Shell Script Printing in table format

Question

I am trying to print set of values in table format using shell script . The table has n rows and 4 columns. I tried the following piece of code.

btfield=`grep -i -r ^"diag.* <string>" *.txt |awk '{print $5}'|cut -d+ -f 1 |sort -u`
ipaddr=''
i=0
format="%10s\t%10s\t%10s   \n"
echo "| Function  " "     |         IP         |"  "    occurences     |"  
for a in $btfield
do
  b=`grep -i -r ^"diag.* <string>" *.txt |grep  -i $a |cut -d: -f 1|cut -d_ -f 2|cut -d't' -f 1`
  noOcc=`grep -i -r ^"diag.* backtrace" *.txt |grep  -i $a|wc -l`
  #echo $b
  ipaddr[i]=${b}
  printf "$format"  $a  ${ipaddr[i]} $noOcc
  i=$((i+1))
  #echo $i
done

The above piece of code finds distinct fields from various files and prints according to the format specifier.

But what I am seeing is misaligned form of output. Is there any way to print the values in table format ? The column width is fixed and if the value in cell exceeds the width it must wrap itself.

Sample output:

|       reason          |        STB IP         |        occurences     |
printf     142.25.1.100.   142.25.1.100.   
142.25.1.102.   142.25.1.105.   192.168.1.100.   
192.168.1.100.  192.168.1.100.  192.168.1.100.   
192.168.1.106.           9                   
class_consol      142.25.1.105.   192.168.1.103.   
         2                                   
getChar   182.168.1.102.           1   
     maindat      142.25.1.103.            1   
_XN52getappdatafrom3EZN2232_iPjj     142.25.1.103.   142.25.1.103.   
182.168.1.106.    

Answer 1

Instead of using echo, use the printf command.

You can use %s for a string and %d for an integer:

printf "%20s %8d %s\n" "$string1" $int1 "$unbounded_string"

You can space according to your screen, by using %20s for using 20 characters for the string, %8d for the integer.

Format control options are also available:

\n: newline

\t: tab (horizontal)

\v: tab (vertical)