I have this pattern:
^u.meta(\.|$)
EXPECTED BEHAVIOUR
^u.meta(\.|$) will match all the roles like:
u.meta
u.meta.admin
u.meta.admin.system
u.meta.*
Where as it should not match something like below:
u.meta_admin
u.meta_admin_system
I have tested this pattern with https://regex101.com/ online regexp tester.
PROBLEM:
I have to implement this pattern with lua script.
but getting invalid escape sequence near '\.':
-- lua script
> return string.match("u.meta.admin", '^u.meta(\.|$)')
stdin:1: invalid escape sequence near '\.'
And I tried adding double \\ as well as removing '\' escape char in that regexp but got nil in return:
-- lua script
> return string.match("u.meta.admin", '^u.meta(\\.|$)')
nil
> return string.match("u.meta.admin", '^u.meta(.|$)')
nil
See Lua regex docs:
The character % works as an escape for those magic characters.
Also, the (...|...) alternation is not supported in Lua. Instead, I guess, you need a word boundary here, like %f[set] frontier pattern:
%f[set], a frontier pattern; such item matches an empty string at any position such that the next character belongs to set and the previous character does not belong to set. The set set is interpreted as previously described. The beginning and the end of the subject are handled as if they were the character \0.
So, you can use
return string.match("u.meta.admin", '^u%.meta%f[%A]')
To only match at the end or before a .:
return string.match("u.meta", '^u%.meta%f[\0.]')
To match only if the admin is not followed with a letter or an underscore, use a negated character class [^%a_]:
return string.match("u.meta_admin", '^u%.meta%f[[^%a_]]')
See IDEONE demo to check the difference between the two expressions.
print(string.match("u.meta", '^u%.meta%f[\0.]')) -- u.meta
print(string.match("u.meta.admin", '^u%.meta%f[\0.]')) -- u.meta
print(string.match("u.meta-admin", '^u%.meta%f[\0.]')) -- nil
print(string.match("u.meta", '^u%.meta%f[%A]')) -- u.meta
print(string.match("u.meta.admin", '^u%.meta%f[%A]')) -- u.meta
print(string.match("u.meta-admin", '^u%.meta%f[%A]')) -- u.meta
-- To exclude a match if `u.admin` is followed with `_`:
print(string.match("u.meta_admin", '^u%.meta%f[[^%a_]]')) -- nil
NOTE To match the end of the string, instead of \0, you can safely use %z (as #moteus noted in his comment) (see this reference):
%z the character with representation 0
Related
how can I extract a few words separated by symbols in a string so that nothing is extracted if the symbols change?
for example I wrote this code:
function split(str)
result = {};
for match in string.gmatch(str, "[^%<%|:%,%FS:%>,%s]+" ) do
table.insert(result, match);
end
return result
end
--------------------------Example--------------------------------------------
str = "<busy|MPos:-750.222,900.853,1450.808|FS:2,10>"
my_status={}
status=split(str)
for key, value in pairs(status) do
table.insert(my_status,value)
end
print(my_status[1]) --
print(my_status[2]) --
print(my_status[3]) --
print(my_status[4]) --
print(my_status[5]) --
print(my_status[6]) --
print(my_status[7]) --
output :
busy
MPos
-750.222
900.853
1450.808
2
10
This code works fine, but if the characters and text in the str string change, the extraction is still done, which I do not want to be.
If the string change to
str = "Hello stack overFlow"
Output:
Hello
stack
over
low
nil
nil
nil
In other words, I only want to extract if the string is in this format: "<busy|MPos:-750.222,900.853,1450.808|FS:2,10>"
In lua patterns, you can use captures, which are perfect for things like this. I use something like the following:
--------------------------Example--------------------------------------------
str = "<busy|MPos:-750.222,900.853,1450.808|FS:2,10>"
local status, mpos1, mpos2, mpos3, fs1, fs2 = string.match(str, "%<(%w+)%|MPos:(%--%d+%.%d+),(%--%d+%.%d+),(%--%d+%.%d+)%|FS:(%d+),(%d+)%>")
print(status, mpos1, mpos2, mpos3, fs1, fs2)
I use string.match, not string.gmatch here, because we don't have an arbitrary number of entries (if that is the case, you have to have a different approach). Let's break down the pattern: All captures are surrounded by parantheses () and get returned, so there are as many return values as captures. The individual captures are:
the status flag (or whatever that is): busy is a simple word, so we can use the %w character class (alphanumeric characters, maybe %a, only letters would also do). Then apply the + operator (you already know that one). The + is within the capture
the three numbers for the MPos entry each get (%--%d+%.%d+), which looks weird at first. I use % in front of any non-alphanumeric character, since it turns all magic characters (such as + into normal ones). - is a magic character, so it is required here to match a literal -, but lua allows to put that in front of any non-alphanumerical character, which I do. So the minus is optional, so the capture starts with %-- which is one or zero repetitions (- operator) of a literal - (%-). Then I just match two integers separated by a dot (%d is a digit, %. matches a literal dot). We do this three times, separated by a comma (which I don't escape since I'm sure it is not a magical character).
