For example suppose I have the following piece of data
ABC,3,4
,,ExtraInfo
,,MoreInfo
XYZ,6,7
,,XyzInfo
,,MoreXyz
ABC,1,2
,,ABCInfo
,,MoreABC
It's trivial to get grep to extract the ABC lines. However if I want to also grab the following lines to produce this output
ABC,3,4
,,ExtraInfo
,,MoreInfo
ABC,1,2
,,ABCInfo
,,MoreABC
Can this be done using grep and standard shell scripting?
Edit: Just to clarify there could be a variable number of lines in between. The logic would be to keep printing while the first column of the CSV is empty.
grep -A 2 {Your regex} will output the two lines following the found strings.
Update:
Since you specified that it could be any number of lines, this would not be possible as grep focuses on matching on a single line see the following questions:
How can I search for a multiline pattern in a file?
Regex (grep) for multi-line search needed
Why can't i match the pattern in this case?
Selecting text spanning multiple lines using grep and regular expressions
You can use this, although a bit hackity due to the grep at the end of the pipeline to mute out anything that does not start with 'A' or ',':
$ sed -n '/^ABC/,/^[^,]/p' yourfile.txt| grep -v '^[^A,]'
Edit: A less hackity way is to use awk:
$ awk '/^ABC/ { want = 1 } !/^ABC/ && !/^,/ { want = 0 } { if (want) print }' f.txt
You can understand what it does if you read out loud the pattern and the thing in the braces.
The manpage has explanations for the options, of which you want to look at -A under Context Line Control.
Related
I am following the example in this post finding contents of one file into another file in unix shell script but want to print out differently.
Basically file "a.txt", with the following lines:
alpha
0891234
beta
Now, the file "b.txt", with the lines:
Alpha
0808080
0891234
gamma
I would like the output of the command is:
alpha
beta
The first one is "incorrect case" and second one is "missing from b.txt". The 0808080 doesn't matter and it can be there.
This is different from using grep -f "a.txt" "b.txt" and print out 0891234 only.
Is there an elegant way to do this?
Thanks.
Use grep with following options:
grep -Fvf b.txt a.txt
The key is to use -v:
-v, --invert-match
Invert the sense of matching, to select non-matching lines.
When reading patterns from a file I recommend to use the -F option as long as you not explicitly want that patterns are treated as regular expressions.
-F, --fixed-strings
Interpret PATTERN as a list of fixed strings (instead of regular expressions), separated by newlines, any of which
is to be matched.
I need to display all the lines using the grep command that contain 2-6 'x's
Also need to know how to display all lines with 3 consecutive 'x's
I have tried grep x{2,6} example.txt but I keep getting an error saying that x6 is not found in the directory. My example file contains 7 lines increasing in the amount of 'x's by one in each line.
The Bash shell uses Brace Expansion to expand:
grep x{2,6} example.txt
into:
grep x2 x6 example.txt
Unless you have a file called x6 in your directory, you will get an error from grep telling you it can't open it.
Rule 1: enclose regular expressions to grep inside quotes — single quotes whenever possible.
Hence, use:
grep 'x{2,6}' example.txt
This deals with getting a regex to grep. Now we need to consider what it means. By default, this means look for the characters x, {, 2, ,, 6, } on a single line. Adding the -E option uses extended regular expressions, and the command looks for anything from 2 to 6 consecutive x's on a single line in the file:
grep -E 'x{2,6}' example.txt
However, it might be worth noting that this is pretty much the same as selecting 'xx' unless you have colouration on, or are selecting 'only' the matched text (the GNU grep extension -o option).
These are all for 2-6 adjacent x's, which is roughly what your proposed regex wanted.
You ask about three adjacent x's:
grep 'xxx' example.txt
The single quotes aren't 100% necessary, but they do no harm and remind you to use them for the regex in general.
Now we face the dilemma that you probably meant "between 2 and 6 x's on a single line, not necessarily adjacent, and not 0 or 1, nor 7 or more".
Rule 2: describe your required result precisely
Imprecise requirements lead to incorrect, or unintended, results. Meeting that requirement needs a more complex regex:
grep -E '^([^x]*x){2,6}[^x]*$' example.txt
That looks for 2-6 occurrences of zero or more non-x's followed by an x at the start of the line, followed by zero or more non-x's up to the end of line.
