Is it possible to have a regex that parses only a1bcdea1 from this line a1bcdea1ABCa1DEFa1 ?
This grep command does not work:
$ cat txtfile
a1bcdea1ABCa1DEFa1
$ grep -oE "[A-Z,a-z]1.*?[A-Z,a-z]1" txtfile
a1bcdea1ABCa1DEFa1
I want the output of grep to be only a1bcdea1.
EDIT:
It is obvious that I can just use grep -o "a1bcdea1" for the above line, but consider if one has several thousands of lines and the goal is to match FIRST [A-Z,a-z]1.*?[A-Z,a-z]1 for each single line.
How about using a ^ start anchor and restricting character set used:
grep -o '^[A-Za-z]1[A-Za-z]*1'
See this Bash demo or Regex Pattern at regex101
If you expect more digits or other characters in between, go with this
grep -oP '^[A-Za-z]1.*?[A-Za-z]1'
The lazy matching requires perl compatible mode. For not at line start, go with this
grep -oP '^.*?\K[A-Za-z]1.*?[A-Za-z]1'
\K resets beginning of the reported match and is a PCRE feature as well.
Here is a gnu awk solution using split function:
awk '(n = split($0, a, /[a-zA-Z]1/, b)) > 1 {print b[1] a[2] b[2]}' file
a1bcdea1
This awk command splits each line on regex /[a-zA-Z]1/ and stores split tokens in array a and delimiters in array b.
Related
grep -E "a|d$$" filename
This is what I have but it does not works.
Can I get some advice how I should approach it?
| is OR, not AND. So your command returns lines that either contain a or end with d (I assume $$ was a typo for $). To match both conditions sequentially, just put one pattern after the other, don't use |.
If the file is one word per line, use:
grep 'a.*d$' filename
If there are multiple words per line, and you're using GNU grep, you can use:
grep -P 'a\w*d\b' filename
\w matches word characters, and \b matches a word boundary after the d.
This will match the whole line containing the word. If you only want to return the word itself, use
grep -P -o '\b\w*a\w*d\b' filename
The -o option means to only show the part of the line that matches the regexp
Using awk:
awk '{for (i=0; i<=NF; i++) {if ($i~/a.*d$/) {print $i}}}'
Using GNU awk, or any awk that implements regex for the record separator (RS):
awk -v RS='[[:space:]]+' '/a.*d$/'
Using GNU grep:
grep -Po '[^[:space:]]*a[^[:space:]]*d(?=[[:space:]]|$)'
I have a log file contains some information like below
"variable1=XXX, emotionType=sad, sentimentType=negative..."
What I want is to grep only the matched string, the string starts with emotionType and ends with the first occurrence of comma.
E.g.
emotionType=sad
emotionType=joy
...
What I have tried is
grep -e "/^emotionType.*,/" file.log -o
but I got nothing. Anyone can tell me what should I do?
You need to use
grep -o "emotionType[^,]*" file.log
Note:
Remove ^ or replace with \<, starting word boundary construct if your matches are not located at the beginning of each line
Remove the / chars on both ends of the regex since grep does not use regex delimiters (like sed)
[^,] is a negated bracket expression that matches any char other than a comma
* is a POSIX BRE quantifier that matches zero or more occurrences.
See an online demo:
#!/bin/bash
s="variable1=XXX, emotionType=sad, sentimentType=negative, emotionType=happy"
grep -o "emotionType=[^,]*" <<< "$s"
Output:
emotionType=sad
emotionType=happy
1st solution: With awk you could try following program. Simple explanation would be using awk's match function capability and using regex to match string emotionType till next occurrence of , and printing all the matches in awk program.
var="variable1=XXX, emotionType=sad, sentimentType=negative, emotionType=happy"
Where var is a shell variable.
echo "$var" |
awk '{while(match($0,/emotionType=[^,]*/)){print substr($0,RSTART,RLENGTH);$0=substr($0,RSTART+RLENGTH)}}'
2nd solution: Or in GNU awk using RS variable try following awk program.
echo "$var" | awk -v RS='emotionType=[^,]*' 'RT{sub(/\n+$/,"",RT);print RT}'
I am trying to do multiple grep pattern to find a number within a grepped string.
