local a = "te\st"
local b = string.gsub(a,'\','\\\\')
assert(false,b)
What am I doing wrong?
When I do assert, I want that to the screen the string te\st will be printed... but it's not working
I have a JSON file, that I want to decode it into Lua table. I don't need to print out nothing, I did the assert just to test a local problem.
So what I need is to keep all data in the JSON file that has '\'.
Use [[]] instead of "" or '' if you don't want backslash to have special meaning.
Read about literal strings in the manual.
Have you tried escaping it with the % character instead of \
I don't know if this will help, but I was having a HELL of a time making Lua's gsub match my string with special characters in it that I wanted treated literally... it turned out that instead of using \ as an escape character, or doubling the character, that I needed to prefix the special character with % to make it be treated literally.
Your question wasn't too clear so I'm not 100% sure what you mean. Do you mean that you want the assert to fire when b is equal to the string "te\st"? If so you can do a simple:
assert(b ~= "te\st")
Or I suppse...
assert(b ~= a)
You don't need the gsub. But here it is anyways.
local a = "te\\st"
local b = string.gsub(a,'\\','\\')
assert(false,b)
Related
So I'm writing a Lua script and I tested it but I got an error that I don't know how to fix:
.\search.lua:10: malformed pattern (missing ']')
Below is my code. If you know what I did wrong, it would be very helpful if you could tell me.
weird = "--[[".."\n"
function readAll(file)
local c = io.open(file, "rb")
local j = c:read("*all")
c:close()
return(j)
end
function blockActive()
local fc = readAll("functions.lua")
if string.find(fc,weird) ~= nil then
require("blockDeactivated")
return("false")
else
return("true")
end
end
print(blockActive())
Edit: first comment had the answer. I changed
weird = "--[[".."\n" to weird = "%-%-%[%[".."\r" The \n to \r change was because it was actually supposed to be that way in the first place.
This errors because string.find uses Lua Patterns.
Most non-alpha-numeric characters, such as "[", ".", "-" etc. convey special meaning.
string.find(fc,weird), or better, fc:find(weird) is trying to parse these special characters, and erroring.
You can use these patterns to cancel out your other patterns, however.
weird = ("--[["):gsub("%W","%%%0") .. "\r?\n"
This is a little daunting, but it will hopefully make sense.
the ("--[[") is the orignal first part of your weird string, working as expected.
:gsub() is a function that replaces a pattern with another one. Once again, see Patterns.
"%W" is a pattern that matches every string that isn't a letter, a number, or an underscore.
%%%0 replaces everything that matches with itself (%0 is a string that represents everything in this match), following a %, which is escaped.
So this means that [[ will be turned into %[%[, which is how find, and similar patterns 'escape' special characters.
The reason \n is now \r?\n refers back to these patterns. This matches it if it ends with a \n, like it did before. However, if this is running on windows, a newline might look like \r\n. (You can read up on this HERE). A ? following a character, \r in this case, means it can optionally match it. So this matches both --[[\n and --[[\r\n, supporting both windows and linux.
Now, when you run your fc:find(weird), it's running fc:find("%-%-%[%[\r?\n"), which should be exactly what you want.
Hope this has helped!
Finished code if you're a bit lazy
weird = ("--[["):gsub("%W","%%%0") .. "\r?\n" // Escape "--[[", add a newline. Used in our find.
// readAll(file)
// Takes a string as input representing a filename, returns the entire contents as a string.
function readAll(file)
local c = io.open(file, "rb") // Open the file specified by the argument. Read-only, binary (Doesn't autoformat things like \r\n)
local j = c:read("*all") // Dump the contents of the file into a string.
c:close() // Close the file, free up memory.
return j // Return the contents of the string.
end
// blockActive()
// returns whether or not the weird string was matched in 'functions.lua', executes 'blockDeactivated.lua' if it wasn't.
function blockActive()
local fc = readAll("functions.lua") // Dump the contents of 'functions.lua' into a string.
if fc:find(weird) then // If it functions.lua has the block-er.
require("blockDeactivated") // Require (Thus, execute, consider loadfile instead) 'blockDeactived.lua'
return false // Return false.
else
return true // Return true.
end
end
print(blockActive()) // Test? the blockActve code.
Hi I've been struggling with this for the last hour and am no closer. How exactly do I strip everything except numbers, commas and decimal points from a rails string? The closest I have so far is:-
rate = rate.gsub!(/[^0-9]/i, '')
This strips everything but the numbers. When I try add commas to the expression, everything is getting stripped. I got the aboves from somewhere else and as far as I can gather:
^ = not
Everything to the left of the comma gets replaced by what's in the '' on the right
No idea what the /i does
I'm very new to gsub. Does anyone know of a good tutorial on building expressions?
Thanks
Try:
rate = rate.gsub(/[^0-9,\.]/, '')
Basically, you know the ^ means not when inside the character class brackets [] which you are using, and then you can just add the comma to the list. The decimal needs to be escaped with a backslash because in regular expressions they are a special character that means "match anything".
Also, be aware of whether you are using gsub or gsub!
gsub! has the bang, so it edits the instance of the string you're passing in, rather than returning another one.
So if using gsub! it would be:
rate.gsub!(/[^0-9,\.]/, '')
And rate would be altered.
If you do not want to alter the original variable, then you can use the version without the bang (and assign it to a different var):
cleaned_rate = rate.gsub!(/[^0-9,\.]/, '')
I'd just google for tutorials. I haven't used one. Regexes are a LOT of time and trial and error (and table-flipping).
