I have a text like "$ $abc $$abc ${a} ${}". I would like to completely disable string interpolation for the string and not to escape each and individual $ from the string. What should I do? In Scala you declare a string where interpolation is enabled with s"$ $abc $$abc ${a} ${}" while the normal string is not interpolated.
String interpolation is available for both regular and raw strings ("""). So you need to escape them, which is easier in a regular String obviously (see here).
"$ \$abc \$\$abc \${a} \${}"
I'm sorry but there's no other way I'm afraid.
Related
I scanned a document in to kotlin and it has words, numbers, values, etc... but I only want the values that start with a $ and have 2 decimal places after the .(so the price) do I use a combination of a substring with other string parses?
Edit: I have looked into Regex and the problem I am having now is I am using this line
val reg = Regex("\$([0-9]*\.[0-9]*)")
to grab all the prices however the portion of *. is saying Invalid escape. However in other languages this works just fine.
You have to use double \ instead of single . It's because the \ is an escape character both in Regex and in Kotlin/Java strings. So when \ appears in a String, Kotlin expects it to be followed by a character that needs to be escaped. But you aren't trying to escape a String's character...you're trying to escape a Regex character. So you have to escape your backslash itself using another backslash, so the backslash is part of the computed String literal and can be understood by Regex.
You also need double \ before your dollar sign for it to behave correctly. Technically, I think it should be triple \ because $ is a special character in both Kotlin and in Regex and you want to escape it in both. However, Kotlin seems smart enough to guess what you're trying to do with a double escape if no variable name or expression follows the dollar sign. Rather than rely on that, I would use the triple escape.
val reg = Regex("\\\$([0-9]*\\.[0-9]*)")
In a normal String I can escape the ${variable} with a backslash:
"You can use \${variable} syntax in Kotlin."
Is it possible to do the same in a String literal? The backslash is no longer an escape character:
// Undesired: Produces "This \something will be substituted.
"""This \${variable} will be substituted."""
So far, the only solutions I see are String concatenation, which is terribly ugly, and nesting the interpolation, which starts to get a bit ridiculous:
// Desired: Produces "This ${variable} will not be substituted."
"""This ${"\${variable}"} will not be substituted."""
From kotlinlang.org:
If you need to represent a literal $ character in a raw string (which doesn't
support backslash escaping), you can use the following syntax:
val price = """
${'$'}9.99
"""
So, in your case:
"""This ${'$'}{variable} will not be substituted."""
As per String templates docs you can represent the $ directly in a raw string:
Templates are supported both inside raw strings and inside escaped strings. If you need to represent a literal $ character in a raw string (which doesn't support backslash escaping), you can use the following syntax:
val text = """This ${'$'}{variable} will be substituted."""
println(text) // This ${variable} will be substituted.
I am looking for a string like this: ">$3.45 in some HTML (I'm screen scraping), using this string as a regular expression: #"\">\\$"
The problem is that since the $ is a Regex character (match at end of line) my target is not found.
How do I write this string expression so NSRegularExpression will find the embedded ">$ in my HTML?
The \ is both the Objective-C string escape character and the regular-expression escape character... so to escape the $ you need to use:
#"\">\\$"
which creates a string containing a single \, and then that backslash is seen by NSRegularExpression and used to escape the $.
Note: At the time of writing this answer the question has been edited by a third party to remove the original problem!
I am trying to take a string like "Hello my name is Nick" and transform it to "Hello+my+name+is+Nick" to be passed through a URL. This would be easily done by replacing all the spaces with a + char however I also need to replace all special characters (. , ! &) with their ASCII values. I have searched the net but cannot find anything. I wonder if anyone knows of existing code to do this as its a fairly common task?
I think you're looking for this: HttpUtility.UrlEncode Method (String)
Handles non-URL compliant characters and spaces.
I have a string that is quite long and complicated, with special characters inside. I want to define this string as a variable, but don't want to escape each of them (because there are so many). I remember that in XML they have a special syntax for that, is there something similar for Objective-C?
Edit: I know I can save the thing in a file and load it easily, but is it possible to do so without a new file? I'm having quite some of them...
No, you have to escape the characters (though you only have to escape ", \, and control characters... is the string mostly control characters and quotes?)
A better idea might be to put the string in a file. Load it using +[NSString stringWithContentsOfFile:encoding:error:].