I am trying to Use the GREP command to list all lines which contain the letter
'd' at least twice in a row (that is, next to each other), from file "test". But i can't get it to work. Can someone please help?
I can't display what I want. And am not sure how this works. I tried grep [dd] but that didn't work.
This is extremely confusing for me
In regex:
^ character means, start of data line.
all characters match themselves.
So, if you are trying to find lines starting with 'dd', try:
grep ^dd <your_file>
However if you just interested in finding out lines with 'dd' pattern, try:
grep dd <your_file>
The following should make it:
grep dd <your_file>
Your grep [dd] was specifying any line with a character from the set (set = []) containing "d" and "d". So simply putting them side-by-side without the square brackets provides the method for all lines containing a "d" with another "d" right after it.
cat yourfile | grep dd
grep dd yourfile
grep dd <yourfile
should all work, assuming you replace "yourfile" with your file's name.
If you really were looking for lines starting with two or more "d"s, using Sagar's "^dd" instead of just "dd". Similarly, you could use "dd$" for lines ending with two or more "d"s.
You might try "man regexp" or http://www.zytrax.com/tech/web/regex.htm for some more info.
Related
I have a large binary file. I want to extract certain strings from it and copy them to a new text file.
For example, in:
D-wM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM-FM MM-[o#^B^#^#^#^#^#E7cacscKLrrok9bwC3Z64NTnZM-^G
I want to take the number '7' (after the #^#^#E) and every character after it stopping at the Z ('ignoring the M-^G).
I want to copy this 7cacscKLrrok9bwC3Z64NTnZ to a new file.
There will be multiple such strings in one file. The end will always be denoted by the M- (which I don't want copied). The start will always be denoted by a 7 (which I do want copied).
Unfortunately, my knowledge of grep, sed, etc, does not extend to this level. Can someone please suggest a viable way to achieve this?
cat -v filename | grep [7][A-Z,a-z] will show all strings with a '7' followed by a letter but that's not much.
Thank you.
I've noticed that my requirements are rather more complicated.
(I've performed the correct - I hope - formatting this time). Thanks to 'tshiono' for his (?) answer to the earlier submission.
I want to check the ending of a string and, if it ends in M-, grep another string that follows it (with junk in between). If the string does not end in M-, then I don't want it copied (let alone any other strings).
So what I would like is:
grep -a -Po "7[[:alnum:]]+(?=M-)" file_name and if the ending is M- then grep -a -Po "5x[[:alnum:]]+(?=\^)" file_name to copy the string that starts with 5x and ends with a ^.
In this example:
D-wM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM-FM MM-[o#^B^#^#^#^#^#E7cacscKLrrok9bwC3Z64NTnZM-^GwM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM5x8w09qewqlkcklwnlkewflewfiewjfoewnflwenfwlkfwelk^89038432nowefe
The outcome would be:
7cacscKLrrok9bwC3Z64NTnZ
5x8w09qewqlkcklwnlkewflewfiewjfoewnflwenfwlkfwelk
However, if the ending is not M- (more precisely, if the ending is ^S), then do not try the second grep and do not record anything at all.
In this example:
D-wM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM-FM MM-[o#^B^#^#^#^#^#E7cacscKLrrok9bwC3Z64NTnZ^SGwM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM5x8w09qewqlkcklwnlkewflewfiewjfoewnflwenfwlkfwelk^89038432nowefe
The outcome would be null (nothing copied) as the 7cacs... string ends in ^S.
Is grep the correct tool? Grep a file and if the condition in the grep command is 'yes' then issue a different grep command but if the condition is 'no' then do nothing.
Thanks again.
I have noticed one addition modification.
Can one add an OR command to the second part? Grep if the second string starts with 5x OR 6x?
In the example below, grep -aPo "7[[:alnum:]]+M-.*?5x[[:alnum:]]+\^" filename | grep -aPo "7[[:alnum:]]+(?=M-)|5x[[:alnum:]]+(?=\^)" will extract the strings starting with 7 and the strings starting with 5x.
How can one change the 5x to 5x or 6x?
