How do we find a particular pattern in files and sub-directories within a directory?
An exact word can be searched for using a find grep and exec.
But how do we find something like below:
'match=true'
where there may or may not be multiple spaces around '='.
This finds match=true with zero or more (*) whitespace characters (\s) around the equals sign (Tab, although uncommon between non-whitespace characters, would also be a valid character around an equals sign in many languages):
grep 'match\s*=\s*true'
If you want to match exactly zero or one whitespace, you can use \? instead of * above.
Related
I am new here. I wanted to ask a question on using REGEX for an entity in DialogFlow
I wanted the entity to accept all text and spaces except for the symbol *
I have tried to use [A-Za-z0-9 ][^*], but it is not working. Any advice. thanks!
In your Regex expression, [^*] means "capture any character at the start of the line." To refer to a literal asterisk rather than matching any character, you need to use \*
If you want to match a line of letters or numbers as in the [A-Za-z0-9] example you give, but only if that string does not include an asterisk, then this expression should work for you:
^[a-zA-Z0-9]+$
This means "match a whole line of text if it only contains one or more of the characters a-z, A-Z, or 0-9".
If you want to match any character or group of characters in a line except for the asterisk, then you could use something like this:
(?!\*)([a-zA-Z0-9]+)(?<!\*)
The first part is called a "negative lookahead," and it looks forward to ensure we're not matching the asterisk. The last part is called a "negative lookbehind," and it looks backwards to make sure we're not matching the asterisk. The middle part is your "capture group," and confirms that you're matching any letters or numbers in a given string, but excluding the * character.
If this Regex gets input like *abc, it will capture abc. If it encounters abc*, it will still capture abc. If it encounters abc*def, it will capture abc and def separately in two capture groups, because it will break around the asterisk.
This link explains the concept of lookarounds in Regex. You can also use this Regex tester to get started practicing your Regular Expressions with explanations of what each block of characters does.
EDITED TO ADD If you're just interested in matching single characters rather than groups of characters, you can use [A-Za-z0-9] and match any upper or lowercase letter and any single digit. You don't need to exclude the * character, because the character group is already exclusive.
This is a slight duplicate of the question below, so responses here may also help you. Hope this helps!
How can I exclude asterisk in a regex expression
[A-Za-z0-9 ][^*]
What you regex will do is match 2 consecutive characters. First, it will look for anything A-Za-z0-9 . Then, it will look at the negated set that includes *, and will match ANY character except *.
You can type your regex into https://regexr.com/ to see a breakdown of how it matches and test some strings.
For example, your regex would match these:
Aa
AA
a&
A1
0_
But would not match these:
A*
a*
1*
And WOULD NOT match anything longer than 2 characters. If you really want to match any string with any characters except *, this should work:
[^\*]+
What that will do is match any number of consecutive characters that are not *. (The + means match 1 or more characters in the set). It is also a good idea to escape * because it is also a reserved character in regex. Even though most regex parsers are smart enough to know that inside a group you probably mean the literal char *, it is still a best practice to escape it. (And by that same token, you would want to use \s instead of the blank space in your original regex.)
I need to rename a lot of files with mmv. I know how to do that but I have a problem with wildcard indexes followed by numbers in the filename.
Basically I need to have an output filename which contains a wildcard followed by numbers.
mmv -n ``\*2\\.3_\*'' ``#11.6#2''
Here, as you can see, I'd like to have an output filename which contains the first wildcard followed by 1.6.
Unfortunately, this way I have #11.6 and the code is interpreted as if I want the 11th wildcard, which of course do not exist.
By reading the documentation you should have been able to find a solution.
Citation from man mmv, see https://ss64.com/bash/mmv.html
To strip any character (e.g. ’*’, ’?’, or ’#’) of its special meaning to mmv, as when the actual replacement name must contain the character ’#’, precede the special character with a ´\’ (and enclose the argument in quotes because of the shell). This also works to terminate a wildcard index when it has to be followed by a digit in the filename, e.g. "a#1\1".
I have the following string in the code at multiple places,
m_cells->a[ Id ]
and I want to replace it with
c(Id)
where the string Id could be anything including numbers also.
A regular expression replace like below should do:
%s/m_cells->a\[\s\(\w\+\)\s\]/c(\1)/g
If you wish to apply the replacement operation on a number of files you could use the :bufdo command.
Full explanation of #BasBossink's answer (as a separate answer because this won't fit in a comment), because regexes are awesome but non-trivial and definitely worth learning:
In Command mode (ie. type : from Normal mode), s/search_term/replacement/ will replace the first occurrence of 'search_term' with 'replacement' on the current line.
The % before the s tells vim to perform the operation on all lines in the document. Any range specification is valid here, eg. 5,10 for lines 5-10.
The g after the last / performs the operation "globally" - all occurrences of 'search_term' on the line or lines, not just the first occurrence.
The "m_cells->a" part of the search term is a literal match. Then it gets interesting.