I need to display all the lines using GREP command that contain 2-6 'x's
grep -P '^(?:[^x]*x[^x]*){2,6}$' file
Also need to know how to display all lines with 3 consecutive 'x's
grep -P 'xxx' file
Everything is in the title. Basicaly let's say I have this pattern
some text lalala
another line
much funny wow grep
I grep funny and I want my output to be "lalala"
Thank you
One possible answer is to use either ed or ex to do this (it is trivial in them):
ed - yourfile <<< 'g/funny/.-2p'
(Or replace ed with ex. You might have red, the restricted editor, too; it can't modify files.) This looks for the pattern /funny/ globally, and whenever it is found, prints the line 2 before the matching line (that's the .-2p part). Or, if you want the most recent line containing 'lalala' before the line matching 'funny':
ed - yourfile <<< 'g/funny/?lalala?p'
The only problem is if you're trying to process standard input rather than a file; then you have to save the standard input to a file and process that file, which spoils the concurrency.
You can't do negative offsets in sed (though GNU sed allows you to do positive offsets, so you could use sed -n '/lalala/,+2p' file to get the 'lalala' to 'funny' lines (which isn't quite what you want) based on finding 'lalala', but you cannot find the 'lalala' lines based on finding 'funny'). Standard sed does not allow offsets at all.
If you need to print just the IP address found on a line 8 lines before the pattern-matching line, you need a slightly more involved ed script, but it is still doable:
ed - yourfile <<< 'g/funny/.-8s/.* //p'
This uses the same basic mechanism to find the right line, then runs a substitute command to remove everything up to the last space on the line and print the modified version. Since there isn't a w command, it doesn't actually modify the file.
Since grep -B only prints each full number of lines before the match, you'll have to pipe the output into something like grep or Awk.
grep -B 2 "funny" file|awk 'NR==1{print $NF; exit}'
You could also just use Awk.
awk -v s="funny" '/[[:space:]]lalala$/{n=NR+2; o=$NF}NR==n && $0~s{print o}' file
For the specific example of an IP address 8 lines before the match as mentioned in your comment:
awk -v s="funny" '
/[[:space:]][0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}$/ {
n=NR+8
ip=$NF
}
NR==n && $0~s {
print ip
}' file
These Awk solutions first find the output field you might want, then print the output only if the word you want exists in the nth following line.
Here's an attempt at a slightly generalized Awk solution. It maintains a circular queue of the last q lines and prints the line at the head of the queue when it sees a match.
#!/bin/sh
: ${q=8}
e=$1
shift
awk -v q="$q" -v e="$e" '{ m[(NR%q)+1] = $0 }
$0 ~ e { print m[((NR+1)%q)+1] }' "${#--}"
Adapting to a different default (I set it to 8) or proper option handling (currently, you'd run it like q=3 ./qgrep regex file) as well as remembering (and hence printing) the entire line should be easy enough.
(I also didn't bother to make it work correctly if you see a match in the first q-1 lines. It will just print an empty line then.)
I'm trying to parse through an application.log that has many lines that follow the same syntax below.
"Error","jrpp-237","10/13/11","02:55:04",,"File not found: /indexUsa~.cfm The specific sequence of files included or processed is: c:\websites\pj7fe4\indexUsa~.cfm '' "
I need to use some type of command to pull out what is listed between c:\websites\ and the next \
e.g. in this case it would be pj7fe4
I thought that the following command would work..
bin/sed -n '/c:\\websites\\/,/\\/p' upload/test.log
Unfortunately from reading further I now understand that this will return the entire line containing c:\websites through the \ and I need to know the in between, not the whole line.
To be more difficult I need to match all of the directory sub paths, not just one particular line as this is for multiple sites.
You're using range patterns incorrectly. You can't use it to limit the command (print in this case) to a part of the line, only to a range of lines. You also don't escape the backspaces.
Try this: sed 's/.*c:\\websites\\\([0-9a-zA-Z]*\)\\.*/\1/'
There's a good sed tutorial here: Sed - An Introduction and Tutorial by Bruce Barnett
grep way:
grep -Po "(?<=c:\\\websites\\\)[^\\\]+(?=\\\)" yourFile
test:
kent$ echo '"Error","jrpp-237","10/13/11","02:55:04",,"File not found: /indexUsa~.cfm The specific sequence of files included or processed is: c:\websites\pj7fe4\indexUsa~.cfm '' "'|grep -Po "(?<=c:\\\websites\\\)[^\\\]+(?=\\\)"
pj7fe4
This question already has answers here:
Colorized grep -- viewing the entire file with highlighted matches
(24 answers)
Closed 7 years ago.
