |sed "s/\.[a-z,a-z]/\.00/" |awk '{print substr($0,0,18)}' |awk '{print $2}'`   export form=$(echo "$file" | cut -f1 -d'.')   export dbform=`echo $form |awk '{print toupper($0)}'`   echo "formname" $form   sqlplus -s /nolog <<eof connect system/xxx@xxxxx set echo on whenever oserror  exit 88 whenever sqlerror exit 1 spool forms.lst select guraobj_current_version bansecr.guraobj guraobj_object = '$dbform'; spool off exit eof   echo $file $current_release done

output

bash-4.1$ ./find_current_release_fmx_db.shl                                                                                                                                                              + ./find_current_release_fmx_db.shl ./find_current_release_fmx_db.shl: line 1: !/bin/bash: no such file or directory searching fmx files audit_trail form version formname peaempl

from database

guraobj_cu ---------- 8.11.2

peaempl.fmx

from compiled form

8.0 8.0 8.0.00 8.0.00 8.1.0. 8.1.0. 8.2.00 8.2.00 8.3 8.3 8.4 8.4 8.7.1 8.7.1 8.7.1. 8.7.1. 8.8.0. 8.8.0. 8.8.1. 8.8.1.  **the 1 need 8.11.2**    8.11.2 8.1.0. 8.8.0. 8.7.1. 8.11.2 8.0.00 8.2.00 8.0 8.3 8.4 8.8.1. 8.7.1

desired output

8.11.2

any appreciated

let's start tidying script not use many pipes , commands, more robust, etc.:

#!/bin/bash printf 'searching fmx files audit_trail form version\n' file in *.fmx   current_release=$(strings "$file" | awk '<something>')   form="${file%%.}"   dbform="${form^^}"   printf 'formname %s\n' "$form"   sqlplus -s /nolog <<eof connect system/xxx@xxxxx set echo on whenever oserror  exit 88 whenever sqlerror exit 1 spool forms.lst select guraobj_current_version bansecr.guraobj guraobj_object = '$dbform'; spool off exit eof   printf '%s %s\n' "$file" "$current_release" done

now need know <something> in awk command should be. couldn't figure out chain of piped commands seem adding stuff , removing again , escaping things shouldn't need escaped, etc. once show sample output of strings , want awk script convert should obvious.


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