Many characters have special meaning in a regex, and if you want to use the character literally, without the special meaning, then you have to "escape" it, by putting a \ in front.
Thus \[ and \] match the literal '[' and ']' characters.
Then we have the opposite case: literal characters that we want to treat as special regex entities.
\s matches white*s*pace (space, tab, etc.).
\w matches "*w*ord" characters (letters, digits, and underscore _).
(. matches any character (except a newline). \d matches digits. There are more...)
If a character is not followed by a quantifier, then exactly one such character matches. Thus, \s will match one space or tab, but not fewer or more.
\+ is a quantifier, and means "one or more". (\? matches 0 or 1; * (with no backslash) matches any number: zero or more. Warning: matching on zero occurrences takes a little getting used to; when you're first learning regexes, you don't always get the results you expected. It's also possible to match on an arbitrary exact number or range of occurrences, but I won't get into that here.)
\( and \) work together to form a "capturing group". This means that we don't just want to match on these characters, we also want to remember them specially so that we can do something with them later. You can have any number of capturing groups, and they can be nested too. You can refer to them later by number, starting at 1 (not 0). Just start counting (escaped) left-parantheses from the left to determine the number.
So here, we are matching a space followed by a group (which we will capture) of at least one "word" character followed by a space, within the square brackets.
Then section between the second and third / is the replacement text.
The "c" is literal.
\1 means the first captured group, which in this case will be the "Id".
In summary, we are finding text that matches the given description, capturing part of it, and replacing the entire match with the replacement text that we have constructed.
Perhaps a final suggestion: c after the final / (doesn't matter whether it comes before or after the 'g') enables *c*onfirmation: vim will highlight the characters to be replaced and will show the replacement text and ask whether you want to go ahead. Great for learning.
Yes, regexes are complicated, but super powerful and well worth learning. Once you have them internalized, they're actually fairly easy. I suggest that, as with learning vim itself, you start with the basics, get fluent in them, and then incrementally add new features to your repertoire.
Good luck and have fun.
I am a beginner at Vim and I've been reading about substitution but I haven't found an answer to this question.
Let's say I have some numbers in a file like so:
1
2
3
And I want to get:
(1)
(2)
(3)
I think the command should resemble something like :s:\d\+:........ Also, what's the difference between :s/foo/bar and :s:foo:bar ?
Thanks
Here is an alternative, slightly less verbose, solution:
:%s/^\d\+/(&)
Explanation:
^ anchors the pattern to the beginning of the line
\d is the atom that covers 0123456789
\+ matches one or more of the preceding item
& is a shorthand for \0, the whole match
Let me address those in reverse.
First: there's no difference between :s/foo/bar and :s:foo:bar; whatever delimiter you use after the s, vim will expect you to use from then on. This can be nice if you have a substitution involving lots of slashes, for instance.
For the first: to do this to the first number on the current line (assuming no commas, decimal places, etc), you could do
:s:\(\d\+\):(\1)
The \(...\) doesn't change what is matched - rather, it tells vim to remember whatever matched what is inside, and store it. The first \(...\) is stored in \1, the second in \2, etc. So, when you do the replacement, you can reference \1 to get the number back.
If you want to change ALL numbers on the current line, change it to
:s:\(\d\+\):(\1):g
If you want to change ALL numbers on ALL lines, change it to
:%s:\(\d\+\):(\1):g
You can do what you want with:
:%s/\([0-9]\)/(\1)/
%s means global search and replace, that is do the search/replace for every line in the file. the \( \) defines a group, which in turn is referenced by \1. So the above search and replace, finds all lines with a single digit ([0-9]), and replaces it with the matched digit surrounded by parentheses.
I'm trying to write a grep (or egrep) command that will find and print any lines in "words.txt" which contain the same lower-case letter three times in a row. The three occurrences of the letter may appear consecutively (as in "mooo") or separated by one or more spaces (as in "x x x") but not separated by any other characters.
words.txt contains:
The monster said "grrr"!
He lived in an igloo only in the winter.
He looked like an aardvark.
Here's what I think the command should look like:
grep -E '\b[^ ]*[[:alpha:]]{3}[^ ]*\b' 'words.txt'
Although I know this is wrong, but I don't know enough of the syntax to figure it out. Using grep, could someone please help me?
Does this work for you?
grep '\([[:lower:]]\) *\1 *\1'
It takes a lowercase character [[:lower:]] and remembers it \( ... \). It than tries to match any number of spaces _* (0 included), the rememberd character \1, any number of spaces, the remembered character. And that's it.
You can try running it with --color=auto to see what parts of the input it matched.
Try this. Note that this will not match "mooo", as the word boundary (\b) occurs before the "m".
grep -E '\b([[:alpha:]]) *\1 *\1 *\b' words.txt
[:alpha:] is an expression of a character class. To use as a regex charset, it needs the extra brackets. You may have already known this, as it looks like you started to do it, but left the open bracket unclosed.