Swift - Remove " character from string - ios

I have a string which is "Optional("5")". I need to remove the "" surrounding the 5. I have removed the 'Optional' by doing:
text2 = text2.stringByReplacingOccurrencesOfString("Optional(", withString: "", options: NSStringCompareOptions.LiteralSearch, range: nil)
I am having difficulties removing the " characters as they designate the end of a string in the code.

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)

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

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})

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.

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)

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

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])

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.

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.

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 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

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 = ""

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.

Related

Swift regex format to remove string contains brackets in iOS

Need to remove part of the string based on brackets.
For ex:
let str = "My Account (_1234)"
I wanted to remove inside brackets (_1234) and the result should be string My Account
Expected Output:
My Account
How can we achieve this with regex format or else using swift default class,need support on this.
Tried with separatedBy but it splits string into array but that is not the expected output.
str.components(separatedBy: "(")
Use replacingOccurrences(…)
let result = str.replacingOccurrences(of: #"\(.*\)"#,
with: "",
options: .regularExpression)
.trimmingCharacters(in: .whitespaces))
Regex ist not necessary. Just get the range of ( and extract the substring up to the lower bound of the range
let str = "My Account (_1234)"
if let range = str.range(of: " (") {
let account = String(str[..<range.lowerBound])
print(account)
}

Arabic Swift String isn't empty after removing all characters

I have this code to replace part of a string and remove white spaces:
let str = "باب ‏".replacingOccurrences(of: "باب", with: "").trimmingCharacters(in: .whitespacesAndNewlines)
print(str.count) /// gives 1 why not 0
But it gives me 1 always while it should be 0. Why?
If it's RTL Mark (and it probably is), that's "\u{200F}" in Swift. If you want to trim it with the whitespace, you'd just add it to your set. That'd be something like:
.trimmingCharacters(in: whitespacesAndNewlines
.union(CharacterSet(charactersIn: "\u{200f}")))
You can also just replace that directly:
.replacingOccurrences(of: "\u{200f}باب", with: "")
Keep in mind the layout rules, since sometimes bidirectional literal strings like this can get confusing in the editor. You may want to separate the Arabic like:
let bab = "باب"
let rtl = "\u{200f}"
string.replacingOccurrences(of: rtl + bab, with: "")
Let's look at the content of your original string:
func hexCharactersArray(_ string: String) -> String {
string.unicodeScalars.map { String(format: "0x%X", $0.value)}.joined(separator: ",")
}
let originalString = "باب ‏"
print(hexCharactersArray(originalString))
The result is [0x628,0x627,0x628,0x20,0x200F]
0x628 - arabic letter beh
0x627 - arabic letter alef
0x628 - arabic letter beh
0x20 - space
0x200F - right-to-left mark
The first three are letters, then some whitespace, but 0x200F is a unicode character in the category of control characters. It's not a letter and it's not whitespace.
When you do:
let replacedString = originalString.replacingOccurrences(of: "باب", with: "").trimmingCharacters(in: .whitespacesAndNewlines)
print(hexCharactersArray(replacedString))
you get [0x200F]
Because you've replaced the letters and trimmed out the whitespace, but you've left behind a control character.
If you want to trim that out too, use:
let replacedString = originalString.replacingOccurrences(of: "باب", with: "").trimmingCharacters(in: .whitespacesAndNewlines.union(.controlCharacters))

How to remove special character from String in swift iOS

I need get value without special character from string
I tried this code but remove special character and letter
example :
var str = ".34!44fgf)(gg#$qwe3"
str.components(separatedBy: CharacterSet.decimalDigits.inverted)//result => 34443
i am want results the following without special character => "3444fgfggqwe3"
Please Advise
You can filter all characters that are letter or digits:
let result = str.filter { $0.isLetter || "0"..."9" ~= $0 }
print(result) // "3444fgfggqwe3"
If you would like to restrict the letters to only lowercase letters from "a" to "z"
"a"..."z" ~= $0
or "A" to "Z"
"A"..."Z" ~= $0
Actually result is a huge array with a lot of empty strings.
This is another approach with Regular Expression
var str = ".34!44fgf)(gg#$qwe3"
str = str.replacingOccurrences(of: "[^[0-9a-zA-z]]", with: "", options: .regularExpression)

Convert string to URL by removing "\"

I want to remove the \" from the begining and the end of the particular string to get the actual URL
Code: let = subString = originalString.replacingOccurrences(of: "\", with: "")
"\"https://api.example.com/deep-link?url=some_url_encoded_string\""
What I want is:
"https://api.example.com/deep-link?url=some_url_encoded_string"
You could trim the string in order to remove all the leading and trailing quote symbols:
let url = "\"https://some-server/some/path\""
let processedString = str.trimmingCharacters(in: .init(charactersIn: "\""))
print(processedString) // https://some-server/some/path
Note that \" is not a string made of two characters, but an escape sequence for the quote symbol.
If you want to remove \" wherever it is occurred in the string then use this:
let subString = originalString.replacingOccurrences(of: "\"", with: "")
if you want to remove it just in the beginning and the end of the string then you will check:
if originalString.first == "\"" {
_ = originalString.removeFirst()
}
if originalString.last == "\"" {
_ = originalString.popLast()
}
You have to remove character ".
like that.
let myString = "\"https://api.example.com/deep-link?url=some_url_encoded_string\""
let newString = myString.replacingOccurrences(of: "\", with: "").replacingOccurrences(of: """, with: "")
print(newString)
Check below code
let originalString = "\"https://api.example.com/deep-link?url=some_url_encoded_string\""
var temp = "\("\"")" as? String
if originalString.hasPrefix(temp!){
originalString.replacingOccurrences(of: "\"", with: "")
}
This backslash is here just because you need double quotes in string which otherwise wouldn't be possible in string declared on one line in Swift 4.2 or earlier.
In Swift 5 you could use raw string literals
let originalString = #""https...""#
So you actually need to remove just the only characters which are actually in string and they're double quotes (simplified: \" = ").
To do this you have to again use backslash before double quote "\" -> "\""
let subString = originalString.replacingOccurrences(of: "\"", with: "")

Remove U\0000fffc unicode scalar from string

I receive an NSAttributedString that contains a NSTextAttachment. I want to remove that attachment, and it looks like it is represented as "\u{ef}" in the string. Printing the unicode scalars of such string, it also seems that unicode scalar for the "\u{ef}" is U\0000fffc.
I tried to do this:
noAttachmentsText = text.replacingOccurrences(of: "\u{ef}", with: "")
with no success, so I'm trying by comparing unicode scalars:
var scalars = Array(text.unicodeScalars)
for scalar in scalars {
// compare `scalar` to `U\0000fffc`
}
but I'm not able either to succeed in the comparison.
How could I do this?
But this code works for me from How do I remove "\U0000fffc" from a string in Swift?
let original = "First part \u{ef} Last part"
let originalRange = Range<String.Index>(start: original.startIndex, end: original.endIndex)
let target = original.stringByReplacingOccurrencesOfString("\u{ef}", withString: "", options: NSStringCompareOptions.LiteralSearch, range: originalRange)
print(target)
Output :
"First part ï Last part"
to
First part Last part
U can use similar code for swift 3 just replace unicode using replacingOccurrences option for exapmle :
func stringTocleanup(str: String) -> String {
var result = str
result = result.replacingOccurrences(of: "\"", with: "\"")
.replacingOccurrences(of: "\u{10}", with: "")
return result
}

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