- Is there a way to ignore header lines in a UNIX sort?
In simple cases, sed can do the job elegantly: or equivalently, The key is in the 1q -- print first line (header) and quit (leaving the rest of the input to sort) For the example given, 2q will do the trick
- text processing - sort but keep header line at the top - Unix . . .
For fixed-width output, use the -b option, as it will make sort ignore leading blanks in the sort key The default field separator is non-blank-to-blank transitions, so fields will start with leading blanks
- sort invocation (GNU Coreutils 9. 7)
GNU sort (as specified for all GNU utilities) has no limit on input line length or restrictions on bytes allowed within lines sort has three modes of operation: sort (the default), merge, and check for order
- sorting - Bash - sort by not first character - Super User
So with your current sort command, sort -k1,1 file uses the first word to the first word as the sort What you want is (for the sort command anyway): sort -k1 5 file | uniq -s 6 -w 5 This will use the fifth character of the first word, which is what you wanted
- Bash: How to Sort Lines in File and Ignore First Line
Often you may want to use Bash to sort the lines in a file but ignore the first line, which is often a header row You can use the following syntax to do so: awk 'NR<2{print;next}{print|"sort"}' players txt
- GNU sort -t : -n numeric -b blanks ignored -f fold . . .
To ignore both leading and trailing white space, use the b modifier to the field-end specifier for the first key, sort -t : -n -k 5,5 b -k 3,3 etc passwd or use the global -b modifier and xplicit n with the second key
- GNU Core-utils: sort invocation
GNU sort (as specified for all GNU utilities) has no limits on input line length or restrictions on bytes allowed within lines In addition, if the final byte of an input file is not a newline, GNU sort silently supplies one A line's trailing newline is not part of the line for comparison purposes Upon any error, sort exits with a status of `2'
|