I have a file temp.txt as below.
a.*,super
I want to grep .* to check whether the value is present in the file or not.
Command used:
grep -i ".*" temp.txt
returns nothing
This is because grep considers the pattern as a regular expression.
To make grep interpret it as a literal, use -F.
grep -F ".*" temp.txt
Also, note -i is not needed, because there is no case distinction to take into account (we for example use it to make grep return AB, aB, Ab and ab when doing grep -i "ab").
As man grep says:
-F, --fixed-strings
Interpret PATTERN as a list of fixed strings, separated by newlines,
any of which is to be matched. (-F is specified by POSIX.)
-i, --ignore-case
Ignore case distinctions in both the PATTERN and the input files. (-i
is specified by POSIX.)
Using awk
awk '/\.\*/' file
or fgrep
fgrep ".*" file
Both ., * have special meaning in regular expression. Escape them to match literally.
$ cat temp.txt
a.*,super
$ grep "\.\*" temp.txt
a.*,super
$ echo $?
0
$ grep "there-is-no-such-string" temp.txt
$ echo $?
1
-i is not need because there's no alphabet in the regular expression.
Related
grep -v mystring myfile.txt
returns ~300KB
grep -v *mystring* myfile.txt
returns ~7GB
....what am I doing wrong here?
Your regular expression is wrong. By default grep takes regular expressions as argument along with the command line flags. The one you have attempted *mystring* is a shell glob expression which expands to a possible set of filenames containing the string mystring. So your grep commands becomes the following; on an assumption that you have filenames containing mystring
grep -v mystring1 foomystring2 foomystring3 myfile.txt
which could produce unexpected results depending on the contents of those files. The right way would be to use the greedy match quantifier .*
grep -v '.*mystring1.*' myfile.txt
My goal is to match a command line argument prefix that looks like:
--abc=
Both of the patterns below (and many others), allow:
--abc==
Somehow, I can't find a grep way to ensure there is just one equal sign.
grep -i '^--[a-z]\{2,\}=\{1,1\}'
grep -i '^--[a-z]\{2,\}='
grep 2.20
CentOS Linux 7.3.1611
ERE:
^--[[:alpha:]]{2,}=[^=]+$
^--[[:alpha:]]{2,}= matches --, then two or more alphabetic characters in your locale, then a literal =
[^=]+$ matches one or more characters that are not = at the end
BRE:
^--[[:alpha:]]\{2,\}=[^=]\+$
Example:
$ grep -E '^--[[:alpha:]]{2,}=[^=]*$' <<<'--foobar=spam'
--foobar=spam
$ grep -E '^--[[:alpha:]]{2,}=[^=]*$' <<<'--foobar=23'
--foobar=23
$ grep -E '^--[[:alpha:]]{2,}=[^=]*$' <<<'--123ad='
$ grep -E '^--[[:alpha:]]{2,}=[^=]+$' <<<'--spamegg='
I am trying to combine grep and grep -v search together.
Output should be display all lines ending with .xml, but to exclude lines starting with $.
Here are the commands I have tried; none worked:
grep *.xml file1.txt | grep -v '$' file1.txt > output
grep *.xml | grep -v '$' file1.txt > output
grep *.xml grep -v '$' file1.txt > output
grep *.xml '$' file1.txt > output
To match a $ at the start of a line, anchor it to the start of the line with ^. Also, $ by itself matches the end of the line (it's a special character, just like ^), and * will not do what you think it does (it works differently in regular expressions compared to in shell globbing patterns). So,
grep -v '^\$'
will filter out all lines starting with a $.
You can do either
grep '\.xml$' file1.txt | grep -v '^\$'
or
grep '^[^$].*\.xml$' file1.txt
to find all lines in the file file1.txt that do not start with $ but that ends with .xml.
Notice that I also escape the dot in .xml as that otherwise matches any character, and that the second command combines both criteria by using a character range ([ ... ]) containing all characters except $ (the .* matches any number of any characters).
The single quotes are necessary so that the shell won't interpret the regular expression as a shell globbing pattern.
You should use "cat" command to direct the output to an file.
And then use regular expression to filter the keyword, in this case all lines start with $ symbol is '^[$]'.
So you can use command cat *.xml | grep -v '^[$]'.
Given the following content:
title="Bar=1; Fizz=2; Foo_Bar=3;"
I'd like to match the first occurrence of Bar value which is 1. Also I don't want to rely on soundings of the word (like double quote in the front), because the pattern could be in the middle of the line.
