Search in specific column for pattern and output entire line

Question

I'm working in HDFS and am trying to get the entire line where the 4th column starts with the number 5:

100|20151010|K|5001
695|20151010|K|1010
309|20151010|R|5005
410|20151010|K|5001
107|20151010|K|1062
652|20151010|K|5001

Hence should output:

100|20151010|K|5001
309|20151010|R|5005
410|20151010|K|5001
652|20151010|K|5001

Answer 1

The simplest approach would probably be awk:

awk -F'|' '$4~/^5/' file

The -F'|' sets the field separator to |. The $4~/^5/ will be true if the 4th field starts with 5. The default action for awk when something evaluates to true is to print the current line, so the script above will print what you want.

Other choices are:

• Perl

  perl -F'\|' -ane 'print if $F[3]=~/^5/' file
  

  Same idea. The -a switch causes perl to split its input fields on the value given by -F into the array @F. We then print if the 4th element (field) of the array (arrays start counting at 0) starts with a 5.

• grep

  grep -E  '^([^|]*\|){3}5' file 
  

  The regex will match a string of non-| followed by a | 3 times, and then a 5.

• GNU or BSD sed

  sed -En '/([^|]*\|){3}5/p' file 
  

  The -E turns on extended regular expressions and the -n suppresses normal output. The regex is the same as the grep above and the p at the end makes sed print only lines matching the regex.

Answer 2

This will print all lines that match |5 and then no more | until the end of the line:

grep '|5[^|]*$' <in >out