I have a text file like this:
This is the first sample line 1
this is the second sample line
another line
total lines: 3 tot
I am trying to find a way to get just the number of total lines. So the output here should be "3"
Here are the things I've tried:
grep "total lines: [0-9]" myfile.txt
grep "total lines" myfile.txt | grep "[0-9]"
You could use sed:
sed -En 's/^total lines: ([0-9]+).*/\1/p' myfile.txt
-E extended regular expressions
-n suppress automatic printing
Match ^total lines: ([0-9]+).* (capture the number)
\1 replace the whole line with the captured number
p print the result
1st solution: Using GNU grep try following. Simply using -o option to print only matched value, -P enables PCRE regex for program. Then in regex portion matching from starting ^total lines: in each line and if a match found then discard matched values by \K option(to remove it from expected output) which is followed by 1 or more digits, using positive look ahead to make sure its followed by space(s) tot here.
grep -oP '^total lines: \K[0-9]+(?=\s+tot)' Input_file
2nd solution: With your shown samples, please try following in awk. This could be done in a single awk itself. Searching line which has string /total lines: / in it then printing 2nd last field of that line.
awk '/total lines: /{print $(NF-1)}' Input_file
3rd solution: Using awk's match function here. Matching total lines: [0-9]+ tot and then substituting everything apart from digits with null in matched values.
awk 'match($0,/total lines: [0-9]+ tot/){val=substr($0,RSTART,RLENGTH);gsub(/[^0-9]+/,"",val);print val}' Input_file
Do you have to use grep?
$ echo myfile.txt | wc -l
If you mean that the file has a line in it formatted as
total lines: 3 tot
Then refer to https://unix.stackexchange.com/questions/13466/can-grep-output-only-specified-groupings-that-match and use something like:
grep -Po 'total lines: \K\d+' myfile.txt
Notes:
Perl regex is not my forte, so the \d\w part might not work.
This may be doable without -P, but I cannot test from this windows computer.
regex101.com helped me test the above line, so it may work.
Problem with relying on pattern of last line and applying grep/sed to find pattern is that if any line in file contains such pattern, then you will have to apply some additional logic to filter that.
e.g. Consider case of below input file.
line001
total lines: 883 tot
This is the first sample line 1
this is the second sample line
another line
total lines: 883 tot
Assuming your file format is constant (i.e. Second last line will be blank and last line will contain total count), instead of using any pattern matching commands you can directly count number of rows using below awk command.
awk 'END { print NR - 2 }' myfile.txt
You can use the following awk to get the third field on a line that starts with total count: and stop processing the file further:
awk '/^total lines:/{print $3; exit}' file
See this online demo.
You can use the following GNU grep:
# Extract a non-whitespace chunk after a certain pattern
grep -oP '^total lines:\s*\K\S+' file
# Extract a number after a pattern
grep -oP '^total lines:\s*\K\d+(?:\.\d+)?' file
See an online demo. Details:
^ - start of string
total lines: - a literal string
\s* - any zero or more whitespace chars
\K - match reset operator discarding all text matched so far
\S+ - one or more non-whitespace chars
\d+(?:\.\d+)? - one or more digits and then an optional sequence of . and one or more digits.
See the regex demo.
I have a file on this form:
X/this is the first match/blabla
X-this is
the second match-
and here we have some fluff.
And I want to extract everything that appears after "X" and between the same markers. So if I have "X+match+", I want to get "match", because it appears after "X" and between the marker "+".
So for the given sample file I would like to have this output:
this is the first match
and then
this is
the second match
I managed to get all the content between X followed by a marker by using:
grep -zPo '(?<=X(.))(.|\n)+(?=\1)' file
That is:
grep -Po '(?<=X(.))(.|\n)+(?=\1)' to match X followed by (something) that gets captured and matched at the end with (?=\1) (I based the code on my answer here).
Note I use (.|\n) to match anything, including a new line, and that I also use -z in grep to match new lines as well.
So this works well, the only problem comes from the display of the output:
$ grep -zPo '(?<=X(.))(.|\n)+(?=\1)' file
this is the first matchthis is
the second match
As you can see, all the matches appear together, with "this is the first match" being followed by "this is the second match" with no separator at all. I know this comes from the usage of "-z", that treats all the file as a set of lines, each terminated by a zero byte (the ASCII NUL character) instead of a newline (quoting "man grep").
So: is there a way to get all these results separately?
I tried also in GNU Awk:
awk 'match($0, /X(.)(\n|.*)\1/, a) {print a[1]}' file
but not even the (\n|.*) worked.
awk doesn't support backreferences within regexp definition.