Here is my attempt:
$ grep -o -m1 'Bar=[ ./0-9a-zA-Z_-]\+' input.txt
Bar=1
Bar=3
I've used -m/--max-count which suppose to stop reading the file after num matches, but it didn't work. Why this option doesn't work as expected?
I could mix with head -n1, but I wondering if it is possible to achieve that with grep?
grep is line-oriented, so it apparently counts matches in terms of lines when using -m[1]
- even if multiple matches are found on the line (and are output individually with -o).
While I wouldn't know to solve the problem with grep alone (except with GNU grep's -P option - see anubhava's helpful answer), awk can do it (in a portable manner):
$ awk -F'Bar=|;' '{ print $2 }' <<<"Bar=1; Fizz=2; Foo_Bar=3;"
1
Use print "Bar=" $2, if the field name should be included.
Also note that the <<< method of providing input via stdin (a so-called here-string) is specific to Bash, Ksh, Zsh; if POSIX compliance is a must, use echo "..." | grep ... instead.
[1] Options -m and -o are not part of the grep POSIX spec., but both GNU and BSD/OSX grep support them and have chosen to implement the line-based logic.
This is consistent with the standard -c option, which counts "selected lines", i.e., the number of matching lines:
grep -o -c 'Bar=[ ./0-9a-zA-Z_-]\+' <<<"Bar=1; Fizz=2; Foo_Bar=3;" yields 1.
Using perl based regex flavor in gnu grep you can use:
grep -oP '^(.(?!Bar=\d+))*Bar=\d+' <<< "Bar=1; Fizz=2; Foo_Bar=3;"
Bar=1
(.(?!Bar=\d+))* will match 0 or more of any characters that don't have Bar=\d+ pattern thus making sure we match first Bar=\d+
If intent is to just print the value after = then use:
grep -oP '^(.(?!Bar=\d+))*Bar=\K\d+' <<< "Bar=1; Fizz=2; Foo_Bar=3;"
1
You can use grep -P (assuming you are on gnu grep) and positive look ahead ((?=.*Bar)) to achieve that in grep:
echo "Bar=1; Fizz=2; Foo_Bar=3;" | grep -oP -m 1 'Bar=[ ./0-9a-zA-Z_-]+(?=.*Bar)'
First use a grep to make the line start with Bar, and then get the Bar at the start of the line:
grep -o "Bar=.*" input.txt | grep -o -m1 "^Bar=[ ./0-9a-zA-Z_-]\+"
When you have a large file, you can optimize with
grep -o -m1 "Bar=.*" input.txt | grep -o -m1 "^Bar=[ ./0-9a-zA-Z_-]\+"
I have redirected some string into one parameter for ex: ab=jyoti,priya, pranit
I have one file : Name.txt which contains -
jyoti
prathmesh
John
Kelvin
pranit
I want to delete the records from the Name.txt file which are contain in ab parameter.
Please suggest if this can be done ?
If ab is a shell variable, you can easily turn it into an extended regular expression, and use it with grep -E:
grep -E -x -v "${ab//,/|}" Name.txt
The string substitution ${ab//,/|} returns the value of $ab with every , substituted with a | which turns it into an extended regular expression, suitable for passing as an argument to grep -E.
The -v option says to remove matching lines.
The -x option specifies that the match needs to cover the whole input line, so that a short substring will not cause an entire longer line to be removed. Without it, ab=prat would cause pratmesh to be removed.
If you really require a sed solution, the transformation should be fairly trivial. grep -E -v -x 'aaa|bbb|ccc' is equivalent to sed '/^\(aaa\|bbb\|ccc)$/d' (with some dialects disliking the backslashes, and others requiring them).
To do an in-place edit (modify Name.txt without a temporary file), try this:
sed -i "/^\(${ab//,/\|}\)\$/d" Name.txt
This is not entirely robust against strings containing whitespace or other shell metacharacters, but if you just need
Try with
sed -e 's/\bjyoti\b//g;s/\bpriya\b//g' < Name.txt
(using \b assuming you need word boundaries)
this will do it:
for param in `echo $ab | sed -e 's/[ ]+//g' -e 's/,/ /g'` ; do res=`sed -e "s/$param//g" < name.txt`; echo $res > name.txt; done
echo $res