I use this text text\r2. And I want to print this in debug and get result:
text\r2
but I get this:
text
2
Try to escape the backslash with another backslash: text\\r2.
The \r will otherwise be interpreted as a line break.
\r in a String literal is a special character and represents a carriage return
See Special Characters in String Literals
String literals can include the following special characters:
* The escaped special characters \0 (null character), \\ (backslash), \t (horizontal tab), \n (line feed), \r (carriage return), \" (double quotation mark) and \' (single quotation mark)
* An arbitrary Unicode scalar, written as \u{n}, where n is a 1–8 digit hexadecimal number with a value equal to a valid Unicode code point (Unicode is discussed in Unicode below)
If you want to use in a String Literal a backslash you have to escape it using \\.
So you'll have to write
print("text\\r2")
to get text\r2
Related
For example, if I were to do this:
print(“\”)
It would say: `unfinished string near: ‘“”’
instead of my expected output of: ‘\’
How would I print this? I have searched on google yet still have yet to find an answer.
The backslash (\) is escaping the following character, being the double quote ("), causing the string to be unfinished.
To include an actual backslash in your string, you escape it with another backslash:
print("\\")
From Lua 5.4 Reference Manual, §3.1 (emphasis mine):
A short literal string can be delimited by matching single or double quotes, and can contain the following C-like escape sequences: '\a' (bell), '\b' (backspace), '\f' (form feed), '\n' (newline), '\r' (carriage return), '\t' (horizontal tab), '\v' (vertical tab), '\\' (backslash), '"' (quotation mark [double quote]), and ''' (apostrophe [single quote]). [...]
I tried to split by "\", but this character is so special in Lua, even if I use escape character "%", the IDE shows an error Unterminated String constant
local index = string.find("lua. is \wonderful", "%\", 1)
To insert backslash \ into a quoted string, escape it with itself: "\\". \ is the escape character in regular quoted strings, so it is escaped with \. Or you can use the long string syntax, which doesn't allow escape sequences, as already pointed out: [[\]].
Percent is only an escape character in a string that is being used as a pattern, so it is used before the magical characters ^$()%.[]*+-? in the second argument to string.find, string.match, string.gmatch, and string.gsub, and %% represents % in the third argument to string.gsub.
The percent is still there in the string that is stored in memory, but backslash escape sequences are replaced with the corresponding character. \\ becomes \ when the string is stored in memory, and if you count the number of backslashes in a string "\\" using string.gsub, it will only find one: select(2, string.gsub("\\", "\\", "")) returns 1.
In Lua (I can only find examples in other languages), how do I remove all punctuation, special characters and whitespace from a string? So, for example, s t!r#i%p^(p,e"d would become stripped?
In Lua patterns, the character class %p represents all punctuation characters, the character class %c represents all control characters, and the character class %s represents all whitespace characters. So you can represent all punctuation characters, all control characters, and all whitespace characters with the set [%p%c%s].
To remove these characters from a string, you can use string.gsub. For a string str, the code would be the following:
str = str:gsub('[%p%c%s]', '')
(Note that this is essentially the same as Egor's code snippet above.)
If you remove all special chars, whitespace, … all that's left is letters and numbers, right? So if str is your string,
str:gsub( "%W", "" )
will be the cleaned string.
%w matches all word characters, upper-case it %W to match all non-word characters. If that's not exactly what you want to match, you can build your own character class -- e.g. if you wanted to include _ as an acceptable character, you could use [^%w_].
This works for me
m=your_string:gsub('%W','')
I am trying to validate a phone number using NSPredicate and regex. The only problem is when setting the regex Swift thinks that I am trying to escape part of it due to the backslashes. How can I get around this?
My code is as follows:
let phoneRegEx = "^((\(?0\d{4}\)?\s?\d{3}\s?\d{3})|(\(?0\d{3}\)?\s?\d{3}\s?\d{4})|(\(?0\d{2}\)?\s?\d{4}\s?\d{4}))(\s?\#(\d{4}|\d{3}))?$"
In Swift regular string literals, you need to double-escape the slashes to define literal backslashes:
let phoneRegEx = "^((\\(?0\\d{4}\\)?\\s?\\d{3}\\s?\\d{3})|(\\(?0\\d{3}\\)?\\s?\\d{3}\\s?\\d{4})|(\\(?0\\d{2}\\)?\\s?\\d{4}\\s?\\d{4}))(\\s?#(\\d{4}|\\d{3}))?$"
Starting from Swift 5, you can use raw string literals and escape regex escapes with a single backslash:
let phoneRegEx = #"^((\(?0\d{4}\)?\s?\d{3}\s?\d{3})|(\(?0\d{3}\)?\s?\d{3}\s?\d{4})|(\(?0\d{2}\)?\s?\d{4}\s?\d{4}))(\s?#(\d{4}|\d{3}))?$"#
Please refer to the Regular Expression Metacharacters table on the ICU Regular Expressions page to see what regex escapes should be escaped this way.
Please mind the difference between the regex escapes (in the above table) and string literal escape sequences used in the regular string literals that you may check, say, at Special Characters in String Literals:
String literals can include the following special characters:
The escaped special characters \0 (null character), \\ (backslash), \t (horizontal tab), \n (line feed), \r (carriage return), \" (double quotation mark) and \' (single quotation mark)
An arbitrary Unicode scalar value, written as \u{n}, where n is a 1–8 digit hexadecimal number (Unicode is discussed in Unicode below)
So, in regular string literals, "\"" is a " string written as a string literal, and you do not have to escape a double quotation mark for the regex engine, so "\"" string literal regex pattern is enough to match a " char in a string. However, "\\\"", a string literal repesenting \" literal string will also match " char, although you can already see how redundant this regex pattern is. Also, "\n" (an LF symbol) matches a newline in the same way as "\\n" does, as "\n" is a literal representation of the newline char and "\\n" is a regex escape defined in the ICU regex escape table.
In raw string literals, \ is just a literal backslash.
What is the \b equivalent in swift? I have to split a string which is received from server with \b?
From "Strings and Characters" in the Swift
Reference:
Special Characters in String Literals
String literals can include the following special characters:
The escaped special characters \0 (null character), \\ (backslash), \t
(horizontal tab), \n (line feed), \r (carriage return), \" (double
quote) and \' (single quote)
An arbitrary Unicode scalar, written as
\u{n}, where n is a 1–8 digit hexadecimal number with a value equal to
a valid Unicode code point
So Swift does not have a special character for the backspace
character, like \b in the C language. You can use the Unicode
special character \u{n}:
let string = "THIS\u{8}IS\u{8}A\u{8}TEST"
or create a string from the Unicode value:
let bs = String(UnicodeScalar(8))
let string = "THIS\(bs)IS\(bs)A\(bs)TEST"