the last entry (FS) works practically the same as the MPos entry
all entries are separated by |, which I simply match with %|
So putting it together:
start of string: %<
status field: (%w+)
separator: %|
MPos (three numbers): MPos:(%--%d+%.%d+),(%--%d+%.%d+),(%--%d+%.%d+)
separator: %|
FS entry (two integers): FS:(%d+),(%d+)
end of string: %>
With this approach you have the data in local variables with sensible names, which you can then put into a table (for example).
If the match failes (for instance, when you use "Hello stack overFlow"), nil` is returned, which can simply be checked for (you could check any of the local variables, but it is common to check the first one.
I have a name spaced class..
"CommonCar::RedTrunk"
I need to convert it to an underscored string "common_car_red_trunk", but when I use
"CommonCar::RedTrunk".underscore, I get "common_car/red_trunk" instead.
Is there another method to accomplish what I need?
Solutions:
"CommonCar::RedTrunk".gsub(':', '').underscore
or:
"CommonCar::RedTrunk".sub('::', '').underscore
or:
"CommonCar::RedTrunk".tr(':', '').underscore
Alternate:
Or turn any of these around and do the underscore() first, followed by whatever method you want to use to replace "/" with "_".
Explanation:
While all of these methods look basically the same, there are subtle differences that can be very impactful.
In short:
gsub() – uses a regex to do pattern matching, therefore, it's finding any occurrence of ":" and replacing it with "".
sub() – uses a regex to do pattern matching, similarly to gsub(), with the exception that it's only finding the first occurrence (the "g" in gsub() meaning "global"). This is why when using that method, it was necessary to use "::", otherwise a single ":" would have been left. Keep in mind with this method, it will only work with a single-nested namespace. Meaning "CommonCar::RedTrunk::BigWheels" would have been transformed to "CommonCarRedTrunk::BigWheels".
tr() – uses the string parameters as arrays of single character replacments. In this case, because we're only replacing a single character, it'll work identically to gsub(). However, if you wanted to replace "on" with "EX", for example, gsub("on", "EX") would produce "CommEXCar::RedTrunk" while tr("on", "EX") would produce "CEmmEXCar::RedTruXk".
Docs:
https://apidock.com/ruby/String/gsub
https://apidock.com/ruby/String/sub
https://apidock.com/ruby/String/tr
This is a pure-Ruby solution.
r = /(?<=[a-z])(?=[A-Z])|::/
"CommonCar::RedTrunk".gsub(r, '_').downcase
#=> "common_car_red_trunk"
See (the first form of) String#gsub and String#downcase.
The regular expression can be made self-documenting by writing it in free-spacing mode:
r = /
(?<=[a-z]) # assert that the previous character is lower-case
(?=[A-Z]) # assert that the following character is upper-case
| # or
:: # match '::'
/x # free-spacing regex definition mode
(?<=[a-z]) is a positive lookbehind; (?=[A-Z]) is a positive lookahead.
Note that /(?<=[a-z])(?=[A-Z])/ matches an empty ("zero-width") string. r matches, for example, the empty string between 'Common' and 'Car', because it is preceeded by a lower-case letter and followed by an upper-case letter.
I don't know Rails but I'm guessing you could write
"CommonCar::RedTrunk".delete(':').underscore
Currently I am getting passed a bunch of strings from a home automation controller to our driver which I am currently developing.
I receive messages such as ZAA and other code which may be AA but the string.match sometimes will match the AA with the ZAA if statement.
This issue is much more wide spread than just those two strings (probably around to 10-15 other similarities).
I understand that I could add more conditions to the if/elseif statements but surely there is an exact match version?
Any ideas would be greatly appreciated.
