Using the standard search function (/) in VIM, is there a way to search using a wildcard (match 0 or more characters)?
Example:
I have an array and I want to find anywhere the array's indices are assigned.
array[0] = 1;
array[i] = 1;
array[index]=1;
etc.
I'm looking for something along the lines of
/array*=
if it's possible.
I think you're misunderstanding how the wildcard works. It does not match 0 or more characters, it matches 0 or more of the preceding atom, which in this case is y. So searching
/array*=
will match any of these:
arra=
array=
arrayyyyyyyy=
If you want to match 0 or more of any character, use the 'dot' atom, which will match any character other than a newline.
/array.*=
If you want something more robust, I would recommend:
/array\s*\[[^\]]\+\]\s*=
which is "array" followed by 0 or more whitespace, followed by anything contained in brackets, followed by 0 or more whitespace, followed by an "equals" sign.
try
array.*
.* makes it mean something or nothing.in array* , * goes to y .
Related
Please suggest a wildcard for below Firstjson list
Firstjson = { p10_7_8 , p10_7_2 , p10_7_3 p10_7_4}
I have tried p10.7.* wildcard for below Secondjson list, it worked. But when I tried p10_7_* for above Firstjson list it did not work
Secondjson = { p10.7.8 , p10.7.2 , p10.7.3 , p10.7.4 }
You are attempting to use wildcard syntax, but Groovy expects regular expression syntax for its pattern matching.
What went wrong with your attempt:
Attempt #1: p10.7.*
A regular expression of . matches any single character and .* matches 0 or more characters. This means:
p10{exactly one character of any kind here}7{zero or more characters of any
kind here}
You didn't realize it, but the . character in your first attempt was acting like a single-character wildcard too. This might match with p10x7abcdefg for example. It also does match p10.7.8 though. But be careful, it also matches p10.78, because the .* expression at the end of your pattern will happily match any sequence of characters, thus any and all characters following p10.7 are accepted.
Attempt #2: p10_7_*
_ matches only a literal underscore. But _* means to match zero or more underscores. It does not mean to match characters of any kind. So p10_7_* matches things like p10_7_______. Literally:
p10_7{zero or more underscores here}
What you can do instead:
You probably want a regular expression like p10_7_\d+
This will match things like p10_7_3 or p10_7_422. It works by matching the literal text p10_7_ followed by one or more digits where a digit is 0 through 9. \d matches any digit, and + means to match one or more of the preceding thing. Literally:
p10_7_{one or more digits here}
I've been trying for about two hours now to write a regular expression which matches a single character that's not preceded or followed by the same character.
This is what I've got: (\d)(?<!\1)\1(?!\1); but it doesn't seem to work! (testing at https://regex101.com/r/whnj5M/6)
For example:
In 1111223 I would expect to match the 3 at the end, since it's not preceded or followed by another 3.
In 1151223 I would expect to match the 5 in the middle, and the 3 at the end for the same reasons as above.
The end goal for this is to be able to find pairs (and only pairs) of characters in strings (e.g. to find 11 in 112223 or 44 in 123544) and I was going to try and match single isolated characters, and then add a {2} to it to find pairs, but I can't even seem to get isolated characters to match!
Any help would be much appreciated, I thought I knew RegEx pretty well!
P.S. I'm testing in JS on regex101.com because it wouldn't let me use variable length lookbacks in Python on there, and I'm using the regex library to allow for this in my actual implementation.
Your regex is close, but by using simply (\d) you are consuming characters, which prevents the other match from occurring. Instead, you can use a positive lookahead to set the capture group and then test for any occurrences of the captured digit not being surrounded by copies of itself:
(?=.*?(.))(?<!\1)\1(?!\1)
By using a lookahead you avoid consuming any characters and so the regex can match anywhere in the string.
Note that in 1151223 this returns 5, 1 and 3 because the third 1 is not adjacent to any other 1s.
Demo on regex101 (requires JS that supports variable width lookbehinds)
The pattern you tried does not match because this part (\d)(?<!\1) can not match.
It reads as:
Capture a digit in group 1. Then, on the position after that captured
digit, assert what is captured should not be on the left.
You could make the pattern work by adding for example a dot after the backreference (?<!\1.) to assert that the value before what you have just matched is not the same as group 1
Pattern
(\d)(?<!\1.)\1(?!\1)
Regex demo | Python demo
Note that you have selected ECMAscript on regex101.
Python re does not support variable width lookbehind.
To make this work in Python, you need the PyPi regex module.
Example code
import regex
pattern = r"(\d)(?<!\1.)\1(?!\1)"
test_str = ("1111223\n"
"1151223\n\n"
"112223\n"
"123544")
matches = regex.finditer(pattern, test_str)
for matchNum, match in enumerate(matches, start=1):
print(match.group())
Output
22
11
22
11
44
#Theforthbird has provided a good explanation for why your regular explanation does not match the characters of interest.
Each character matched by the following regular expression is neither preceded nor followed by the same character (including characters at the beginning and end of the string).
r'^.$|^(.)(?!\1)|(?<=(.))(?!\2)(.)(?!\3)'
Demo
Python's re regex engine performs the following operations.
^.$ match the first char if it is the only char in the line
| or
^ match beginning of line
(.) match a char in capture group 1...
(?!\1) ...that is not followed by the same character
| or
(?<=(.)) save the previous char in capture group 2...
(?!\2) ...that is not equal to the next char
(.) match a character and save to capture group 3...
(?!\3) ...that is not equal to the following char
Suppose the string were "cat".
The internal string pointer is initially at the beginning of the line.