D-wM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM-FM MM-[o#^B^#^#^#^#^#E7cacscKLrrok9bwC3Z64NTnZM-^GwM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM5x8w09qewqlkcklwnlkewflewfiewjfoewnflwenfwlkfwelk^89038432nowefe
D-wM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM-FM MM-[o#^B^#^#^#^#^#E7AAAAAscKLrrok9bwC3Z64NTnZM-^GwM-^?^#^#^#^#^#^#^#^Y^#^#^#^#^#^#^#M-lM-FM-MM-[o#^B^#M-lM6x8w09qewqlkcklwnlkewflewfiewjfoewnflwenfwlkfwelk^89038432nowefe
In this example, the desired outcome would be:
7cacscKLrrok9bwC3Z64NTnZ
5x8w09qewqlkcklwnlkewflewfiewjfoewnflwenfwlkfwelk
7AAAAAscKLrrok9bwC3Z64NTnZ
6x8w09qewqlkcklwnlkewflewfiewjfoewnflwenfwlkfwelk
UPDATE MARCH 09:
I need to create a series of complex grep (or perl) commands to extract strings from a series of binary files.
I need two strings from the binary file.
The first string will always start with a 1.
The first string will end with a letter or number. The next letter will always be a lower case k. I do not want this k character.
The difficulty is that the ending k will not always be the first k in the string. It might be the first k but it might not.
After the k, there is a second string. The second string will always start with an A or a B.
The ending of the second string will be in one of two forms:
a) it will end with a space then display the first three characters from the first string in lower case followed by a )
b) it will end with a ^K then display the first three characters from the first string in lower case.
For example:
1pppsx9YPar8Rvs75tJYWZq3eo8PgwbckB4m4zT7Yg042KIDYUE82e893hY ppp)
Should be:
1pppsx9YPar8Rvs75tJYWZq3eo8Pgwbc and B4m4zT7Yg042KIDYUE82e893hY - delete the k and the space then ppp.
For example:
1zzzsx9YPkr8Rvs75tJYWZq3eo8PgwbckA2m4zT7Yg042KIDYUE82e893hY^Kzzz
Should be:
1zzzsx9YPkar8Rvs75tJYWZq3eo8Pgwbc and A4m4zT7Yg042KIDYUE82e893hY - delete the second k and the ^Kzzz.
In the second example, we see that the first k is part of the first string. It is the k before the A that breaks up the first and second strings.
I hope there is a super grep expert who can help! Many thanks!
If your grep supports -P option, would you please try:
grep -a -Po "7[[:alnum:]]+(?=M-)" file
The -a option forces grep to read the input as a text file.
The -P option enables the perl-compatible regex.
The -o option tells grep to print only the matched substring(s).
The pattern (?=M-) is a zero-width lookahead assertion (introduced in
Perl) without including it in the result.
Alternatively you can also say with sed:
sed 's/M-/\n/g' file | sed -n 's/.*\(7[[:alnum:]]\+\).*/\1/p'
The first sed command splits the input file into miltiple lines by
replacing the substring M- with a newline.
It has two benefits: it breaks the lines to allow multiple matches with
sed and excludes the unnecessary portion M- from the input.
The next sed command extracts the desired pattern from the input.
It assumes your sed accepts \n in the replacement, which is
a GNU extension (not POSIX compliant). Otherwise please try (in case you are working on bash):
sed 's/M-/\'$'\n''/g' file | sed -n 's/.*\(7[[:alnum:]]\+\).*/\1/p'
[UPDATE]
(The requirement has been updated by the OP and the followings are solutions according to it.)
Let me assume the string which starts with 7 and ends with M- is always followed
by another (no more and no less than one) string which starts with 5x and ends
with ^ (ascii caret character) with junks in between.
Then would you please try the following:
grep -aPo "7[[:alnum:]]+M-.*?5x[[:alnum:]]+\^" file | grep -aPo "7[[:alnum:]]+(?=M-)|5x[[:alnum:]]+(?=\^)"
It executes the task in two steps (two cascaded greps).