Example; Even though the string is "AA" it will match "ZAA"
stringInput = "AA"
if string.match("ZAA", stringInput) then
print("I matched: ZAA")
elseif string.match("AA", stringInput) then
print("I matched: AA")
end
If you want an exact match, just use ==.
if stringInput == 'ZAA' then
print('I matched: ZAA')
elseif stringInput == 'AA' then
print('I matched: AA')
end
From the Lua 5.3. Reference Manual: string.match
string.match (s, pattern [, init])
Looks for the first match of pattern (see §6.4.1) in the string s. If it finds one, then match
returns the captures from the pattern; otherwise it returns nil. If
pattern specifies no captures, then the whole match is returned. A
third, optional numeric argument init specifies where to start the
search; its default value is 1 and can be negative.
so
local inputString = "AA"
string.match("ZAA", inputString)
will match because "AA" is in "ZAA".
You confused the function arguments s and pattern.
local inputString = "AA"
inputString:match("ZAA")
will not match because the pattern contains more characters than inputString.
But as
local inputString = "ZAA"
will both match
inputString:match("AA") and inputString:match("ZAA") you'll probably have to add more constraints.
Please read the manual!
I want to run two different lua string find on the same string " (55)"
Pattern 1 "[^%w_](%d+)", should match any number
Pattern 2 "[%(|%)|%%|%+|%=|%-|%{%|%}|%,|%:|%*|%^]", should match any of these ( ) % + = - { } , : * ^ characters.
Both of these patterns return 2, why? Also if I run a string match, they return ( and 55 respectivly (as expected).
It seems you are using the patterns with string.find that finds the first occurrence of the pattern in the string passed. If an instance of the pattern is found a pair of values representing the start and end of the string is returned. If the pattern cannot be found nil is returned.
Both patterns find a match at Position 2: [^%w_](%d+) finds ( because it is matched with [^%w_] (a char other than letter, digit or _), and [%(|%)|%%|%+|%=|%-|%{%|%}|%,|%:|%*|%^] matches the ( because it is part of the character set.
However, the first pattern can be re-written using a frontier pattern, %f[%w_]%d+, that will match 1+ digits if not preceded with letters, digits or underscore, and the second pattern does not require such heavy escaping, [()%%+={},:*^-] is enough (only % needs escaping here, as the - is placed at the end of the character set and is thus treated as a literal hyphen).
See this Lua demo:
a = " (55)"
for word in string.gmatch(a, "%f[%w_]%d+") do print(word) end
-- 55
for word in string.gmatch(a, "[()%%+={},:*^-]+") do print(word) end
-- (, )
Suppose I have string variables like following:
s1="10$"
s2="10$ I am a student"
s3="10$Good"
s4="10$ Nice weekend!"
As you see above, s2 and s4 have white space(s) after 10$ .
Generally, I would like to have a way to check if a string start with 10$ and have white-space(s) after 10$ . For example, The rule should find s2 and s4 in my above case. how to define such rule to check if a string start with '10$' and have white space(s) after?
What I mean is something like s2.RULE? should return true or false to tell if it is the matched string.
---------- update -------------------
please also tell the solution if 10# is used instead of 10$
You can do this using Regular Expressions (Ruby has Perl-style regular expressions, to be exact).
# For ease of demonstration, I've moved your strings into an array
strings = [
"10$",
"10$ I am a student",
"10$Good",
"10$ Nice weekend!"
]
p strings.find_all { |s| s =~ /\A10\$[ \t]+/ }
The regular expression breaks down like this:
The / at the beginning and the end tell Ruby that everything in between is part of the regular expression
\A matches the beginning of a string
The 10 is matched verbatim
\$ means to match a $ verbatim. We need to escape it since $ has a special meaning in regular expressions.
[ \t]+ means "match at least one blank and/or tab"
So this regular expressions says "Match every string that starts with 10$ followed by at least one blank or tab character". Using the =~ you can test strings in Ruby against this expression. =~ will return a non-nil value, which evaluates to true if used in a conditional like if.
Edit: Updated white space matching as per Asmageddon's suggestion.
this works:
"10$ " =~ /^10\$ +/
and returns either nil when false or 0 when true. Thanks to Ruby's rule, you can use it directly.
Use a regular expression like this one:
/10\$\s+/
EDIT
If you use =~ for matching, note that
The =~ operator returns the character position in the string of the
start of the match
So it might return 0 to denote a match. Only a return of nil means no match.
See for example http://www.regular-expressions.info/ruby.html on a regular expression tutorial for ruby.
If you want to proceed to cases with $ and # then try this regular expression:
/^10[\$#] +/