I can't figure out how to get Lua to return ALL matches for a particular pattern match.
I have the following regex which works and is so basic:
.*\n
This just splits a long string per line.
The equivelent of this in Lua is:
.-\n
If you run the above in a regex website against the following text it will find three matches (if using the global flag).
Hello
my name is
Someone
If you do not use the global flag it will return only the first match. This is the behaviour of LUA; it's as if it does not have a global switch and will only ever return the first match.
The exact code I have is:
local test = {string.match(string_variable_here, ".-\n")}
If I run it on the above test for example, test will be a table with only one item (the first row). I even tried using capture groups but the result is the same.
I cannot find a way to make it return all occurrences of a match, does anyone know if this is possible in LUA?
Thanks,
You can use string.gmatch(s, pattern) / s:gmatch(pattern):
This returns a pattern finding iterator. The iterator will search through the string passed looking for instances of the pattern you passed.
See the online Lua demo:
local a = "Hello\nmy name is\nSomeone\n"
for i in string.gmatch(a, ".*\n") do
print(i)
end
Note that .*\n regex is equivalent to .*\n Lua pattern. - in Lua patterns is the equivalent of *? non-greedy ("lazy") quantifier.
isnumber(search("-tr",right(j2,3
))),isnumber(search("-trus",right(j2,5))),isnumber(search(" ll",right(j2,3))),isnumber(search(" homes",right(j2,6))),isnumber(search("the ",left(j2,4))),isnumber(search(" hoa",right(j2,4))),isnumber(search("b ch",right(j2,4))),isnumber(search(" ch",right(j2,3))),isnumber(search("-trs",right(j2,4))),isnumber(search(" prop",right(j2,5))),isnumber(search(" st",right(j2,3))),isnumber(search(" av",right(j2,3))),isnumber(search(" ave",right(j2,4))),isnumber(search(" servi",right(j2,6))),isnumber(search(" maint",right(j2,6))),isnumber(search(" home",right(j2,5))),isnumber(search(" tr",right(j2,3))),isnumber(search(" assn",right(j2,5))),isnumber(search(" co",right(j2,3))),isnumber(search(" trus",right(j2,5))),isnumber(search(" trs",right(j2,4))),isnumber(search("-trs",right(j2,4))),isnumber(search(" tru",right(j2,4))),isnumber(search("jtrs",right(j2,4))),isnumber(search(" est of",right(j2,7))),isnumber(search(" trs",right(j2,4))),isnumber(value(LEFT(j2,1))),isnumber(search(" apts",right(j2,5))),isnumber(value(right(j2,3))),isnumber(search(" grp",right(j2,4))),isnumber(value(left(right(j2,4),1))),isnumber(search(" mgmt",right(j2,5))),isnumber(search(" props",right(j2,6))),isnumber(search(" tr",right(j2,3))),isnumber(search(" dev",right(j2,4))),isnumber(search(" tr",right(j2,3))),isnumber(search(" fdn",right(j2,4))),isnumber(search(" ent",right(j2,4))),isnumber(search(" PRPTS",right(j2,6))),isnumber(search(" ARPTS",right(j2,6))),isnumber(search(" univ",right(j2,5)))
So I have this giant =OR() statement containing a bunch of isnumner(search() statements checking to see if the string in a cell ends in these phrases. It is for the purpose of identifying company names in lists that contain both peoples names and company names. I feel like there must be a more efficient way. Adding them all together in one isnumber(search() in this format {item1|item2|item3} does not work.
I feel like there must be a more efficient way.
Building on the answer provided here, matching the end of the string can be done by using the $-sign (which means 'end of the string in regular expressions). Matching the beginning of the string on the other hand is done by providing a pattern after a caret (^), indicating the beginning of a string.
So, if you'd want to add both the the formula provided in the other thread
(LP|JT/RS)$ : match LP OR JT/RS at the end of the string
^(ABC|DEF) : match ABC OR DEF at the beginning of the string
That would make the formula look something like:
=REGEXMATCH(A2, "(?i)LLC|CORPORATION|COMPANY|HOLDINGS|PARTNERS|EQUITY|(LP|JT/RS)$|^(ABC|DEF)")
REFERENCE:
REGEXMATCH()
RE2 SYNTAX
I'm new to GREP in BBEdit. I need to find a string inside an XML file. Such string is enclosed in quotes. I need to replace only what's inside the quotes.
