How would i for an example remove the h from "helloh" so it will look like "ello"
I have no idea what else to write but it looks like i need to write some more text and maybe add some code so this is just junk.
print("junk 1")
print("junk 2")
print("junk 4")
You can used string.gsub to replace characters, gsub stands for global subtitusion.
print(("helloh"):gsub("h", "")) -- replace all instances of `h` with empty string
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 working on a project, in which you type your input sentence, and I need to be able to use " and ' in the sentence, such as Input = "I said, "Hi what's up?" print(Input) in which I get an error. If anyone knows how to fix this that would be great.
See https://www.lua.org/pil/2.4.html. Lua has very interesting feature to declare string with square brackets:
input = [[I said, "Hi what's up?"]]
input = "I said, \"Hi what's up?\""
input = 'I said, "Hi what\'s up?"'
I will tell some things in addition to what #Darius told above
When you tried to add a quatation mark inside a string, the lua interpreter get confused and break your string after the next quation mark without reaching the end of the line. That's the reason for the error.
Try to understand it by the following code
str = "Hello I"m somebody" -- here the interpreter will think str equals to "Hello I" at first, and then it will find some random characters after which may make it confused (as m somebody is neither a variable nor a keyword)"
-- you can also see the way it got confused by looking at the highlighted code
--What you can do to avoid this is escaping the quotes
str = "Hello I\"m somebody" -- here the interpreter will treat \" as a raw character (") and parse the rest.
You can also use the escape character () with others such as \', \", \[, \n (newline character), \t (tab) and so on.
how to make sure my string format must be like this :
locker_number=3,email=ucup#gmail.com,mobile_phone=091332771331,firstname=ucup
i want my string format `"key=value,"
how to make regex for check my string on ruby?
This regex will find what you're after.
\w+=.*?(,|$)
If you want to capture each pairing use
(\w+)=(.*?)(?:,|$)
http://rubular.com/r/A2ernIzQkq
The \w+ is one or more occurrences of a character a-z, 1-9, or an underscore. The .*? is everything until the first , or the end of the string ($). The pipe is or and the ?: tells the regex no to capture that part of the expression.
Per your comment it would be used in Ruby as such,
(/\w+=.*?(,|$)/ =~ my_string) == 0
You can use a regex like this:
\w+=.*?(,|$)
Working demo
You can use this code:
"<your string>" =~ /\w+=.*?(,|$)/
What about something like this? It's picky about the last element not ending with ,. But it doesn't enforce the need for no commas in the key or no equals in the value.
'locker_number=3,email=ucup#gmail.com,mobile_phone=091332771331,firstname=ucup' =~ /^([^=]+=[^,]+,)*([^=]+=[^,]+)$/
Data stored in the database is like this:
This is a line
This is another line
How about this line
When I output it to the view, I want to convert that to:
This is a line\n\nThis is another line\n\nHow about this line
with no new lines and the actual \n characters printed out. How can I do that?
> s = "hi\nthere"
> puts s
hi
there
> puts s.gsub(/\n/, "\\n")
hi\nthere
I would personally use gsub if you only want newlines specifically converted. However, if you want to generally inspect the contents of the string, do this:
str = "This is a line\n\nThis is another line\n\nHow about this line"
puts str.inspect[1..-2]
#=> This is a line\n\nThis is another line\n\nHow about this line
The String#inspect method escapes various 'control' characters in your string. It also wraps the string with ", which I've stripped off above. Note that this may produce undesirable results, e.g. the string My name is "Phrogz" will come out as My name is \"Phrogz\".
> s = "line 1\n\nline2"
=> "line 1\n\nline2"
> puts s
line 1
line2
> puts s.gsub("\n", "\\n")
line 1\n\nline2
The key is to escape the single backslash.
I'm trying to remove any lines that begin with the character '>' in a long string (i.e. replies to an email).
In PHP I'd iterate over each line with an if statement, in linux I'd try and use sed or awk.
What's the most elegant rails approach?
You can try this:
your_string.gsub(/^\>.+\n/,'')
Your question is implying that the input is one string, containing multiple lines.
Do you want the output to be just one string with multiple lines as well? I'm assuming yes.
either using String and Array operations:
str.lines.reject{|x| x =~ /^>/}.join # this will return a new string, without those ">" lines
or using Regular Expressions:
str.gsub(/^>.+\n*/. '')
Better Solution:
You will need to use non-greedy multi-line matching mode for your Regular Expression:
str.gsub(/^>.*?$\n*/m, '') # by using gsub!() you can modify the string in place
^> matches your ">" character at the start of a line
.*?$ matches any characters after the start character until the end of the line (non-greedy)
\n* matches the newline character itself if any (you want to remove that as well)
the "m" at the end of the regular expressions indicates multi-line matching , which will apply the RegExp for each line in the string.
It should work as you expect:
your_string.lines.to_a.reject{|line| line[0] == '>'}.join