The 1st grep narrows down the input data into the candidate substring
which will include the desired two sequences and junks in between.
The regex .*? in between matches any (ascii or binary) characters
except for a newline character.
The trailing ? enables the shortest match
which avoids the overrun due to the greedy nature of regex. The regex is intended to match junks in between.
The 2nd grep includes two regex's merged with a pipe | meaning logical OR.
Then it extracts two desired sequences.
A potential problem of grep solution is that grep is a line oriented command
and cannot include the newline character in the matched string.
If a newline character is included in the junks in between (I'm not sure about the possibility), the above solution will fail.
As a workaround, perl will provide flexible manipulations with binary data.
perl -0777 -ne '
while (/(7[[:alnum:]]+)M-.*?(5x[[:alnum:]]+)\^/sg) {
printf("%s\n%s\n", $1, $2);
}
' file
The regex is mostly same as that of grep because the -P option of grep means
perl-compatible.
It can capture multiple patterns at once in variables $1 and $2 hence just one regex is enough.
The -0777 option to the perl command tells perl to slurp all data
at once.
The s option at the end the regex makes a dot match a newline character.
The g option enables the global (multiple) match.
[UPDATE2]
In order to make the regex match either 5x or 6x, replace 5x with (5|6)x.
Namely:
grep -aPo "7[[:alnum:]]+M-.*?(5|6)x[[:alnum:]]+\^" file | grep -aPo "7[[:alnum:]]+(?=M-)|(5|6)x[[:alnum:]]+(?=\^)"
As mentioned before, the pipe | means OR. The OR operator has the lowest priority in the evaluation, hence you need to enclose them with parens in this case.
If there is a possibility any other number than 5 or 6 may appear, it will be safer to put [[:digit:]] instead, which matches any one digit betweeen 0 and 9:
grep -aPo "7[[:alnum:]]+M-.*?[[:digit:]]x[[:alnum:]]+\^" file | grep -aPo "7[[:alnum:]]+(?=M-)|[[:digit:]]x[[:alnum:]]+(?=\^)"
[UPDATE3]
(Answering the OP's requirement on March 9th)
Let me start with a perl code which regex will be relatively easier
to explain.
perl -0777 -ne 'while (/(1(.{3}).+)k([AB].*)[\013 ]\2/g){print "$1 $3\n"}' file
Output:
1pppsx9YPar8Rvs75tJYWZq3eo8Pgwbc B4m4zT7Yg042KIDYUE82e893hY
1zzzsx9YPkr8Rvs75tJYWZq3eo8Pgwbc A2m4zT7Yg042KIDYUE82e893hY
[Explanation of regex]
(1(.{3}).+)k([AB].*)[\013 ]\2
( start of the 1st capture group referred by $1 later
1 literal "1"
( start of the 2nd capture group referred by \2 later
.{3} a sequence of the identical three characters such as ppp or zzz
) end of the 2nd capture group
.+ followed by any characters with "greedy" match which may include the 1st "k"
) end of the 1st capture group
k literal "k"
( start of the 3rd capture group referred by $3 later
[AB].* the character "A" or "B" followed by any characters
) end of the 3rd capture group
[\013 ] followed by ^K or a whitespace
\2 followed by the capture group 2 previously assigned
When implementing it with grep, we will encounter a limitation of grep.
Although we want to extract multiple patterns from the input file,
the -e option (which can specify multiple search patterns) does not
work with -P option. Then we need to split the regex into two patterns
such as:
grep -Po "(1(.{3}).+)(?=k([AB].*)[\013 ]\2)" file
grep -Po "(1(.{3}).+)k\K([AB].*)(?=[\013 ]\2)" file
And the result will be:
1pppsx9YPar8Rvs75tJYWZq3eo8Pgwbc
1zzzsx9YPkr8Rvs75tJYWZq3eo8Pgwbc
B4m4zT7Yg042KIDYUE82e893hY
A2m4zT7Yg042KIDYUE82e893hY
Please be noted the order of output is not same as the order of appearance in the original file.
