For some reason I cannot get this to output just the version of this line. I suspect it has something to do with how grep interprets the dash.
This command:
admin#DEV:~/TEMP$ sendemail
Yields the following:
sendemail-1.56 by Brandon Zehm
More output below omitted
The first line is of interest. I'm trying to store the version to variable.
TESTVAR=$(sendemail | grep '\s1.56\s')
Does anyone see what I am doing wrong? Thanks
TESTVAR is just empty. Even without TESTVAR, the output is empty.
I just tried the following too, thinking this might work.
sendemail | grep '\<1.56\>'
I just tried it again, while editing and I think I have another issue. Perhaps im not handling the output correctly. Its outputting the entire line, but I can see that grep is finding 1.56 because it highlights it in the line.
$ TESTVAR=$(echo 'sendemail-1.56 by Brandon Zehm' | grep -Eo '1.56')
$ echo $TESTVAR
1.56
The point is grep -Eo '1.56'
from grep man page:
-E, --extended-regexp
Interpret PATTERN as an extended regular expression (ERE, see below). (-E is specified by POSIX.)
-o, --only-matching
Print only the matched (non-empty) parts of a matching line, with each such part on a separate output
line.
Your regular expression doesn't match the form of the version. You have specified that the version is surrounded by spaces, yet in front of it you have a dash.
Replace the first \s with the capitalized form \S, or explicit set of characters and it should work.
I'm wondering: In your example you seem to know the version (since you grep for it), so you could just assign the version string to the variable. I assume that you want to obtain any (unknown) version string there. The regular expression for this in sed could be (using POSIX character classes):
sendemail |sed -n -r '1 s/sendemail-([[:digit:]]+\.[[:digit:]]+).*/\1/ p'
The -n suppresses the normal default output of every line; -r enables extended regular expressions; the leading 1 tells sed to only work on line 1 (I assume the version appears in the first line). I anchored the version number to the telltale string sendemail- so that potential other numbers elsewhere in that line are not matched. If the program name changes or the hyphen goes away in future versions, this wouldn't match any longer though.
Both the grep solution above and this one have the disadvantage to read the whole output which (as emails go these days) may be long. In addition, grep would find all other lines in the program's output which contain the pattern (if it's indeed emails, somebody might discuss this problem in them, with examples!). If it's indeed the first line, piping through head -1 first would be efficient and prudent.
jayadevan#jayadevan-Vostro-2520:~$ echo $sendmail
sendemail-1.56 by Brandon Zehm
jayadevan#jayadevan-Vostro-2520:~$ echo $sendmail | cut -f2 -d "-" | cut -f1 -d" "
1.56
Related
So here's my issue. I need to develop a small bash script that can grep a file containing account names (let's call it file.txt). The contents would be something like this:
accounttest
account2
account
accountbtest
account.test
Matching an exact line SHOULD be easy but apparently it's really not.
I tried:
grep "^account$" 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 $.
fgrep is the same as grep -F. grep also has the -x option which matches against whole lines only. You can combine these to get what you want:
grep -Fx account.test file.txt
Im starting to code bash and not the best but i have a situation. I have an output like:
Configuration file 'hello2.conf' is in use by process 735.
Ending
I want to extract the process ID 735.
I seen answers were to extract ONLY numbers from outputs but then i am left with 2735?
How can i go about extracting 735 from the output? I was thinking search for process then grab number after perhaps?
Thanks!
Use GNU grep with its Perl Compatible Regular Expression capabilities enabled with the -P flag and print only the matching entry using -o flag.
grep -Po 'process \K[0-9]+' <<<"Configuration file 'hello2.conf' is in use by process 735."
735
Use it in a command line as
.. | grep -Po 'process \K[0-9]+'
where the \K escape sequence stands for
\K: This sequence resets the starting point of the reported match. Any previously matched characters are not included in the final matched sequence.
RegEx Demo
You might want to use a regular expressions:
[[ "$line" =~ ([0-9]+)\.$ ]] && echo "${BASH_REMATCH[1]}"
This should match any number at the end of the line, select the number part, and print it!
Good Luck!
If you line remains the same, use cut -d" " -f 9
sed can extract only the numbers at the specific location of the message (using \(...\) match grouping and \1 replacement).
... | sed "s#^Configuration file '.*' is in use by process \([0-9]*\)\.#\1#"
I've never used sed apart from the few hours trying to solve this. I have a config file with parameters like:
test.us.param=value
test.eu.param=value
prod.us.param=value
prod.eu.param=value
I need to parse these and output this if REGIONID is US:
test.param=value
prod.param=value
Any help on how to do this (with sed or otherwise) would be great.
