bash shell how to print line in file including pattern taken from user searching with data type for pattern -

i want take pattern user , print line includes (the whole record) in terminal searching process data types id integer (pk) , name string , salary integer

more explanation: if searching pattern 2 , id have display id , not salary contains 2000 , that's problem... didn't face problem string names in id , salary.

the code incorrectly formatted.

#! /bin/bash     #################taking tablename "database name"###############         echo "enter table name";         read tablename;         if [ -f ./tables/$tablename ];             #############check table exists######################         echo "file $tablename exists."         echo "1)search id";         echo "2)search name";         echo "3)search salary";         echo "enter choice";         read input         echo "enter pattern";         read pattern;                 if [ $input -eq 1 ]                                               test=  cut -d ',' -f1 ./tables/$tablename                         if [[ ${#test[$pattern]} ]];                                                 ###############problem here################                          fi                 elif [ $input -eq 2 ]                                  grep $pattern ./tables/$tablename                 elif [ $input -eq 3 ]                                  ##########################problem here######################                 else                  echo "error in input";                 fi         else             echo "table $tablename not exist "         fi     ##########################code ends################# 

and file has records:

id:pk,name:str,salary:int, 2,tony,2000, 3,tony,2000, 4,sara,3000, 

you use case different operations depending of input choice. , in cases, values file.

echo "enter table name"; read tablename; if [ ! -f ./tables/$tablename ];   #try avoid looooong if/then blocks if can handle error right away    echo "table $tablename not exist"    exit 1 fi #############check table exists######################     echo "file $tablename exists."     echo "1)search id";     echo "2)search name";     echo "3)search salary";     echo "enter choice";     read input     echo "enter pattern";     read pattern; entries=()        #initialize entries empty array case $input in      1) #case id        entries=($(grep -p "^${pattern}," "./tables/$tablename"))        ;;     2) #case name        entries=($(grep -p "^\d*,${pattern}," "./tables/$tablename"))        ;;     3) #case salary        entries=($(grep -p ",${pattern}$" "./tables/$tablename"))        ;;     *) #choice not found        echo "error in input" esac  if [[ "${#entries[@]}" = "0" ]]; #entries array count 0    echo "no entries found matching pattern"    exit 1 fi  #to noted here, entries useless if want print match #(use grep directly , print output).     #thank's jonathan leffler trick allowing print whole array in 1 line printf "%s\n" "${entries[@]}"