This is a cool tool to use with a mini cheat-sheet on it for ruby that allows you to quickly edit and test your expression:
http://rubular.com/
You can just add the comma and period in the square-bracketed expression:
rate.gsub(/[^0-9,.]/, '')
You don't need the i for case-insensitivity for numbers and symbols.
There's lots of info on regular expressions, regex, etc. Maybe search for those instead of gsub.
You can use this:
rate = rate.gsub!(/[^0-9\.\,]/g,'')
Also check this out to learn more about regular expressions:
http://www.regexr.com/
I have two strings - each string has many lines like the following:
string1 = " DEFAULT-VLAN | Manual 10.1.1.3 255.255.255.0 "
string2 = " 1 DEFAULT-VLAN | Port-based No No"
The first string I split into the following strings: "DEFAULT-VLAN", "|", "Manual"...
Then I want to look up the ID ("1") in string2 for the vlanName ("DEFAULT-VLAN") from string1.
I use this code to find the correct substring:
vpos1, vpos2 = vlan:find("%d-%s-" .. vlanName .. "%s-|")
But vpos1 and vpos2 are nil; When the hyphen ("-") is deleted from the vlanName it is working.
Shouldn't Lua take care to escape the special characters in such strings? The string is handed over from my C++ application to Lua and there may be lots of special characters.
Is there an easy way to solve this?
Thanks!
Lua is not magic. All the expression "%d-%s-" .. vlanName .. "%s-|" does is concatenate some strings, producing a final string. It has no idea what that string is intended to be used for. Only string.find knows that, and it can't have any affect on how the parameter it is given will be used.
So yes, vlanName will be interpreted as a Lua pattern. And if you want to use special characters, you will need to escape them. I would suggest using string.gsub for that. It'd be something like this:
vlanName:gsub("[%-...]", "%%%0")
Where ... are any other characters you want to escape.
I read a file:
local logfile = io.open("log.txt", "r")
data = logfile:read("*a")
print(data)
output:
...
"(\.)\n(\w)", r"\1 \2"
"\n[^\t]", "", x, re.S
...
Yes, logfile looks awful as it's full of various commands
How can I call gsub and remove i.e. "(\.)\n(\w)", r"\1 \2" line from data variable?
Below snippet, does not work:
s='"(\.)\n(\w)", r"\1 \2"'
data=data:gsub(s, '')
I guess some escaping needs to be done. Any easy solution?
Update:
local data = [["(\.)\n(\w)", r"\1 \2"
"\n[^\t]", "", x, re.S]]
local s = [["(\.)\n(\w)", r"\1 \2"]]
local function esc(x)
return (x:gsub('%%', '%%%%')
:gsub('^%^', '%%^')
:gsub('%$$', '%%$')
:gsub('%(', '%%(')
:gsub('%)', '%%)')
:gsub('%.', '%%.')
:gsub('%[', '%%[')
:gsub('%]', '%%]')
:gsub('%*', '%%*')
:gsub('%+', '%%+')
:gsub('%-', '%%-')
:gsub('%?', '%%?'))
end
print(data:gsub(esc(s), ''))
This seems to works fine, only that I need to escape, escape character %, as it wont work if % is in matched string. I tried :gsub('%%', '%%%%') or :gsub('\%', '\%\%') but it doesn't work.
Update 2:
OK, % can be escaped this way if set first in above "table" which I just corrected
:terrible experience:
Update 3:
Escaping of ^ and $
As stated in Lua manual (5.1, 5.2, 5.3)
A caret ^ at the beginning of a pattern anchors the match at the beginning of the subject string. A $ at the end of a pattern anchors the match at the end of the subject string. At other positions, ^ and $ have no special meaning and represent themselves.
So a better idea would be to escape ^ and $ only when they are found (respectively) and the beginning or the end of the string.
Lua 5.1 - 5.2+ incompatibilities
string.gsub now raises an error if the replacement string contains a % followed by a character other than the permitted % or digit.
There is no need to double every % in the replacement string. See lua-users.
According to Programming in Lua:
The character `%´ works as an escape for those magic characters. So, '%.' matches a dot; '%%' matches the character `%´ itself. You can use the escape `%´ not only for the magic characters, but also for all other non-alphanumeric characters. When in doubt, play safe and put an escape.
Doesn't this mean that you can simply put % in front of every non alphanumeric character and be fine. This would also be future proof (in the case that new special characters are introduced). Like this:
function escape_pattern(text)
return text:gsub("([^%w])", "%%%1")
end
It worked for me on Lua 5.3.2 (only rudimentary testing was performed). Not sure if it will work with older versions.
Why not:
local quotepattern = '(['..("%^$().[]*+-?"):gsub("(.)", "%%%1")..'])'
string.quote = function(str)
return str:gsub(quotepattern, "%%%1")
end
to escape and then gsub it away?
try
line = '"(\.)\n(\w)", r"\1 \2"'
rx = '\"%(%\.%)%\n%(%\w%)\", r\"%\1 %\2\"'
print(string.gsub(line, rx, ""))
escape special characters with %, and quotes with \
Try s=[["(\.)\n(\w)", r"\1 \2"]].
Use stringx.replace() from Penlight Lua Libraries instead.
Reference: https://stevedonovan.github.io/Penlight/api/libraries/pl.stringx.html#replace
Implementation (v1.12.0): https://github.com/lunarmodules/Penlight/blob/1.12.0/lua/pl/stringx.lua#L288
Based on their implementation:
function escape(s)
return (s:gsub('[%-%.%+%[%]%(%)%$%^%%%?%*]','%%%1'))
end
function replace(s,old,new,n)
return (gsub(s,escape(old),new:gsub('%%','%%%%'),n))
end
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.