I have the following IP addresses in a file
3.3.3.1
3.3.3.11
3.3.3.111
I am using this file as input file to another program. In that program it will grep each IP address. But when I grep the contents I am getting some wrong outputs.
like
cat testfile | grep -o 3.3.3.1
but I am getting output like
3.3.3.1
3.3.3.1
3.3.3.1
I just want to get the exact output. How can I do that with grep?
Use the following command:
grep -owF "3.3.3.1" tesfile
-o returns the match only and not the whole line.-w greps for whole words, meaning the match must be enclosed in non word chars like <space>, <tab>, ,, ; the start or the end of the line etc. It prevents grep from matching 3.3.3.1 out of 3.3.3.111.
-F greps for fixed strings instead of patterns. This prevents the . in the IP address to be interpreted as any char, meaning grep will not match 3a3b3c1 (or something like this).
To match whole words only, use grep -ow 3.3.3.1 testfile
UPDATE: Use the solution provided by hek2mgl as it is more robust.
You may use anhcors.
grep '^3\.3\.3\.1$' file
Since by default grep uses regex, you need to escape the dots in-order to make grep to match literal dot character.
Related
i found many similar questions about my issue but i still don't find the correct one for me.
I need to grep for the content of a variable plus a dot but it doesn't run escaping the dot after the variable. For example:
The file content is
item.
newitem.
My variable content is item. and i want to grep for the exact word, therefore I must use -w and not -F but with the command I can't obtain the correct output:
cat file | grep -w "$variable\."
Do you have suggestions please?
Hi, I have to rectify my scenario. My file contains some FQDN and for some reasons I have to look for hostname. with the dot.
Unfortunatelly the grep -wF doesn't run:
My file is
hostname1.domain.com
hostname2.domain.com
and the command
cat file | grep -wF hostname1.
doesn't show any output. I have to find another solution and I'm not sure that grep could help.
If $variable contains item., you're searching for item.\. which is not what you want. In fact, you want -F which interprets the pattern literally, not as a regular expression.
var=item.
echo $'item.\nnewitem.' | grep -F "$var"
Try:
grep "\b$word\."
\b: word boundary
\.: the dot itself is a word boundary
Following awk solution may help you in same.
awk -v var="item." '$0==var' Input_file
You are dereferencing variable and append \. to it, which results in calling
cat file | grep -w "item.\.".
Since grep accepts files as parameter, calling grep "item\." file should do.
from man grep
-w, --word-regexp
Select only those lines containing matches that form whole words. The test is that the matching substring must either be at the beginning of the line, or preceded by a non-word constituent
character. Similarly, it must be either at the end of the line or followed by a non-word constituent character. Word-constituent characters are letters, digits, and the underscore.
and
The Backslash Character and Special Expressions
The symbols \< and \> respectively match the empty string at the beginning and end of a word. The symbol \b matches the empty string at the edge of a word, and \B matches the empty string
provided it's not at the edge of a word. The symbol \w is a synonym for [[:alnum:]] and \W is a synonym for [^[:alnum:]].
as the last character is a . it must be followed by a non word [A-Za-z0-9_] however the next character is d
grep '\<hostname1\.'
should work as \< ensures previous chracter is not a word constituent.
You can dynamically construct the search pattern and then call grep
rexp='^hostname1\.'
grep "$rexp" file.txt
The single quotes tell bash not to interpret special characters in the variable. Double quotes tell bash to allow replacing $rexp with its value. The caret ( ^ ) in the expression tells grep to look for lines starting with 'hostname1.'
So here's my issue. I need to develop a small bash script that can grep a file containing account names (let's call it file.txt). The contents would be something like this:
accounttest
account2
account
accountbtest
account.test
Matching an exact line SHOULD be easy but apparently it's really not.
I tried:
grep "^account$" file.txt
The output is:
account
So in this situation the output is OK, only "account" is displayed.
But if I try:
grep "^account.test$" file.txt
The output is:
accountbtest
account.test
So the next obvious solution that comes to mind, in order to stop interpreting the dot character as "any character", is using fgrep, right?
fgrep account.test file.txt
The output, as expected, is correct this time:
account.test
But what if I try now:
fgrep account file.txt
Output:
accounttest
account2
account
accountbtest
account.test
This time the output is completely wrong, because I can't use the beginning/end line characters with fgrep.