Another option will be to introduce ripgrep or rg which is a fast
and versatile version of grep. You may need to install ripgrep with
sudo apt install ripgrep or using other package handling tool.
An advantage of ripgrep is it supports -r (replace) option in which
you can make use of the backreferences:
rg -N -Po "(1(.{3}).+)k([AB].*)[\013 ]\2" -r '$1 $3' file
The -r '$1 $3' option prints the 1st and the 3rd capture groups and the result will be the same as perl.
In the general case, you can use the strings utility to pluck out ASCII from binary files; then of course you can try to grep that output for patterns that you find interesting.
Many traditional Unix utilities like grep have internal special markers which might get messed up by binary input. For example, the character \xFF was used for internal purposes by some versions of GNU grep so you can't grep for that character even if you can figure out a way to represent it in the shell (Bash supports $'\xff' for example).
A traditional approach would be to run hexdump or a similar utility, and then grep that for patterns. However, more modern scripting languages like Perl and Python make it easy to manipulate arbitrary binary data.
perl -ne 'print if m/\xff\xff/' </dev/urandom
This might work for you (GNU sed):
sed -En '/\n/!{s/M-\^G/\n/;s/7[^\n]*\n/\n&/};/^7[^\n]*/P;D' file
Split each line into zero or more lines that begin with 7 and end just before M-^G and only print such lines.
I have an issue with string manipulation in bash. I have a list of names, each name being composed of two parts, chars and numbers: for example
abcdef01234
I want to cut the last character before the numeric part starts, in this case
f
I think there is a regular expression to help me with this but just can't figure it out. AWK/sed solutions are accepted too. Hope someone can help.
Thank you.
In bash it can be done with parameter expansion with substring removal and string indexes, e.g.,
a=abcdef01234 # your string
tmp=${a%%[0-9]*} # remove all numbers from right
echo ${tmp:(-1)} # output last of remaining chars
Output: f
You can use a regexp like [a-zA-Z]+([a-zA-Z])[0-9]+. If you know how to use sed is pretty easy.
Check https://regex101.com/r/XCkKM5/1
The match will be the letter you want.
^\w+([a-zA-Z])\d+$
As a sed command (on OSX) this will be :
echo "abcdef12345" | sed -E "s#^[a-zA-Z]+([a-zA-Z])[0-9]+\$#\1#"
try following too once.
echo "abcdef01234" | awk '{match($0,/[a-zA-Z]+/);print substr($0,RLENGTH,1)}'
I have a list of names I assume is a file, file. Using grep's PCRE and (positive) lookahead:
$ grep -oP "[a-z](?=[^a-z])" file
f
It prints out the first (lowercase) letter followed by a non-(lowercase)-letter.
I want to print lines that contains single word only.
For example:
this is a line
another line
one
more
line
last one
I want to get the ones with single word only
one
more
line
EDIT: Guys, thank you for answers. Almost all of the answers work for my test file. However I wanted to list single lines in bash history. When I try your answers like
history | your posted commands
all of them below fails. Some only prints some numbers (might line numbers?)
You want to get all those commands in history that contain just one word. Considering that history prints the number of the command as a first column, you need to match those lines consisting in two words.
For this, you can say:
history | awk 'NF==2'
If you just want to print the command itself, say:
history | awk 'NF==2 {print $2}'
To rehash your problem, any line containing a space or nothing should be removed.
grep -Ev '^$| ' file
Your problem statement is unspecific on whether lines containing only punctuation might also occur. Maybe try
grep -Ex '[A-Za-z]+' file
to only match lines containing only one or more alphabetics. (The -x option implicitly anchors the pattern -- it requires the entire line to match.)
In Bash, the output from history is decorated with line numbers; maybe try
history | grep -E '^ *[0-9]+ [A-Za-z]+$'
to match lines where the line number is followed by a single alphanumeric token. Notice that there will be two spaces between the line number and the command.
In all cases above, the -E selects extended regular expression matching, aka egrep (basic RE aka traditional grep does not support e.g. the + operator, though it's available as \+).