This works for me:
sed -n 's/\.us\././p'
i.e. if the ".us." can be replaced by a dot, print the result.
If there are hundreds and hundreds of lines it might be more efficient to first search for lines containing .us. and then do the string replacement... AWK is another good choice or pipe grep into sed
cat INPUT_FILE | grep "\.us\." | sed 's/\.us\./\./g'
Of course if '.us.' can be in the value this isn't sufficient.
You could also do with with the address syntax (technically you can embed the second sed into the first statement as well just can't remember syntax)
sed -n '/\(prod\|test\).us.[^=]*=/p' FILE | sed 's/\.us\./\./g'
We should probably do something cleaner. If the format is always environment.region.param we could look at forcing this only to occur on the text PRIOR to the equal sign.
sed -n 's/^\([^,]*\)\.us\.\([^=]\)=/\1.\2=/g'
This will only work on lines starting with any number of chars followed by '.' then 'us', then '.' and then anynumber prior to '=' sign. This way we won't potentially modify '.us.' if found within a "value"
For some reason I can't seem to find a straightforward answer to this and I'm on a bit of a time crunch at the moment. How would I go about inserting a choice line of text after the first line matching a specific string using the sed command. I have ...
CLIENTSCRIPT="foo"
CLIENTFILE="bar"
And I want insert a line after the CLIENTSCRIPT= line resulting in ...
CLIENTSCRIPT="foo"
CLIENTSCRIPT2="hello"
CLIENTFILE="bar"
Try doing this using GNU sed:
sed '/CLIENTSCRIPT="foo"/a CLIENTSCRIPT2="hello"' file
if you want to substitute in-place, use
sed -i '/CLIENTSCRIPT="foo"/a CLIENTSCRIPT2="hello"' file
Output
CLIENTSCRIPT="foo"
CLIENTSCRIPT2="hello"
CLIENTFILE="bar"
Doc
see sed doc and search \a (append)
Note the standard sed syntax (as in POSIX, so supported by all conforming sed implementations around (GNU, OS/X, BSD, Solaris...)):
sed '/CLIENTSCRIPT=/a\
CLIENTSCRIPT2="hello"' file
Or on one line:
sed -e '/CLIENTSCRIPT=/a\' -e 'CLIENTSCRIPT2="hello"' file
(-expressions (and the contents of -files) are joined with newlines to make up the sed script sed interprets).
The -i option for in-place editing is also a GNU extension, some other implementations (like FreeBSD's) support -i '' for that.
Alternatively, for portability, you can use perl instead:
perl -pi -e '$_ .= qq(CLIENTSCRIPT2="hello"\n) if /CLIENTSCRIPT=/' file
Or you could use ed or ex:
printf '%s\n' /CLIENTSCRIPT=/a 'CLIENTSCRIPT2="hello"' . w q | ex -s file
Sed command that works on MacOS (at least, OS 10) and Unix alike (ie. doesn't require gnu sed like Gilles' (currently accepted) one does):
sed -e '/CLIENTSCRIPT="foo"/a\'$'\n''CLIENTSCRIPT2="hello"' file
This works in bash and maybe other shells too that know the $'\n' evaluation quote style. Everything can be on one line and work in
older/POSIX sed commands. If there might be multiple lines matching the CLIENTSCRIPT="foo" (or your equivalent) and you wish to only add the extra line the first time, you can rework it as follows:
sed -e '/^ *CLIENTSCRIPT="foo"/b ins' -e b -e ':ins' -e 'a\'$'\n''CLIENTSCRIPT2="hello"' -e ': done' -e 'n;b done' file
(this creates a loop after the line insertion code that just cycles through the rest of the file, never getting back to the first sed command again).
You might notice I added a '^ *' to the matching pattern in case that line shows up in a comment, say, or is indented. Its not 100% perfect but covers some other situations likely to be common. Adjust as required...
These two solutions also get round the problem (for the generic solution to adding a line) that if your new inserted line contains unescaped backslashes or ampersands they will be interpreted by sed and likely not come out the same, just like the \n is - eg. \0 would be the first line matched. Especially handy if you're adding a line that comes from a variable where you'd otherwise have to escape everything first using ${var//} before, or another sed statement etc.
This solution is a little less messy in scripts (that quoting and \n is not easy to read though), when you don't want to put the replacement text for the a command at the start of a line if say, in a function with indented lines. I've taken advantage that $'\n' is evaluated to a newline by the shell, its not in regular '\n' single-quoted values.