So my question is, how can I properly grep a whole line, including the beginning and end of line special characters, while also matching exactly the "." character?
EDIT: Please note that I do know that the "." character needs to be escaped, but in my situation, escaping is not an option, because of further processing that needs to be done to the account name, which would make things too complicated.
The . is a special character in regex notation which needs to be escaped to match it as a literal string when passing to grep, so do
grep "^account\.test$" file.txt
Or if you cannot afford to modify the search string use the -F flag in grep to treat it as literal string and not do any extra processing in it
grep -Fx 'account.test' file.txt
From man grep
-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.
-x, --line-regexp
Select only those matches that exactly match the whole line. For a regular expression pattern, this is like parenthesizing the pattern and then surrounding it with ^ and $.
fgrep is the same as grep -F. grep also has the -x option which matches against whole lines only. You can combine these to get what you want:
grep -Fx account.test file.txt
Here is my problem : I use grep to find a string into multiple files.
Let's say I am looking for the word "balloon". grep is returning lines containing balloon like "Here is a balloon", "loginxxballoonx", "balloon123" etc. This is not a problem except for one case : I want to ignore the line if it finds "/balloon/".
How can I look for every "balloon" strings in multiple files, but ignore those with / before and after (ignore "/balloon/")
EDIT : I will precise my problem a bit more : my strings to search for are stored in a file. I use grep -f mytokenfile to search for every strings stored in my "mytokenfile" file. For example, my file "mytokenfile" looks like this :
balloon
avion
car
bus
I would like to get all the lines containing these strings, with or without prefixes/suffixes, except if the prefix and suffix are "/".
Should work by using the negation sign ^
grep [^/]balloon[^/] ballonfile
Edit:
But this doesn't work if there is a 'balloon' not prefixed or suffixed by any other characters.
Use the following approach(considering that there could be a line with multiple occurrences of search keyword such as loginxxballoonx, sme text /balloon/ text):
cat testfile | grep '[^/]balloon[^/]' | grep -v '/balloon/'
-v (--invert-match)
Invert the sense of matching, to select non-matching lines. (-v is
specified by POSIX.)
I have an input file
RAKESH_ONE
RAKESH-TWO
RAKESH123
RAKESHTHREE
/RAKESH/
FIVERAKESH
456RAKESH
WELCOME123
This is RAKESH
I would like to get the output
RAKESH_ONE
RAKESH-TWO
/RAKESH/
This is RAKESH
I want to print the line matching the pattern RAKESH. If the pattern is prefixed or suffixed with alphanumeric we should avoid it.
([^a-zA-Z0-9]+|^)RAKESH([^a-zA-Z0-9]+|$)
This will match patterns on the lines without alphanumeric prefixes or suffixes. It will not match the whole line, but if used with grep or sed you can output just the lines you need.
UPDATE
As requested, here's the full grep command. Use the -E option to use extended regex:
grep -E "([^a-zA-Z0-9]+|^)RAKESH([^a-zA-Z0-9]+|$)" file.txt
I have a webpage source in a text doc, there's a few lines like so:
"rid" : 'http://web.site/urlhere',
How do I use Linux/terminal to grep just the http://web.site/urlhere portion?
You can pass the -o option to grep to tell it to only display the matching pattern.
grep -o http://web.site/urlhere somefile.txt
Assuming you're looking for generic URLs, you could start with something like this (and probably improve it):
grep -o "'http.*'" someFile.txt | sed "s/'//g"
This will search for the text http after a single quote and will include all the characters from that line until the last single quote. It will then pipe the result (only the matching pattern) to sed and remove the single quotes.
Note: You could run into trouble if you have more single quotes after the url (but your question doesn't mention that)...
Since you're question is very non-specific, there are probably many other input conditions that could cause problems, but the above should be a good starting point.
More info on grep: http://unixhelp.ed.ac.uk/CGI/man-cgi?grep