shell convert tab chars to n spaces where n=8

leniviy · 11-29-2012, 09:06 AM

Try this script:

Code:

printf "a\tb \tc  \td   \te\n"; printf "a\tb\tc\td\te\n"

the output in terminal is:

Code:

a       b       c       d       e
a       b       c       d       e

but the number of characters is different.
How to convert the tab characters in the piped input to spaces, so the result looks the same?
I'm thinking of awk:

Code:

( printf "a\tb \tc  \td   \te\n"; printf "a\tb\tc\td\te\n" ) | awk '{
  s="";
  col=0;
  for (i=1; i<=length;i++) {
    c=substr($0,i,1);
    #print "col:", col, "\"" c "\""
    if (c == "\t") {
      j = 8 - (col % 8);
      #print "remainder:", j;
      s = s substr("        ", 1, j );
      col += j;
    } else {
      col++;
      s = s c;
    }
  }
  print s;
}'

colucix · 11-29-2012, 10:01 AM

You can try the expand command:

Code:

$ (printf "a\tb \tc  \td   \te\n"; printf "a\tb\tc\td\te\n") | expand -
a       b       c       d       e
a       b       c       d       e

The output has no tabs, but equally spaced fields.