grep - List out items ending with '(' from the file - grep

I have a list and need to extract the below patterns
funcName(
funcNameOther(
all function names ending with '( succeeded by a newline'

egrep -o '.*\($' does the trick for me, if that's what you're looking after.

grep '($' file
If that doesn't do what you want, provide some more useful sample input and expected output.

Related

Grep Individual Commands not working when combined in Multi Pattern grep command

I have a need to perform multiple grep matches as part of the same grep command. When I run them individually, they work fine. But not when together. I hope someone could either show me a solution or perhaps can help me find a work-around. Here is sample stream:
(string start..) RollUp:"V" Enzyme:"ENZA ENZB ENZD ENZE" (..string end)
In the first command I am needing to isolate all RollUp substrings.Value is always A or V:
grep -o "RollUp:\"[AV]\""
In the second command I am needing to isolate all combinations of Enzyme values (1-20 total, spaces in between, don't know values names). This command works:
grep -oE 'Enzyme:[[:space:]]*"[^"]+"'
However, I need to match both patterns as part of same stream. When I try:
grep -oE "RollUp:\"[AV]\""\|Enzyme:[[:space:]]*"[^"]+""
, nothing is returned. I would be grateful for any ideas for getting this double grep pattern match to work. Thank you!
regex someting[^"]+ : this means string something followed by anything till next " is seen. Here + sign means , at least one or more match.
grep -oE 'RollUp:"[^"]+|Enzyme:[[:space:]]*"[^"]+"' file

Match pattern ending with a certain character in grep

This is a common problem I encounter when using grep. Say the pattern is 'chr1' in a third column of a file, when I do the following:
grep 'chr1' file
How can I avoid getting the results including chr10, chr11, chr13 etc as well?
Thanks!
It seems this works:
grep -w 'chr1' file
Since you're interested in values in specific columns, you're much better off using awk:
awk '$3 == "chr1"' file

ksh - search for multiple strings and write lines to file

Any help would be greatly appreciated. I can read code and figure it out, but I have trouble writing from scratch.
I need help starting a ksh script that would search a file for multiple strings and write each line containing one of those strings to an output file.
If I use the following command:
$ grep "search pattern" file >> output file
...that does what I want it to. But I need to search multiple strings, and write the output in the order listed in the file.
Again... any help would be great! Thank you in advance!
Have a look at the regular expression manuals. You can specify multiple strings in the search expression such as grep "John|Bill"
Man grep will teach you a lot about regular expressions, but there are several online sites where you try them out, such as regex101 and (more colorful) regexr.
Sometimes you need egrep.
egrep "first substring|second substring" file
When you have a lot substrings you can put them in a variable first
findalot="first substring|second substring"
findalot="${findalot}|third substring"
findalot="${findalot}|find me too"
skipsome="notme"
skipsome="${skipsome}|dirty words"
egrep "${findalot}" file | egrep -v "${skipsome}"
Use "-f" in grep .
Write all the strings you want to match in a file ( lets say pattern_file , the list of strings should be one per line)
and use grep like below
grep -f pattern_file file > output_file

grep from beginning of found word to end of word

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.

Opposite of "only-matching" in grep?

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: $&"

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