I need to search for a PHP variable $someVar. However, Grep thinks that I am trying to run a regex and is complaining:
$ grep -ir "Something Here" * | grep $someVar
Usage: grep [OPTION]... PATTERN [FILE]...
Try `grep --help' for more information.
$ grep -ir "Something Here" * | grep "$someVar"
<<Here it returns all rows with "someVar", not only those with "$someVar">>
I don't see an option for telling grep not to interpret the string as a regex, but to include the $ as just another string character.
Use fgrep (deprecated), grep -F or grep --fixed-strings, to make it treat the pattern as a list of fixed strings, instead of a regex.
For reference, the documentation mentions (excerpts):
-F --fixed-strings Interpret the pattern as a list of fixed
strings (instead of regular expressions), separated by newlines, any
of which is to be matched. (-F is specified by POSIX.)
fgrep is the same as grep -F. Direct invocation as fgrep is
deprecated, but is provided to allow historical applications that rely
on them to run unmodified.
For the complete reference, check:
https://www.gnu.org/savannah-checkouts/gnu/grep/manual/grep.html
grep -F is a standard way to tell grep to interpret argument as a fixed string, not a pattern.
You have to tell grep you use a fixed-string, instead of a pattern, using '-F' :
grep -ir "Something Here" * | grep -F \$somevar
In this question, the main issue is not about grep interpreting $ as a regex. It's about the shell substituting $someVar with the value of the environment variable someVar, likely the empty string.
So in the first example, it's like calling grep without any argument, and that's why it gives you a usage output. The second example should not return all rows containing someVar but all lines, because the empty string is in all lines.
To tell the shell to not substitute, you have to use '$someVar' or \$someVar. Then you'll have to deal with the grep interpretation of the $ character, hence the grep -F option given in many other answers.
So one valid answer would be:
grep -ir "Something Here" * | grep '$someVar'
+1 for the -F option, it shall be the accepted answer.
Also, I had a "strange" behaviour while searching for the -I.. pattern in my files, as the -I was considered as an option of grep ; to avoid such kind of errors, we can explicitly specify the end of the arguments of the command using --.
Example:
grep -HnrF -- <pattern> <files>
Hope that'll help someone.
Escape the $ by putting a \ in front of it.
Related
I am trying to see if my nohup file contains the words that I am looking for. If it does, then I need to put that into tmp file.
So I am currently using:
if grep -q "Started|missing" $DIR3/$dirName/nohup.out
then
grep -E "Started|missing" "$DIR3/$dirName/nohup.out" > tmp
fi
But it never goes into the if statement even if there are words that I am looking for.
How can I fix this?
Since basic sed uses BRE, regex alternation operator is represented by \| . | matches a literal | symbol. And you don't need to touch | symbol in the grep which uses ERE.
if grep -q "Started\|missing" $DIR3/$dirName/nohup.out
You should use egrep instead of grep (Avinash Raj has explained that in other words already in his answer).
I would generally recommend using egrep as a default for everyday use (even though many expressions only contain the basic regular expression syntax). From a practical point the standard grep is only interesting for performance reasons.
Details about the advantages of grep vs. egrep can be found in that superuser question.
When you only put the grep results into the tmp-file, you do not want to grep the file twice.
You can not use
egrep "Started|missing" $DIR3/$dirName/nohup.out > tmp
since that would create an empty tmp file when nothing is found.
You can remove empty files with if [ ! -s tmp ] or use another solution:
Redirectong the grep results without grepping again can be done with
rm -f tmp 2>/dev/null
egrep "Started|missing" $DIR3/$dirName/nohup.out | while read -r strange_line; do
echo "${strange_line}" >> tmp
done
Say for instance I'm searching a line that is like this:
Color asdf
and I use grep to find that line, like grep asdf file.txt
How would I then display Color? Learning linux is hard.
With the command line tool sed you can replace stings by using regular expressions:
echo "Color asdf" | sed 's/\([^ ]*\).*/\1/'
This part: \([^ ]*\).* is a regular expresion. The first part of the regex: [^ ]*, matches any character except a space as many times as possible and what's between the \( and \) is being captured in the variable \1. Then you also match the remaining part of the string with .* and replace all of that with only the first word which was captured by \([^ ]*\) by using \1 in the replace part of the sed command.
Here some more info about sed:
http://linux.about.com/od/commands/a/Example-Uses-Of-Sed-Cmdsedxa.htm
You could use sed:
sed -n 's/[[:space:]][[:space:]]*asdf$//p' file.txt
Details:
The -n option tells sed not to print the pattern space automatically. Basically, it doesn't output anything unless you tell it to.
The s command of sed replaces text. Here, if a line ends with asdf, preceded by at least one whitespace character, we replace all of that with nothing and then print the line (notice the p flag at the end of the s command). The printing is only done if something was actually replaced. More information about the s command can be found e. g. in the GNU sed manual.
Edit for clarity: When using single quotes, parameter expansion does not work and thus, variables won't be replaced. To use variables, use double quotes:
search=asdf
sed -n "s/[[:space:]][[:space:]]*${search}\$//p" file.txt
If you'd really like to use grep here, you could pipe the output from grep into cut:
grep -h asdf *.txt | cut -s -d -f 1
Note that there have to be two spaces after the -d option to cut - the first tells cut to use a blank as the field delimiter (I'm assuming your fields are blank-delimited rather than tab-delimited), while the second separates the -d option from the following option (-f).