When using grep, it will highlight any text in a line with a match to your regular expression.
What if I want this behaviour, but have grep print out all lines as well? I came up empty after a quick look through the grep man page.
Use ack. Checkout its --passthru option here: ack. It has the added benefit of allowing full perl regular expressions.
$ ack --passthru 'pattern1' file_name
$ command_here | ack --passthru 'pattern1'
You can also do it using grep like this:
$ grep --color -E '^|pattern1|pattern2' file_name
$ command_here | grep --color -E '^|pattern1|pattern2'
This will match all lines and highlight the patterns. The ^ matches every start of line, but won't get printed/highlighted since it's not a character.
(Note that most of the setups will use --color by default. You may not need that flag).
You can make sure that all lines match but there is nothing to highlight on irrelevant matches
egrep --color 'apple|' test.txt
Notes:
egrep may be spelled also grep -E
--color is usually default in most distributions
some variants of grep will "optimize" the empty match, so you might want to use "apple|$" instead (see: https://stackoverflow.com/a/13979036/939457)
EDIT:
This works with OS X Mountain Lion's grep:
grep --color -E 'pattern1|pattern2|$'
This is better than '^|pattern1|pattern2' because the ^ part of the alternation matches at the beginning of the line whereas the $ matches at the end of the line. Some regular expression engines won't highlight pattern1 or pattern2 because ^ already matched and the engine is eager.
Something similar happens for 'pattern1|pattern2|' because the regex engine notices the empty alternation at the end of the pattern string matches the beginning of the subject string.
[1]: http://www.regular-expressions.info/engine.html
FIRST EDIT:
I ended up using perl:
perl -pe 's:pattern:\033[31;1m$&\033[30;0m:g'
This assumes you have an ANSI-compatible terminal.
ORIGINAL ANSWER:
If you're stuck with a strange grep, this might work:
grep -E --color=always -A500 -B500 'pattern1|pattern2' | grep -v '^--'
Adjust the numbers to get all the lines you want.
The second grep just removes extraneous -- lines inserted by the BSD-style grep on Mac OS X Mountain Lion, even when the context of consecutive matches overlap.
I thought GNU grep omitted the -- lines when context overlaps, but it's been awhile so maybe I remember wrong.
You can use my highlight script from https://github.com/kepkin/dev-shell-essentials
It's better than grep cause you can highlight each match with it's own color.
$ command_here | highlight green "input" | highlight red "output"
Since you want matches highlighted, this is probably for human consumption (as opposed to piping to another program for instance), so a nice solution would be to use:
less -p <your-pattern> <your-file>
And if you don't care about case sensitivity:
less -i -p <your-pattern> <your-file>
This also has the advantage of having pages, which is nice when having to go through a long output
You can do it using only grep by:
reading the file line by line
matching a pattern in each line and highlighting pattern by grep
if there is no match, echo the line as is
which gives you the following:
while read line ; do (echo $line | grep PATTERN) || echo $line ; done < inputfile
If you want to print "all" lines, there is a simple working solution:
grep "test" -A 9999999 -B 9999999
A => After
B => Before
If you are doing this because you want more context in your search, you can do this:
cat BIG_FILE.txt | less
Doing a search in less should highlight your search terms.
Or pipe the output to your favorite editor. One example:
cat BIG_FILE.txt | vim -
Then search/highlight/replace.
If you are looking for a pattern in a directory recursively, you can either first save it to file.
ls -1R ./ | list-of-files.txt
And then grep that, or pipe it to the grep search
ls -1R | grep --color -rE '[A-Z]|'
This will look of listing all files, but colour the ones with uppercase letters. If you remove the last | you will only see the matches.
I use this to find images named badly with upper case for example, but normal grep does not show the path for each file just once per directory so this way I can see context.
Maybe this is an XY problem, and what you are really trying to do is to highlight occurrences of words as they appear in your shell. If so, you may be able to use your terminal emulator for this. For instance, in Konsole, start Find (ctrl+shift+F) and type your word. The word will then be highlighted whenever it occurs in new or existing output until you cancel the function.