Workarounds:
$ grep -zPo '(?s)(?<=X(.)).+(?=\1)' ip.txt | tr '\0' '\n'
this is the first match
this is
the second match
# with ripgrep, which supports multiline matching
$ rg -NoUP '(?s)(?<=X(.)).+(?=\1)' ip.txt
this is the first match
this is
the second match
Can also use (?s)X(.)\K.+(?=\1) instead of (?s)(?<=X(.)).+(?=\1). Also, you might want to use non-greedy quantifier here to avoid matching match+xyz+foobaz for an input X+match+xyz+foobaz+
With perl
$ perl -0777 -nE 'say $& while(/X(.)\K.+(?=\1)/sg)' ip.txt
this is the first match
this is
the second match
Here is another gnu-awk solution making use of RS and RT:
awk -v RS='X.' 'ch != "" && n=index($0, ch) {
print substr($0, 1, n-1)
}
RT {
ch = substr(RT, 2, 1)
}' file
this is the first match
this is
the second match
With GNU awk for multi-char RS, RT, and gensub() and without having to read the whole file into memory:
$ awk -v RS='X.' 'NR>1{print "<" gensub(end".*","",1) ">"} {end=substr(RT,2,1)}' file
<this is the first match>
<this is
the second match>
Obviously I added the "<" and ">" so you could see where each output record starts/ends.
The above assumes that the character after X isn't a non-repetition regexp metachar (e.g. ., ^, [, etc.) so YMMV
The use case is kind of problematic, because as soon as you print the matches, you lose the information about where exactly the separator was. But if that's acceptable, try piping to xargs -r0.
grep -zPo '(?<=X(.))(.|\n)+(?=\1)' file | xargs -r0
These options are GNU extensions, but then so is grep -z and (mostly) grep -P, so perhaps that's acceptable.
GNU grep -z terminates input/output records with null characters (useful in conjunction with other tools such as sort -z). pcregrep will not do that:
pcregrep -Mo2 '(?s)X(.)(.+?)\1' file
-onumber used instead of lookarounds. ? lazy quantifier added (in case \1 occurs later).
I need to get X to Y in the file with multiple occurrences, each time it matches an occurrence it will save to a file.
Here is an example file (demo.txt):
\x00START how are you? END\x00
\x00START good thanks END\x00
sometimes random things\x00\x00 inbetween it (ignore this text)
\x00START thats nice END\x00
And now after running a command each file (/folder/demo1.txt, /folder/demo2.txt, etc) should have the contents between \x00START and END\x00 (\x00 is null) in addition to 'START' but not 'END'.
/folder/demo1.txt should say "START how are you? ", /folder/demo2.txt should say "START good thanks".
So basicly it should pipe "how are you?" and using 'echo' I can prepend the 'START'.
It's worth keeping in mind that I am dealing with a very large binary file.
I am currently using
sed -n -e '/\x00START/,/END\x00/ p' demo.txt > demo1.txt
but that's not working as expected (it's getting lines before the '\x00START' and doesn't stop at the first 'END\x00').
If you have GNU awk, try:
awk -v RS='\0START|END\0' '
length($0) {printf "START%s\n", $0 > ("folder/demo"++i".txt")}
' demo.txt
RS='\0START|END\0' defines a regular expression acting as the [input] Record Separator which breaks the input file into records by strings (byte sequences) between \0START and END\0 (\0 represents NUL (null char.) here).
Using a multi-character, regex-based record separate is NOT POSIX-compliant; GNU awk supports it (as does mawk in general, but seemingly not with NUL chars.).
Pattern length($0) ensures that the associated action ({...}) is only executed if the records is nonempty.
{printf "START%s\n", $0 > ("folder/demo"++i)} outputs each nonempty record preceded by "START", into file folder/demo{n}.txt", where {n} represent a sequence number starting with 1.
You can use grep for that:
grep -Po "START\s+\K.*?(?=END)" file
how are you?
good thanks
thats nice
Explanation:
-P To allow Perl regex
-o To extract only matched pattern
-K Positive lookbehind
(?=something) Positive lookahead
EDIT: To match \00 as START and END may appear in between:
echo -e '\00START hi how are you END\00' | grep -aPo '\00START\K.*?(?=END\00)'
hi how are you
EDIT2: The solution using grep would only match single line, for multi-line it's better use perl instead. The syntax will be very similar:
echo -e '\00START hi \n how\n are\n you END\00' | perl -ne 'BEGIN{undef $/ } /\A.*?\00START\K((.|\n)*?)(?=END)/gm; print $1'
hi
how
are
you
What's new here:
undef $/ Undefine INPUT separator $/ which defaults to '\n'
(.|\n)* Dot matches almost any character, but it does not match
\n so we need to add it here.
/gm Modifiers, g for global m for multi-line
I would translate the nulls into newlines so that grep can find your wanted text on a clean line by itself:
tr '\000' '\n' < yourfile.bin | grep "^START"
from there you can take it into sed as before.