-- print the first non-empty line
repeat
line = io.read()
until line ~= ""
print(line)
So, the logic is like, keep reading the characters until the next character is empty? Thanks.
Did you read the comment?
-- print the first non-empty line
It prints the first non-empty line.
This is achieved by reading the input (keyboard, unless redirected) until a non-empty line has been entered. This line is then printed.
You read a line, check wether it is not an empty string "". If it is not, you read the next line. If it is you won't read again but print what you've read last.
Related
Let's say I have a string with the contents
local my_str = [[
line1
line2
line4
]]
I'd like to get the following table:
{"line1","line2","","line4"}
In other words, I'd like the blank line 3 to be included in my result. I've tried the following:
local result = {};
for line in string.gmatch(my_str, "[^\n]+") do
table.insert(result, line);
end
However, this produces a result which will not include the blank line 3.
How can I make sure the blank line is included? Am I just using the wrong regex?
Try this instead:
local result = {};
for line in string.gmatch(my_str .. "\n", "(.-)\n") do
table.insert(result, line);
end
If you don't want the empty fifth element that gives you, then get rid of the blank line at the end of my_str, like this:
local my_str = [[
line1
line2
line4]]
(Note that a newline at the beginning of a long literal is ignored, but a newline at the end is not.)
You can replace the + with *, but that won't work in all Lua versions; LuaJIT will add random empty strings to your result (which isn't even technically wrong).
If your string always includes a newline character at the end of the last line like in your example, you can just do something like "([^\n]*)\n" to prevent random empty strings and the last empty string.
In Lua 5.2+ you can also just use a frontier pattern to check for either a newline or the end of the string: [^\n]*%f[\n\0], but that won't work in LuaJIT either.
If you need to support LuaJIT and don't have the trailing newline in your actual string, then you could just add it manually:
string.gmatch(my_str .. "\n", "([^\n]*)\n")
I am using Lua in World of Warcraft.
I have this string:
"This\nis\nmy\nlife."
So when printed, the output is this:
This
is
my
life.
How can I store the entire string except the last line in a new variable?
So I want the output of the new variable to be this:
This
is
my
I want the Lua code to find the last line (regardless of how many lines in the string), remove the last line and store the remaining lines in a new variable.
Thank you.
So I found that Egor Skriptunoff's solutions in the comments worked very well indeed but I am unable to mark his comments as an answer so I'll put his answers here.
This removes the last line and stores the remaining lines in a new variable:
new_str = old_str:gsub("\n[^\n]*$", "")
If there is a new line marker at the end of the last line, Egor posted this as a solution:
new_str = old_str:gsub("\n[^\n]*(\n?)$", "%1")
While this removes the first line and stores the remaining lines in a new variable:
first_line = old_str:match("[^\n]*")
Thanks for your help, Egor.
Most efficient solution is plain string.find.
local s = "This\nis\nmy\nlife." -- string with newlines
local s1 = "Thisismylife." -- string without newlines
local function RemoveLastLine(str)
local pos = 0 -- start position
while true do -- loop for searching newlines
local nl = string.find(str, "\n", pos, true) -- find next newline, true indicates we use plain search, this speeds up on LuaJIT.
if not nl then break end -- We didn't find any newline or no newlines left.
pos = nl + 1 -- Save newline position, + 1 is necessary to avoid infinite loop of scanning the same newline, so we search for newlines __after__ this character
end
if pos == 0 then return str end -- If didn't find any newline, return original string
return string.sub(str, 1, pos - 2) -- Return substring from the beginning of the string up to last newline (- 2 returns new string without the last newline itself
end
print(RemoveLastLine(s))
print(RemoveLastLine(s1))
Keep in mind this works only for strings with \n-style newlines, if you have \n\r or \r\n easier solution would be a pattern.
This solution is efficient for LuaJIT and for long strings.
For small strings string.sub(s1, 1, string.find(s1,"\n[^\n]*$") - 1) is fine (Not on LuaJIT tho).
