I have a file where I want to grep for lines that start with either -rwx or drwx AND end in any number.
I've got this, but it isnt quite right. Any ideas?
grep [^.rwx]*[0-9] usrLog.txt
The tricky part is a regex that includes a dash as one of the valid characters in a character class. The dash has to come immediately after the start for a (normal) character class and immediately after the caret for a negated character class. If you need a close square bracket too, then you need the close square bracket followed by the dash. Mercifully, you only need dash, hence the notation chosen.
grep '^[-d]rwx.*[0-9]$' "$#"
See: Regular Expressions and grep for POSIX-standard details.
It looks like you were on the right track... The ^ character matches beginning-of-line, and $ matches end-of-line. Jonathan's pattern will work for you... just wanted to give you the explanation behind it
It should be noted that not only will the caret (^) behave differently within the brackets, it will have the opposite result of placing it outside of the brackets. Placing the caret where you have it will search for all strings NOT beginning with the content you placed within the brackets. You also would want to place a period before the asterisk in between your brackets as with grep, it also acts as a "wildcard".
grep ^[.rwx].*[0-9]$
This should work for you, I noticed that some posters used a character class in their expressions which is an effective method as well, but you were not using any in your original expression so I am trying to get one as close to yours as possible explaining every minor change along the way so that it is better understood. How can we learn otherwise?
You probably want egrep. Try:
egrep '^[d-]rwx.*[0-9]$' usrLog.txt
are you parsing output of ls -l?
If you are, and you just want to get the file name
find . -iname "*[0-9]"
If you have no choice because usrLog.txt is created by something/someone else and you absolutely must use this file, other options include
awk '/^[-d].*[0-9]$/' file
Ruby(1.9+)
ruby -ne 'print if /^[-d].*[0-9]$/' file
Bash
while read -r line ; do case $line in [-d]*[0-9] ) echo $line; esac; done < file
Many answers provided for this question. Just wanted to add one more which uses bashism-
#! /bin/bash
while read -r || [[ -n "$REPLY" ]]; do
[[ "$REPLY" =~ ^(-rwx|drwx).*[[:digit:]]+$ ]] && echo "Got one -> $REPLY"
done <"$1"
#kurumi answer for bash, which uses case is also correct but it will not read last line of file if there is no newline sequence at the end(Just save the file without pressing 'Enter/Return' at the last line).
Related
I'm trying to isolate lines that contain the following: '[Homosapiens]' from a file.
My file looks something like this:
br
blabla
>blabldi[Homosapiens]
>skadlfjkl[Musmusculus]
I only want to isolate the third line.
I have tried the following:
grep -F '\*[Homosapiens]' mytext.txt
and
fgrep '\*[Homosapiens]' mytext.txt
but both are not working.
Can anyone solve this problem?
What about the following:
fgrep "[Homosapiens]" mytext.txt
Or:
grep "\[Homosapiens\]" mytext.txt
Two remarks:
grep (or whatever of its family members fgrep, egrep, ...) search for an entry inside a line of text, so there is no need to try to fit the whole line inside your grep expression.
The square brackets have a special meaning (grep [a-e] means a search for all letters from 'a' to 'e'). Using a backslash in front of a square bracket disables that feature and gives you the opportunity to look for a square bracket.
I'm writing a script which can parse an HTML document. I would like to remove two lines, how does sed work with newlines? I tried
sed 's/<!DOCTYPE.*\n<h1.*/<newstring>/g'
which didn't work. I tried this statement but it removes the whole document because it seems to remove all newlines:
sed ':a;N;$!ba;s/<!DOCTYPE.*\n<h1.*\n<b.*/<newstring>/g'
Any ideas? Maybe I should work with awk?
For the simple task of removing two lines if each matches some pattern, all you need to do is:
sed '/<!DOCTYPE.*/{N;/\n<h1.*/d}'
This uses an address matching the first line you want to delete. When the address matches, it executes:
Next - append the next line to the current pattern-space (including \n)
Then, it matches on an address for the contents of the second line (following \n). If that works it executes:
delete - discard current input and start reading next unread line
If d isn't executed, then both lines will print by default and execution will continue as normal.
To adjust this for three lines, you need only use N again. If you want to pull in multiple lines until some delimiter is reached, you can use a line-pump, which looks something like this:
/<!DOCTYPE.*/{
:pump
N
/some-regex-to-stop-pump/!b pump
/regex-which-indicates-we-should-delete/d
}
However, writing a full XML parser in sed or awk is a Herculean task and you're likely better off using an existing solution.