The problem is that the replacement string starts with a number thus confuses BBEdit when I put together the replacement pattern. Example:
Original string in XML looks like this:
What I need to replace it with:
01 new file name.png
My grep search and replace patterns:
Using the replacement pattern above, BBEdit wrongly thinks that the first backreference is "\101" when what I really need it understand is that I mean "\01".
TIA for any help.
Your example is highly artificial because in fact there is no need for your \1 or \3 as you know their value: it is " and you can just type that directly to get the desired result.
"01 new file name.png"
However, just for the sake of completeness, the answer to your actual question (how to write a replacement group number followed by a number) is that you write this:
\0101 new file name.png\3
The reason that works is that there can only be 99 capture groups, so \0101 is parsed as \01 (the first capture group) followed by literal 01.
I have the following strings as input for scheduler file
Z:\cnt_development\cnt\test\Test-cases-blr\v80-WM\scheduler\FRQ\AUTO\sml-hr454\SRISM.xml
Z:\cnt_development\cnt\test\Test-cases-blr\v80-WM\scheduler\FRQ\AUTO\sml-lr454\Swap_MUL.xml
Z:\cnt_development\cnt\test\Test-cases-blr\v80-WM\scheduler\FRQ\AUTO\sml-lr456\Swap_MU.xml
I need to extract the complete part from v80-WM
i.e The regex must be able to select the following string
v80-WM\scheduler\FRQ\AUTO\sml-hr454\SRISM.xml
v80-WM\scheduler\FRQ\AUTO\sml-lr454\Swap_MUL.xml
v80-WM\scheduler\FRQ\AUTO\sml-lr456\Swap_MU.xml
Currently I am using the following regex where the regex finds the last occurence of "Q" in the above string and trimming for there and using workardoung to construct the above mentioned results.
<echo message="runpART ... Scheduler File ${schedulerFile}"/>
<propertyregex property="cfg.arg" input="${schedulerFile}" regexp="([^Q]*).xml" select="\1" casesensitive="false"/>
Need help in extracting string from "v80-WM....xml".
Some inputs will be helpful
That's good. The v80-WM gives you a fixed "starting point"
Using this as your regular expression should do it.
^.(v80-WM.)
What it means:
^.* match anything until you get to *the caret isn't really necessary, but I like making the reg exp more strict)
v80-WM
= .* then match the rest
The parens include the v80-WM name and everything that comes after so you don't have to reconstruct it.
Could anybody help me make a proper regular expression from a bunch of text in Ruby. I tried a lot but I don't know how to handle variable length titles.
The string will be of format <sometext>title:"<actual_title>"<sometext>. I want to extract actual_title from this string.
I tried /title:"."/ but it doesnt find any matches as it expects a closing quotation after one variable from opening quotation. I couldn't figure how to make it check for variable length of string. Any help is appreciated. Thanks.
. matches any single character. Putting + after a character will match one or more of those characters. So .+ will match one or more characters of any sort. Also, you should put a question mark after it so that it matches the first closing-quotation mark it comes across. So:
/title:"(.+?)"/
The parentheses are necessary if you want to extract the title text that it matched out of there.
/title:"([^"]*)"/
The parentheses create a capturing group. Inside is first a character class. The ^ means it's negated, so it matches any character that's not a ". The * means 0 or more. You can change it to one or more by using + instead of *.
I like /title:"(.+?)"/ because of it's use of lazy matching to stop the .+ consuming all text until the last " on the line is found.
It won't work if the string wraps lines or includes escaped quotes.
In programming languages where you want to be able to include the string deliminator inside a string you usually provide an 'escape' character or sequence.
If your escape character was \ then you could write something like this...
/title:"((?:\\"|[^"])+)"/
This is a railroad diagram. Railroad diagrams show you what order things are parsed... imagine you are a train starting at the left. You consume title:" then \" if you can.. if you can't then you consume not a ". The > means this path is preferred... so you try to loop... if you can't you have to consume a '"' to finish.
I made this with https://regexper.com/#%2Ftitle%3A%22((%3F%3A%5C%5C%22%7C%5B%5E%22%5D)%2B)%22%2F
but there is now a plugin for Atom text editor too that does this.