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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
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 $F[3]=~/^5/' file
Same idea. The -a switch causes perl to split its input fields on the value given by -F into the array @F. We then print if the 4th element (field) of the array (arrays start counting at 0) starts with a 5.
grep
grep -E '^([^|]*\|){3}5' file
The regex will match a string of non-| followed by a | 3 times, and then a 5.
GNU or BSD sed
sed -En '/([^|]*\|){3}5/p' file
The -E turns on extended regular expressions and the -n suppresses normal output. The regex is the same as the grep above and the p at the end makes sed print only lines matching the regex.
-E isn't mentioned in either man or info. It does activate ERE, right?
– terdon♦
Dec 2 '15 at 23:33
-Extended regexp is slated for the next POSIX version, too, so might as well just get used to it. plus, -r doesn't make any sense. and yeah, it's a synonym - they both do exactly the same thing. almost the same - i think w/ -r you can switch back -re ... -Ge ... or something, but who would?
– mikeserv
Dec 2 '15 at 23:34
This will print all lines that match |5 and then no more | until the end of the line:
grep '|5[^|]*$' <in >out