"c" is not at the end of the line so the first part of the alternation fails and the second part is considered.
"c" is matched and saved to capture group 1.
The negative lookahead asserting that "c" is not followed by the content of capture group 1 succeeds, so "c" is matched and the internal string pointer is advanced to a position between "c" and "a".
"a" fails the first two parts of the assertion so the third part is considered.
The positive lookbehind (?<=(.)) saves the preceding character ("c") in capture group 2.
The negative lookahead (?!\2), which asserts that the next character ("a") is not equal to the content of capture group 2, succeeds. The string pointer remains just before "a".
The next character ("a") is matched and saved in capture group 3.
The negative lookahead (?!\3), which asserts that the following character ("t") does not equal the content of capture group 3, succeeds, so "a" is matched and the string pointer advances to just before "t".
The same steps are performed when evaluating "t" as were performed when evaluating "a". Here the last token ((?!\3)) succeeds, however, because no characters follow "t".
I'm looking for a little help on some Lua. I need some code to match this exact line:
efs.test efs.test.gpg
Here's what I have so far, which matches "efs.test":
if string.match(a.message, "%a+%a+%a+.%%a+%a+%a+%a+") then
print(a.message)
else
print ("Does not match")
end
I've also tried this, which matches:
if string.match(a.message, "efs.test") then
print(a.message)
else
print ("Does not match")
end
But when I try to add the extra text my compiler errors with "Number expected, got string" when running this code:
if string.match(a.message, "efs.test", "efs") then
print(a.message)
else
print ("Does not match")
end
Any pointers would be great!
Thanks.
if string.match(a.message, "%a+%a+%a+.%%a+%a+%a+%a+") then
Firstly, this is a wrong use of quantifiers. From PiL 20.2:
+ 1 or more repetitions
* 0 or more repetitions
- also 0 or more repetitions
? optional (0 or 1 occurrence)
In words, you try to match for unlimited %a+ after you already matched the full word with unlimited %a+
To match efs.test efs.test.gpg - we have 2 filenames I suppose, in a strict sense file names may contain only %w - alphanumeric characters (A-Za-z0-9). This would correctly match efs.test:
string.match(message, "%w+%.%w+")
Going one step further, match efs.test as filename and the following filename:
string.match(message, "%w+%.%w+ %w+%.%w+%.gpg")
While this would match both filenames, you would need to check if matched filenames are the same. We can go one step further yet:
local file, gpgfile = string.match(message, "(%w+%.%w+) (%1%.gpg)")
This pattern will return any <filename> <filename>.gpg where the filenames are equal.
With the use of capture-groups, we capture the filename: it will be returned as the first variable and further represented as %1. Then after the space char, we try to match for %1 (captured filename) followed by .gpg. Since it's also enclosed in brackets, it will become the second captured group and returned as the second variable. Done!
PS: You may want to grab ".gpg" by case-insensitive [Gg][Pp][Gg] pattern.
PPS: File names may contain spaces, dashes, UTF-8 characters etc. E.g. ext4 only forbids \0 and / characters.
string.match optional third argument is the index of the given string to start searching at. If you are looking for exactly efs.test efs.test.gpg in that order with that given spacing, why not just use:
string.match(a.message, "efs%.test efs%.test%.gpg")
If you want to match the entire line containing that substring:
string.match(a.message, ".*efs%.test efs%.test%.gpg.*")
For reference
If you are trying to match that exact line its way easier to just use:
if "efs.test efs.test.gpg" = a.message then
print(a.message)
else
print("string does not match!")
end
Of course this wouldn't find any other strings than this.
Another interpretation I see for your question is that you want to know if it has efs.test in the string, which you should be able to accomplish by doing:
if string.match(a.message, "%w+%.%w+") == "efs.test" then
...
end
Also, look into regex, it's basically the language Lua used to match strings with some exceptions.
I'm trying to find exact matches of strings in Lua including, special characters. I want the example below to return that it is an exact match, but because of the - character it returns nil
index = string.find("test-string", "test-string")
returns nil
index = string.find("test-string", "test-")
returns 1
index = string.find("test-string", "test")
also returns 1
How can I get it to do full matching?
- is a pattern operator in a Lua string pattern, so when you say test-string, you're telling find() to match the string test as few times as possible. So what happens is it looks at test-string, sees test in there, and since - isn't an actual minus sign in this case, it's really looking for teststring.
Do as Mike has said and escape it with the % character.
I found this helpful for better understanding patterns.
You can also ask for a plain substring match that ignores magic characters:
string.find("test-string", "test-string",1,true)
you need to escape special characters in the pattern with the % character.
so in this case you are looking for
local index = string.find('test-string', 'test%-string')
I have a string of numbers separated by spaces and I need to store them in a table but for some reason negative symbol is not getting recognize.
cord = "-53 2 -21"
map = {}
for num in cord:gmatch("%w+") do
table.insert(map, num)
end
map[1], map[2], map[3] = tonumber(map[1]), tonumber(map[2]), tonumber(map[3])
print(map[1])
print(map[2])
print(map[3])
This is the output I'm getting:
53
2
21
I think the problem is with the pattern I'm using, what should I change?
The pattern "%w" is for alphanumeric characters, which doesn't include -, use this pattern instead:
"%-?%w+"
or better:
"%-?%d+"
since numbers are all you need.
%w+ does not attempt to mach only numbers, so try %S+ to get all "words", that is, all sequences of non-zero characters.
If you want to match only numbers, try %-?%d+. Note the optional minus sign in the pattern. Note also that you must escape the minus sign.