But, yeah, sed or awk are probably your friends here... :-)
you can color pattern in the line using grep
grep --colour -o 'asdf' file.txt
edit: the -o option will print only the patterns
I need to search in a directory of files which has pattern1 but not pattern2.
look at the -v flag to grep. You can pipe multiple calls to grep together, which is probably the simplest approach here. One to look for pattern1, and another to grep -v pattern2.
grep pattern1 $(grep -L pattern2 *)
is probably the easiest way to do it, if I understand correctly what you want. -L means "print just the names of all files that do not contain this pattern"; it's the inverse of -l. This will not work correctly if you have files with whitespace or some other shell metacharacters in their names.
You can add a grep to the first grep:
grep -r "this pattern" /path | grep -v "not this patten"
HTH
Francisco
I'm trying to write a script with a file as an argument that greps the text file to find any word that starts with a capital and has 8 letters following it. I'm bad with syntax so I'll show you my code, I'm sure it's an easy fix.
grep -o '[A-Z][^ ]*' $1
I'm not sure how to specify that:
a) it starts with a capital letter, and
b)that it's a 9 letter word.
Cheers
EDIT:
As an edit I'd like to add my new code:
while read p
do
echo $p | grep -Eo '^[A-Z][[:alpha:]]{8}'
done < $1
I still can't get it to work, any help on my new code?
'[A-Z][^ ]*' will match one character between A and Z, followed by zero or more non-space characters. So it would match any A-Z character on its own.
Use \b to indicate a word boundary, and a quantifier inside braces, for example:
grep '\b[A-Z][a-z]\{8\}\b'
If you just did grep '[A-Z][a-z]\{8\}' that would match (for example) "aaaaHellosailor".
I use \{8\}, the braces need to be escaped unless you use grep -E, also known as egrep, which uses Extended Regular Expressions. Vanilla grep, that you are using, uses Basic Regular Expressions. Also note that \b is not part of the standard, but commonly supported.
If you use ^ at the beginning and $ at the end then it will not find "Wiltshire" in "A Wiltshire pig makes great sausages", it will only find lines which just consist of a 9 character pronoun and nothing else.
This works for me:
$ echo "one-Abcdefgh.foo" | grep -o -E '[A-Z][[:alpha:]]{8}'
$ echo "one-Abcdefghi.foo" | grep -o -E '[A-Z][[:alpha:]]{8}'
Abcdefghi
$
Note that this doesn't handle extensions or prefixes. If you want to FORCE the input to be a 9-letter capitalized word, we need to be more explicit:
$ echo "one-Abcdefghij.foo" | grep -o -E '\b[A-Z][[:alpha:]]{8}\b'
$ echo "Abcdefghij" | grep -o -E '\b[A-Z][[:alpha:]]{8}\b'
$ echo "Abcdefghi" | grep -o -E '\b[A-Z][[:alpha:]]{8}\b'
Abcdefghi
$
I have a test file named 'testfile' with the following content:
Aabcdefgh
Babcdefgh
cabcdefgh
eabcd
Now you can use the following command to grep in this file:
grep -Eo '^[A-Z][[:alpha:]]{8}' testfile
The code above is equal to:
cat testfile | grep -Eo '^[A-Z][[:alpha:]]{8}'
This matches
Aabcdefgh
Babcdefgh
How can I make use of grep in cygwin to find all files that contain BOTH words.
This is what I use to search all files in a directory recursively for one word:
grep -r "db-connect.php" .
How can I extend the above to look for files that contain both "db-connect.php" AND "version".
I tried this: grep -r "db-connect.php\|version" . but this is an OR i.e. it gets file that contain one or the other.
Thanks all for any help
grep -r db-connect.php . | grep version
If you want to grep for several strings in a file which have different lines, use the following command:
grep -rl expr1 | xargs grep -l expr2 | xargs grep -l expr3
This will give you a list of files that contain expr1, expr2, and expr3.
Note that if any of the file names in the directory contains spaces, these files will produce errors. This can be fixed by adding -0 I think to grep and xargs.
grep "db-connect.php" * | cut -d: -f1 | xargs grep "version"
I didn't try it in recursive mode but it should be the same.
To and together multiple searches, use multiple lookahead assertions, one per thing looked for apart from the last one:
instead of writing
grep -P A * | grep B
you write
grep -P '(?=.*A)B' *
grep -Pr '(?=.*db-connect\.php)version' .
Don’t write
grep -P 'A.*B|B.*A' *
because that fails on overlaps, whereas the (?=…)(?=…) technique does not.
You can also add in NOT operators as well. To search for lines that don’t match X, you normally of course use -v on the command line. But you can’t do that if it is part of a larger pattern. When it is, you add (?=(?!X).)*$) to the pattern to exclude anything with X in it.
So imagine you want to match lines with all three of A, B, and then either of C or D, but which don’t have X or Y in them. All you need is this:
grep -P '(?=^.*A)(?=^.*B)(?=^(?:(?!X).)*$)(?=^(?:(?!Y).)*$)C|D' *
In some shells and in some settings. you’ll have to escape the ! if it’s your history-substitution character.
There, isn’t that pretty cool?
In my cygwin the given answers didn't work, but the following did:
grep -l firststring `grep -r -l secondstring . `
Do you mean "string1" and "string2" on the same line?
grep 'string1.*string2'
On the same line but in indeterminate order?
grep '(string1.*string2)|(string2.*string1)'
Or both strings must appear in the file anywhere?
grep -e string1 -e string2
The uses PCRE (Perl-Compatible Regular Expressions) with multiline matching and returns the filenames of files that contain both strings (AND rather than OR).
grep -Plr '(?m)db-connect\.php(.*\n)*version|version(.*\n)*db-connect\.php' .
Why to stick to only grep:
perl -lne 'print if(/db-connect.php/&/version/)' *