How to Grep a String After a Specified Line Number

How to grep a string after a specified line number?

You can tail it, then grep:

tail -n +50000 myfile.txt | grep -in "time spent"

How to grep out specific line ranges of a file

Try using sed as mentioned on
http://linuxcommando.blogspot.com/2008/03/using-sed-to-extract-lines-in-text-file.html. For example use

sed '2,4!d' somefile.txt

to print from the second line to the fourth line of somefile.txt. (And don't forget to check http://www.grymoire.com/Unix/Sed.html, sed is a wonderful tool.)

Find the line number where a specific word appears with grep

Use grep -n to get the line number of a match.

I don't think there's a way to get grep to start on a certain line number. For that, use sed. For example, to start at line 10 and print the line number and line for matching lines, use:

sed -n '10,$ { /regex/ { =; p; } }' file

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To get only the line numbers, you could use

grep -n 'regex' | sed 's/^\([0-9]\+\):.*$/\1/'

Or you could simply use sed:

sed -n '/regex/=' file

Combining the two sed commands, you get:

sed -n '10,$ { /regex/= }' file

How to grep for contents after pattern?

grep 'potato:' file.txt | sed 's/^.*: //'

grep looks for any line that contains the string potato:, then, for each of these lines, sed replaces (s/// - substitute) any character (.*) from the beginning of the line (^) until the last occurrence of the sequence : (colon followed by space) with the empty string (s/...// - substitute the first part with the second part, which is empty).

or

grep 'potato:' file.txt | cut -d\   -f2

For each line that contains potato:, cut will split the line into multiple fields delimited by space (-d\ - d = delimiter, \ = escaped space character, something like -d" " would have also worked) and print the second field of each such line (-f2).

or

grep 'potato:' file.txt | awk '{print $2}'

For each line that contains potato:, awk will print the second field (print $2) which is delimited by default by spaces.

or

grep 'potato:' file.txt | perl -e 'for(<>){s/^.*: //;print}'

All lines that contain potato: are sent to an inline (-e) Perl script that takes all lines from stdin, then, for each of these lines, does the same substitution as in the first example above, then prints it.

or

awk '{if(/potato:/) print $2}' < file.txt

The file is sent via stdin (< file.txt sends the contents of the file via stdin to the command on the left) to an awk script that, for each line that contains potato: (if(/potato:/) returns true if the regular expression /potato:/ matches the current line), prints the second field, as described above.

or

perl -e 'for(<>){/potato:/ && s/^.*: // && print}' < file.txt

The file is sent via stdin (< file.txt, see above) to a Perl script that works similarly to the one above, but this time it also makes sure each line contains the string potato: (/potato:/ is a regular expression that matches if the current line contains potato:, and, if it does (&&), then proceeds to apply the regular expression described above and prints the result).

Use grep to report back only line numbers

try:

grep -n "text to find" file.ext | cut -f1 -d:

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