How to print a specific column of characters using awk

Question

I would like to print the 55-60th character on each line of a file. I have to do this because I have a FORTRAN formatted file, and I don't have the same number of field separators on each line:

HETATM 3109  O   HOH B 999      10.307  26.441  12.306  0.26 30.00           O
HETATM 3110  O   HOH B1000      10.905  26.874  14.064  0.20 30.00           O

字符55-60是顶行的第10个字段,底行的第9个字段.如何55-60使用awk 打印字符？

Answer 1

使用-c来自cut:

$ cut -c55-60 file
  0.26
  0.20

来自man cut:

-c, - characters = LIST

只选择这些字符

或者sed:

$ sed -r 's/.{54}(.{6}).*/\1/' file
  0.26
  0.20

或awk用substr():

$ awk '{print substr($0,55,6)}' file
  0.26
  0.20