Try this:
grep -E '^\s*\S+\s*$' file
With the above input, it will output:
one
more
line
If your test strings are in a file called in.txt, you can try the following:
grep -E "^\w+$" in.txt
What it means is:
^ starting the line with
\w any word character [a-zA-Z0-9]
+ there should be at least 1 of those characters or more
$ line end
And output would be
one
more
line
Assuming your file as texts.txt and if grep is not the only criteria; then
awk '{ if ( NF == 1 ) print }' texts.txt
If your single worded lines don't have a space at the end you can also search for lines without an empty space :
grep -v " "
I think that what you're looking for could be best described as a newline followed by a word with a negative lookahead for a space,
/\n\w+\b(?! )/g
example
I've UTF-8 plain text lists of usernames, 1 per line, in list1.txt and list2.txt. Note, in case pertinent, that usernames may contain regex characters e.g. ! ^ . ( and such as well as spaces.
I want to get and save to matches.txt a list of all unique values occurring in both lists. I've little command line expertise but this almost gets me there:
grep -Ff list1.txt list2.txt > matches.txt
...but that is treating "jdoe" and "jdoe III" as a match, returning "jdoe III" as the matched value. This is incorrect for the task. I need the per-line pattern match to be the whole line, i.e. from ^ to $. I've tried adding the -x flag but that gets no matches at all (edit: see comment to accepted answer - I got the flag order wrong).
I'm on OS X 10.9.5 and I don't have to use grep - another command line (tool) solving the problem will do.
All you need to do is add the -x flag to your grep query:
grep -Fxf list1.txt list2.txt > matches.txt
The -x flag will restrict matches to full line matches (each PATTERN becomes ^PATTERN$). I'm not sure why your attempt at -x failed. Maybe you put it after the -f, which must be immediately followed by the first file?
This awk will be handy than grep here:
awk 'FNR==NR{a[$0]; next} $0 in a' list1.txt list2.txt > matches.txt
$0 is the line, FNR is the current line number of the current file, NR is the overall line number (they are only the same when you are on the first file). a[$0] is a associative array (hash) whose key is the line. next will ensure that further clauses (the $0 in a) will not run if the current clause (the fact that this is the first file) did. $0 in a will be true when the current line has a value in the array a, thus only lines present in both will be displayed. The order will be their order of occurence in the second file.
A very simple and straightforward way to do it that doesn't require one to do all sorts of crazy things with grep is as follows
cat list1.txt list2.txt|grep match > matches.txt
Not only that, but it's also easier to remember, (especially if you regularly use cat).
grep -Fwf file1 file2 would match word to word !!
I am trying to extract a single string out of a line having many segments in a key-value order, but I don't get it as it matches much more than I want to.
This is my example line:
|SEGA~1~MAGIC~DESCRIPTION~~~M~TEST~|SEGB~34~12.11.2011~3~M~O~|SEGC~HELLO~WORLD~|
This lines is a kind concatenation of many segments into one line. Now I want to extract the the string at index 2 in the segment starting with SEGA.
So what I do is grep for this:
egrep -o 'SEGA(.*?)\~\|'
But it gives me the whole line, sometimes it gives me only the segment I am looking for. With the match I would split that segment by using the ~ character and take the third one.
Since I use .*? with the question mark I expected egrep to only match the content between SEGA and the very first occurrence of ~| which is right before SEGB and not the one at the end of SEGC or SEGB.
How can I tell grep to search for SEGA and give the whole content starting right after SEGA until THE VERY FIRST occurrence of ~|
You can use the -P(--perl-regexp) option in grep:
grep -oP '(?<=SEGA).*?(?=~\|)' file
If you want to include the trailing ~|, please remove the lookahead (?=...).
I think .*? (lazy) does not exit in egrep.
I'd suggest you break the line into lines on | and then grep from those:
$ echo "|SEGA~1~MAGIC~DESCRIPTION~~~M~TEST~|SEGB~34~12.11.2011~3~M~O~|SEGC~HELLO~WORLD~|" | sed -e 's/|/\n/g' | grep ^SEGA
SEGA~1~MAGIC~DESCRIPTION~~~M~TEST~