从字符串中删除“字符

Swift uses backslash to escape double quotes. Here is the list of escaped special characters in Swift:

• \0 (null character)
• \\ (backslash)
• \t (horizontal tab)
• \n (line feed)
• \r (carriage return)
• \" (double quote)
• \' (single quote)

This should work:

text2 = text2.replacingOccurrences(of: "\\", with: "", options: NSString.CompareOptions.literal, range: nil)

As Martin R says, your string "Optional("5")" looks like you did something wrong.

dasblinkenlight answers you so it is fine, but for future readers, I will try to add alternative code as:

if let realString = yourOriginalString {
text2 = realString
} else {
text2 = ""
}

text2 in your example looks like String and it is maybe already set to "" but it looks like you have an yourOriginalString of type Optional(String) somewhere that it wasn't cast or use correctly.

I hope this can help some reader.

To remove the optional you only should do this

println("\(text2!)")

cause if you dont use "!" it takes the optional value of text2

And to remove "" from 5 you have to convert it to NSInteger or NSNumber easy peasy. It has "" cause its an string.

I've eventually got this to work in the playground, having multiple characters I'm trying to remove from a string:

var otherstring = "lat\" : 40.7127837,\n"
var new = otherstring.stringByTrimmingCharactersInSet(NSCharacterSet.init(charactersInString: "la t, \n \" ':"))
count(new) //result = 10
println(new)
//yielding what I'm after just the numeric portion 40.7127837

Swift 3 and Swift 4:

text2 = text2.textureName.replacingOccurrences(of: "\"", with: "", options: NSString.CompareOptions.literal, range:nil)

Latest documents updated to Swift 3.0.1 have:

• Null Character (\0)
• Backslash (\\)
• Horizontal Tab (\t)
• Line Feed (\n)
• Carriage Return (\r)
• Double Quote (\")
• Single Quote (\')
• Unicode scalar (\u{n}), where n is between one and eight hexadecimal digits

If you need more details you can take a look to the official docs here

You've instantiated text2 as an Optional (e.g. var text2: String?). This is why you receive Optional("5") in your string. take away the ? and replace with:

var text2: String = ""

Here is the swift 3 updated answer

var editedText = myLabel.text?.replacingOccurrences(of: "\"", with: "")

Null Character (\0)
Backslash (\\)
Horizontal Tab (\t)
Line Feed (\n)
Carriage Return (\r)
Double Quote (\")
Single Quote (\')
Unicode scalar (\u{n})

Let's say you have a string:

var string = "potatoes + carrots"

And you want to replace the word "potatoes" in that string with "tomatoes"

string = string.replacingOccurrences(of: "potatoes", with: "tomatoes", options: NSString.CompareOptions.literal, range: nil)

If you print your string, it will now be: "tomatoes + carrots"

If you want to remove the word potatoes from the sting altogether, you can use:

string = string.replacingOccurrences(of: "potatoes", with: "", options: NSString.CompareOptions.literal, range: nil)

If you want to use some other characters in your sting, use:

• Null Character (\0)
• Backslash (\)
• Horizontal Tab (\t)
• Line Feed (\n)
• Carriage Return (\r)
• Double Quote (\")
• Single Quote (\')

Example:

string = string.replacingOccurrences(of: "potatoes", with: "dog\'s toys", options: NSString.CompareOptions.literal, range: nil)

Output: "dog's toys + carrots"

If you want to remove more characters for example "a", "A", "b", "B", "c", "C" from string you can do it this way:

someString = someString.replacingOccurrences(of: "[abc]", with: "", options: [.regularExpression, .caseInsensitive])

If you are getting the output Optional(5) when trying to print the value of 5 in an optional Int or String, you should unwrap the value first:

if let value = text {
print(value)
}

Now you've got the value without the "Optional" string that Swift adds when the value is not unwrapped before.

Replacing for Removing is not quite logical. String.filter allows to iterate a string char by char and keep only true assertion.

Swift 4 & 5

var aString = "Optional(\"5\")"


aString = aString.filter { $0 != "\"" }


> Optional(5)

Or to extend

var aString = "Optional(\"5\")"


let filteredChars = "\"\n\t"


aString = aString.filter { filteredChars.range(of: String($0)) == nil }


> Optional(5)

If you are getting the output Optional(5) when trying to print the value of 5 in an optional Int or String, you should unwrap the value first:

if value != nil
{ print(value)
}

or you can use this:

if let value = text {
print(value)
}

or in simple just 1 line answer:

print(value ?? "")

The last line will check if variable 'value' has any value assigned to it, if not it will print empty string

Swift 5 (working). Only 1 line code.

For removing single / multiple characters.

trimmingCharacters(in: CharacterSet)

In action:

var yourString:String = "(\"This Is: Your String\")"
yourString = yourString.trimmingCharacters(in: ["("," ",":","\"",")"])
print(yourString)

Output:

ThisIsYourString

You are entering a Set that contains characters you're required to trim.