Its getting long enough though that I think perl/even awk might win due to being more readable.
A POSIX compliant one using the s command:
sed '/CLIENTSCRIPT="foo"/s/.*/&\
CLIENTSCRIPT2="hello"/' file
Maybe a bit late to post an answer for this, but I found some of the above solutions a bit cumbersome.
I tried simple string replacement in sed and it worked:
sed 's/CLIENTSCRIPT="foo"/&\nCLIENTSCRIPT2="hello"/' file
& sign reflects the matched string, and then you add \n and the new line.
As mentioned, if you want to do it in-place:
sed -i 's/CLIENTSCRIPT="foo"/&\nCLIENTSCRIPT2="hello"/' file
Another thing. You can match using an expression:
sed -i 's/CLIENTSCRIPT=.*/&\nCLIENTSCRIPT2="hello"/' file
Hope this helps someone
The awk variant :
awk '1;/CLIENTSCRIPT=/{print "CLIENTSCRIPT2=\"hello\""}' file
I had a similar task, and was not able to get the above perl solution to work.
Here is my solution:
perl -i -pe "BEGIN{undef $/;} s/^\[mysqld\]$/[mysqld]\n\ncollation-server = utf8_unicode_ci\n/sgm" /etc/mysql/my.cnf
Explanation:
Uses a regular expression to search for a line in my /etc/mysql/my.cnf file that contained only [mysqld] and replaced it with
[mysqld]
collation-server = utf8_unicode_ci
effectively adding the collation-server = utf8_unicode_ci line after the line containing [mysqld].
I had to do this recently as well for both Mac and Linux OS's and after browsing through many posts and trying many things out, in my particular opinion I never got to where I wanted to which is: a simple enough to understand solution using well known and standard commands with simple patterns, one liner, portable, expandable to add in more constraints. Then I tried to looked at it with a different perspective, that's when I realized i could do without the "one liner" option if a "2-liner" met the rest of my criteria. At the end I came up with this solution I like that works in both Ubuntu and Mac which i wanted to share with everyone:
insertLine=$(( $(grep -n "foo" sample.txt | cut -f1 -d: | head -1) + 1 ))
sed -i -e "$insertLine"' i\'$'\n''bar'$'\n' sample.txt
In first command, grep looks for line numbers containing "foo", cut/head selects 1st occurrence, and the arithmetic op increments that first occurrence line number by 1 since I want to insert after the occurrence.
In second command, it's an in-place file edit, "i" for inserting: an ansi-c quoting new line, "bar", then another new line. The result is adding a new line containing "bar" after the "foo" line. Each of these 2 commands can be expanded to more complex operations and matching.
I'm running sed as a part of a shell script to clean up bind logs for insertion into a database.
One of the sed commands is the following:
sed -i 's/-/:/g' $DPath/named.query.log
This turns out to be problematic as it disrupts any resource requests that also include a dash (I'm using : as a delimiter for an awk statement further down).
My question is how do I limit the sed command above to only the first ten characters of the line? I haven't seen a specific switch that does this, and I'm nowhere near good enough with RegEx to even start on developing one that works. I can't just use regex to match the preceding numbers because it's possible that the pattern could be part of a resource request. Heck, I can't even use pattern matching for ####-##-## because, again, it could be part of the resource.
Any ideas are much appreciated.
It's [almost always] simpler with awk:
awk '{target=substr($0,1,10); gsub(/-/,":",target); print target substr($0,11)}' file
I think the shortest solution, and perhaps the simplest, is provided by sed itself, rather than awk[ward]:
sed "h;s/-/:/g;G;s/\(..........\).*\n........../\1/"
Explanation:
(h) copy everything to the hold space
(s) do the substitution (to the entire pattern space)
(G) append the hold space, with a \n separator
(s) delete the characters up to the tenth after the \n, but keep the first ten.
Some test code:
echo "--------------------------------" > foo
sed -i "h;s/-/:/g;G;s/\(..........\).*\n........../\1/" foo
cat foo
::::::::::----------------------
I'm not sure how make sed do it per se, however, I do know that you can feed sed the first 10 characters then paste the rest back in, like so:
paste -d"\0" <(cut -c1-10 $DPath/named.query.log | sed 's/\-/:/g') <(cut -c11- $DPath/named.query.log)
You can do the following:
cut -c 1-10 $DPath/named.query.log | sed -i 's/-/:/g'
The cut statemnt takes only the first 10 chars of each line in that file. The output of that should be piped in a file. As of now it will just output to your terminal