If an xml parsing tool is definitely not an option, awk maybe an option:
awk '/<!DOCTYPE/ { lne=NR+1;next } NR==lne && /<h1/ { next }1' file
When we encounter a line with "<!DOCTYPE" set the variable lne to the line number + 1 (NR+1) and then skip to the next line. Then when the line is equal to lne (NR==lne) and the line contains "<h1", skip to the next line. Print all other lines by using 1.
My solution for a document like this:
<b>...
<first...
<second...
<third...
<a ...
this awk command works well:
awk -v RS='<first[^\n]*\n<second[^\n]*\n<third[^\n]*\n' '{printf "%s", $0}'
that's all.
This might work for you (GNU sed):
sed 'N;/<!DOCTYPE.*\n<h1.*/d;P;D' file
Append the following line and if the pattern matches both lines in the pattern space delete them.
Otherwise, print then delete the first of the two lines and repeat.
To replace the two lines with another string, use:
sed 'N;s/<!DOCTYPE.*\n<h1.*/another string/;P;D'
There is a possibility to search using grep in TextWrangler
I want to find and replace the following word: bauvol, but not bauvolumen.
I tried typing ^bauvol$ into the search field but that didn't do the trick, it didn't find anything, although the word is clearly there.
I think it's because, in grep, the ^and $signify start and end of line, not a word?!
You want to use \b as word boundaries, as #gromi08 said:
\bbauvol\b
If you want to copy any portion of this word (so you can replace it, modify it, change the case, etc.) it is usually best to wrap it in ( and ) braces so you can reference them in the Replace box:
Find:
(\bbauvol\b)
Replace:
<some_tag>\1</some_tag>
Did you have anything specific you were trying to do with the result once you found it (cut it, duplicate it, etc.)?
Use the -w option of grep (see grep man-page.
This option searches for the expression as a word.
Therefore the command will be:
cat file.txt | grep -w bauvol
And yes, ^ and $ are for start and end of line.
I am trying to grep the output of a command that outputs unknown text and a directory per line. Below is an example of what I mean:
.MHuj.5.. /var/log/messages
The text and directory may be different from time to time or system to system. All I want to do though is be able to grep the directory out and send it to a variable.
I have looked around but cannot figure out how to grep to the end of a word. I know I can start the search phrase looking for a "/", but I don't know how to tell grep to stop at the end of the word, or if it will consider the next "/" a new word or not. The directories listed could change, so I can't assume the same amount of directories will be listed each time. In some cases, there will be multiple lines listed and each will have a directory list in it's output. Thanks for any help you can provide!
If your directory paths does not have spaces then you can do:
$ echo '.MHuj.5.. /var/log/messages' | awk '{print $NF}'
/var/log/messages
It's not clear from a single example whether we can generalize that e.g. the first occurrence of a slash marks the beginning of the data you want to extract. If that holds, try
grep -o '/.*' file
To fetch everything after the last space, try
grep -o '[^ ]*$' file
For more advanced pattern matching and extraction, maybe look at sed, or Awk or Perl or Python.
Your line can be described as:
^\S+\s+(\S+)$
That's assuming whitespace is your delimiter between the random text and the directory. It simply separates the whitespace from the non-whitespace and captures the second part.
Or you might want to look into the word boundary character class: \b.
I know you said to use grep, but I can't help to mention that this is trivially done using awk:
awk '{ print $NF }' input.txt
This is assuming that a whitespace is the delimiter and that the path does not contain any whitespaces.
Is there any way to do the opposite of showing only the matching part of strings in grep (the -o flag), that is, show everything except the part that matches the regex?
That is, the -v flag is not the answer, since that would not show files containing the match at all, but I want to show these lines, but not the part of the line that matches.
EDIT: I wanted to use grep over sed, since it can do "only-matching" matches on multi-line, with:
cat file.xml|grep -Pzo "<starttag>.*?(\n.*?)+.*?</starttag>"
This is a rather unusual requirement, I don't think grep would alternate the strings like that. You can achieve this with sed, though:
sed -n 's/$PATTERN//gp' file
EDIT in response to OP's edit:
You can do multiline matching with sed, too, if the file is small enough to load it all into memory:
sed -rn ':r;$!{N;br};s/<starttag>.*?(\n.*?)+.*?<\/starttag>//gp' file.xml
You can do that with a little help from sed:
grep "pattern" input_file | sed 's/pattern//g'
I don't think there is a way in grep.
If you use ack, you could output Perl's special variables $` and $' variables to show everything before and after the match, respectively:
ack string --output="\$`\$'"
Similarly if you wanted to output what did match along with other text, you could use $& which contains the matched string;
ack string --output="Matched: $&"