How can I use the grep command to extract line which is the next line of greping line?

Question

I have text file like (the specific part of the whole file);

!        lat             long          elev (m)
!      44.5199005      11.6468000       50.4276

I need to extract second line and save them in another text file. The numbers are variable and "!" syntax is also exist several times in the text. For the first line there is always 8 spaces between "!" and "lat" but this is not the case for the second line. Is there any command available to allow me to extract next line after grep "! lat" text

score 8 · Accepted Answer · edited Oct 23 '15 at 08:51

8

You can use grep -A1 'expression'

From the manual:

-A num
--after-context=num
    Print num lines of trailing context after matching lines.

In your case:

grep -A1 "!        lat" foo | grep -v "!        lat"

should work to extract the second line only.

Note:
You can use grep -P '^! {8}lat' instead of typing many . This is less error prone. The -P flag enables perl style regular expressions and {8} (There is a space before the '{') matches exactly 8 spaces.

edited Oct 23 '15 at 08:51

A.B.

90,397

answered Oct 23 '15 at 07:51

Wayne_Yux

4,873

How can I save only the second line to another text file? – deepblue_86 Oct 23 '15 at 07:56
@deepblue_86: I edited my answer – Wayne_Yux Oct 23 '15 at 08:00
Can be simplified: grep -A1 "! lat" text | grep -v "! lat". No need for cat. (Comments remove multiple spaces, unfortunately!) – Byte Commander Oct 23 '15 at 08:01
true, but I think it is easier to read and understand as your "working through" the commands from left to right – Wayne_Yux Oct 23 '15 at 08:08
{} is an extended Regex pattern (use grep -E), PCRE supporting it does not imply that its a PCRE only pattern.. – heemayl Oct 23 '15 at 09:33
1

Note -E (extended regexp) suffices for the {8}, there is no need to use -P (perl regexp). – fedorqui Oct 23 '15 at 11:44
The second filter (instead of grep -v) could be: tail -n+2 (from the second line to the end) or tail -n1 (last line). Which is 1. simpler, 2. without need to repeat the regex, 3. a little bit faster – pabouk - Ukraine stay strong Oct 23 '15 at 13:22
but doesn't work, if you have multiple matches of the first grep ;-) – Wayne_Yux Oct 23 '15 at 13:24
Also please note that neither -P nor -A are options specified by POSIX (http://pubs.opengroup.org/onlinepubs/9699919799/utilities/grep.html) so they may not work on some Unix-like systems. – pabouk - Ukraine stay strong Oct 23 '15 at 13:46

score 5 · Answer 2 · answered Oct 23 '15 at 09:28

You can get the desired output in one go of grep using -z option to treat the lines of input file to be separated by NUL characters rather than newline charaters so that newline characters can be matched literally. You can do:

grep -Pzo '!\s{8}lat[^\n]*\n\K[^\n]+' file.txt

-P will enable us to use PCRE, -o will get us the desired portion only
!\s{8}lat[^\n]*\n will match the line with lat after ! and 8 spaces, also it will match the trailing newline character, the \K will discard this match
[^\n]+ will then match the next line i.e. what we want as output.

Here i also assume that there is no space after lat, if there must be at least a space you can do:

grep -Pzo '!\s{8}lat\s+[^\n]*\n\K[^\n]+' file.txt

Example:

$ cat file.txt 
!        lat             long          elev (m)
!      44.5199005      11.6468000       50.4276
!      30.5199005      2445.6468000       23.89

$ grep -Pzo '!\s{8}lat[^\n]*\n\K[^\n]+' file.txt 
!      44.5199005      11.6468000       50.4276

A.B. · Answer 3 · 2015-10-23T09:27:05.550

1

awk

awk '/! {8}lat/ {a=1; next} a {a=0; print};' foo

sed
```
sed -n '/! \{8\}lat/{n;p}' foo
```

edited Oct 23 '15 at 09:27

answered Oct 23 '15 at 09:00

A.B.

90,397

2

It is more idiomatic to say something like awk 'f{print; exit} /^! {8}lat/ {f=1}' file. Note getline is not recommended in general. – fedorqui Oct 23 '15 at 09:25

How can I use the grep command to extract line which is the next line of greping line?

3 Answers3