I scan it backward because it more easier to remove thing from back with backward scanning rather than forward it would be more complex if you scan forward and much simpler scanning backward
I succeed it in one take
function removeLastLine(str) --It will return empty string when there just 1 line
local letters = {}
for let in string.gmatch(str, ".") do --Extract letter by letter to a table
table.insert(letters, let)
end
local i = #letters --We're scanning backward
while i >= 0 do --Scan from bacward
if letters[i] == "\n" then
letters[i] = nil
break
end
letters[i] = nil --Remove letter from letters table
i = i - 1
end
return table.concat(letters)
end
print("This\nis\nmy\nlife.")
print(removeLastLine("This\nis\nmy\nlife."))
How the code work
The letters in str argument will be extracted to a table ("Hello" will become {"H", "e", "l", "l", "o"})
i local is set to the end of the table because we scan it from the back to front
Check if letters[i] is \n if it newline then goto step 7
Remove entry at letters[i]
Minus i with 1
Goto step 3 until i is zero if i is zero then goto step 8
Remove entry at letters[i] because it havent removed when checking for newline
Return table.concat(letters). Won't cause error because table.concat return empty string if the table is empty
#! /usr/bin/env lua
local serif = "Is this the\nreal life?\nIs this\njust fantasy?"
local reversed = serif :reverse() -- flip it
local pos = reversed :find( '\n' ) +1 -- count backwards
local sans_serif = serif :sub( 1, -pos ) -- strip it
print( sans_serif )
you can oneline it if you want, same results.
local str = "Is this the\nreal life?\nIs this\njust fantasy?"
print( str :sub( 1, -str :reverse() :find( '\n' ) -1 ) )
Is this the
real life?
Is this
lets suppose that i have this .txt file:
this is line one
hello world
line three
in Lua, i want to creat a string only with the content of line two something like
i want to get a specific line from this file and put into a string
io.open('file.txt', 'r')
-- reads only line two and put this into a string, like:
local line2 = "hello world"
Lua files has the same methods as io library.
That means files have read() with all options as well.
Example:
local f = io.open("file.txt") -- 'r' is unnecessary because it's a default value.
print(f:read()) -- '*l' is unnecessary because it's a default value.
f:close()
If you want some specific line you can call f:read() and do nothing with it until you begin reading required line.
But more proper solution will be f:lines() iterator:
function ReadLine(f, line)
local i = 1 -- line counter
for l in f:lines() do -- lines iterator, "l" returns the line
if i == line then return l end -- we found this line, return it
i = i + 1 -- counting lines
end
return "" -- Doesn't have that line
end
In BASIC I know of two instructions to print to the screen, PRINT, and WRITE, both of which automatically print strings with a newline at the end. I want to print a string without a newline. How can I do this? I'm using GW-BASIC.
Using PRINT with a semicolon will not print a new line:
10 REM The trailing semicolon prevents a newline
20 PRINT "Goodbye, World!";
Source: Rosettacode
When I try to output some data into a text file using FasterCSV, sometimes it adds the quotes to the concatenated string and sometimes it does not.
For instance:
FasterCSV.generate do |csv|
csv << ["E"+company_code]
csv << ["A"+company_name]
end
Both company_code and company_name are Strings and contains data but the output will show:
EtheCompanyCode
"AtheCompanyName"
I found how to force quoting in FasterCSV's docs but I need exactly the opposite and can not figure out why it quotes one line and not the other when they are both strings...
If anybody has the solution, I'll be deeply grateful for a lead :)
Thanks
If the real input is 'theCompanyName' and 'theCompanyCode' then I would also be confused by one line being quoted and the other not. But I suspect your real input is something else.
Most likely, the quoted line has some character that needs quoting, such as a comma; while the unquoted line doesn't. (Other characters that typically need quoting in Excel-style CSVs are quotation marks and newlines.)