What's the , Lua equivalent of pythons endswith()? - lua

I want to convert this python code to lua .
for i in range(1000,9999):
if str(i).endswith('9'):
print(i)
I've come this far ,,
for var=1000,9000 then
if tostring(var).endswith('9') then
print (var)
end
end
but I don't know what's the lua equivalent of endswith() is ,,, im writing an nmap script,,
working 1st time with lua so pls let me know if there are any errors ,, on my current code .

The python code is not great, you can get the last digit by using modulo %
# python code using modulo
for i in range(1000,9999):
if i % 10 == 9:
print(i)
This also works in Lua. However Lua includes the last number in the loop, unlike python.
-- lua code to do this
for i=1000, 9998 do
if i % 10 == 9 then
print(i)
end
end
However in both languages you could iterate by 10 each time
for i in range(1009, 9999, 10):
print(i)
for i=9, 9998, 10 do
print(i)

for var = 1000, 9000 do
if string.sub(var, -1) == "9" then
-- do your stuff
end
end

XY-Problem
The X problem of how to best port your code to Lua has been answered by quantumpro already, who optimized it & cleaned it up.
I'll focus on your Y problem:
What's the Lua equivalent of Python endswith?
Calling string functions, OOP-style
In Lua, strings have a metatable that indexes the global string library table. String functions are called using str:func(...) in Lua rather than str.func(...) to pass the string str as first "self" argument (see "Difference between . and : in Lua").
Furthermore, if the argument to the call is a single string, you can omit the parentheses, turning str:func("...") into str:func"...".
Constant suffix: Pattern Matching
Lua provides a more powerful pattern matching function that can be used to check whether a string ends with a suffix: string.match. str.endswith("9") in Python is equivalent to str:match"9$" in Lua: $ anchors the pattern at the end of the string and 9 matches the literal character 9.
Be careful though: This approach doesn't work with arbitrary, possibly variable suffices since certain characters - such as $ - are magic characters in Lua patterns and thus have a special meaning. Consider str.endswith("."); this is not equivalent to string:match".$" in Lua, since . matches any character.
I'd say that this is the lua-esque way of checking whether a string ends with a constant suffix. Note that it does not return a boolean, but rather a match (the suffix, a truthy value) if successful or nil (a falsey value) if unsuccessful; it can thus safely be used in ifs. To convert the result into a boolean, you could use not not string:match"9$".
Variable suffix: Rolling your own
Lua's standard library is very minimalistic; as such, you often need to roll your own functions even for basic things. There are two possible implementations for endswith, one using pattern matching and another one using substrings; the latter approach is preferable because it's shorter, possibly faster (Lua uses a naive pattern matching engine) and doesn't have to take care of pattern escaping:
function string:endswith(suffix)
return self:sub(-#suffix) == suffix
end
Explanation: self:sub(-#suffix) returns the last suffix length characters of self, the first argument. This is compared against the suffix.
You can then call this function using the colon (:) syntax:
str = "prefixsuffix"
assert(str:endswith"suffix")
assert(not str:endswith"prefix")

Related

How to capture a string between signs in lua?

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.

Lua Pattern matching only returning first match

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.

calculating string length causes confustion

I am confused by the following output:
local a = "string"
print(a.len) -- function: 0xc8a8f0
print(a.len(a)) -- 6
print(len(a))
--[[
/home/pi/test/wxlua/wxLua/ZeroBraneStudio/bin/linux/armhf/lua: /home/pi/Desktop/untitled.lua:4: attempt to call global 'len' (a nil value)
stack traceback:
/home/pi/Desktop/untitled.lua:4: in main chunk
[C]: ?
]]
What is the proper way to calculate a string length in Lua?
Thank you in advance,
You can use:
a = "string"
string.len(a)
Or:
a = "string"
a:len()
Or:
a = "string"
#a
EDIT: your original code is not idiomatic but is also working
> a = "string"
> a.len
function: 0000000065ba16e0
> a.len(a)
6
The string a is linked to a table (named metatable) containing all the methods, including len.
A method is just a function, taking the string as the first parameter.
function a.len (string) .... end
You can call this function, a.len("test") just like a normal function. Lua has a special syntax to make it easier to write. You can use this special syntax and write a:len(), it will be equivalent to a.len(a).
print(a.len) -- function: 0xc8a8f0
This prints a string representation of a.len which is a function value. All strings share a common metatable.
From Lua 5.4 Reference Manual: 6.4 String Manipulation:
The string library provides all its functions inside the table string.
It also sets a metatable for strings where the __index field points to
the string table. Therefore, you can use the string functions in
object-oriented style. For instance, string.byte(s,i) can be written
as s:byte(i).
So given that a is a string value, a.len actually refers to string.len
For the same reason
print(a.len(a))
is equivalent to print(string.len(a)) or print(a:len()). This time you called the function with argument a instead of printing its string representation so you print its return value which is the length of string a.
print(len(a))
on the other hand causes an error because you attempt to call a global nil value. len does not exist in your script. It has never been defined and is hence nil. Calling nil values doesn't make sense so Lua raises an error.
According to Lua 5.4 Reference Manual: 3.4.7 Length Operator
The length of a string is its number of bytes. (That is the usual
meaning of string length when each character is one byte.)
You can also call print(#a) to print a's length.
The length operator was introduced in Lua 5.1,

Is there a way to specify the argument positions in the format string for `string.format`?

In C, I can tell printf to print the arguments in an order different than the order they are passed in:
printf("%2$d %1$d\n", 10, 20);
//prints 20 10
However, if I try to do the same in Lua I get an error:
print(string.format("%2$d %1$d\n", 10, 20))
invalid option '%$' to 'format'
Is there a way to create a Lua format string that causes string.format to write the second argument before the first? I am working with an internationalization and changing the format string is easy but changing the argument order is much more tricky.
I would have expected the technique that I used in C to work with Lua because, according to the manual, string.format should receive the same parameters as sprintf. Are %2$ directives not part of ANSI C or is the Lua manual just forgetting to mention that they are not supported?
In short, no. %2$ directives are a POSIX extension, thus not part of ANSI C or Lua. This has been brought up on the Lua mailing list before, and according to lhf, the feature was around in versions prior to Lua 5 but was removed with that version's release. In the same thread, a wiki page of alternatives was suggested.
If you really want the %2$ style, then it's not too difficult to cook up your own fix either.
local function reorder(fmt, ...)
local args, order = {...}, {}
fmt = fmt:gsub('%%(%d+)%$', function(i)
table.insert(order, args[tonumber(i)])
return '%'
end)
return string.format(fmt, table.unpack(order))
end
print(reorder('%2$d %1$d\n', 10, 20))
You cannot do this with string.format, but you can actually achieve almost the same result with string.gsub. The caveat here is that the last argument of string.gsub can be either string or table (with multiple values to replace)
So this code would do the trick:
local output = string.gsub("%2 %1 %2 %1\n", '%S+', {['%1'] = 10, ['%2'] = 20})
print(output)
> 20 10 20 10

Escaping strings for gsub

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

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