This question already has answers here:
Find number of spaces in a string in Swift
(3 answers)
Closed 5 years ago.
How do you get the count of the empty space within text?
It would be more helpful to me if explained with an example.
You can either use componentsSeparatedBy or filter function like
let array = string.components(separatedBy:" ")
let spaceCount = array.count - 1
or
let spaceCount = string.filter{$0 == " "}.count
If you want to consider other whitespace characters (not only space) use regular expression:
let string = "How to get count of the empty space in text,Like how we get character count like wise i need empty space count in a text, It would be more helpful if explained with an example."
let regex = try! NSRegularExpression(pattern: "\\s")
let numberOfWhitespaceCharacters = regex.numberOfMatches(in: string, range: NSRange(location: 0, length: string.utf16.count))
Regular expression \\s considers tab, cr, lf and space
Easiest way is to do something like this:
let emptySpacesCount = yourString.characters.filter { $0 == " " }.count
What this does is it takes characters from your string, filter out everything that is not space and then counts number of remaining elements.
You can try this example;
let string = "Whitespace count in a string swift"
let spaceCount = string.characters.filter{$0 == " "}.count
Related
A string:
"jim#domain.com, bill#domain.com, chad#domain.com, tom#domain.com"
Through gesture recognizer, I am able to get the character the user tapped on (happy to provide code, but don't see the relevance at this point).
Let's say the User tapped on o in "chad#domain.com" and the character index is 39
Given 39 the index of o, I would like to get the string start index of c where "chad#domain.com" begins, and an end index for m from "com" where "chad#domain.com" ends.
In another words, given an index of a character in a String, I need to get the index on the left and right right before we encounter a space in a String on the left and a comma on the right.
Tried, but this only provides the last word in the String:
if let range = text.range(of: " ", options: .backwards) {
let suffix = String(text.suffix(from: range.upperBound))
print(suffix) // tom#domain.com
}
I am not sure where to go from here?
You can call range(of:) on two slices of the given string:
text[..<index] is the text preceding the given character position,
and text[index...] is the text starting at the given position.
Example:
let text = "jim#domain.com, bill#domain.com, chad#domain.com, tom#domain.com"
let index = text.index(text.startIndex, offsetBy: 39)
// Search the space before the given position:
let start = text[..<index].range(of: " ", options: .backwards)?.upperBound ?? text.startIndex
// Search the comma after the given position:
let end = text[index...].range(of: ",")?.lowerBound ?? text.endIndex
print(text[start..<end]) // chad#domain.com
Both range(of:) calls return nil if no space (or comma) has
been found. In that case the nil-coalescing operator ?? is used
to get the start (or end) index instead.
(Note that this works because Substrings share a common index
with their originating string.)
An alternative approach is to use a "data detector",
so that the URL detection does not depend on certain separators.
Example (compare How to detect a URL in a String using NSDataDetector):
let text = "jim#domain.com, bill#domain.com, chad#domain.com, tom#domain.com"
let index = text.index(text.startIndex, offsetBy: 39)
let detector = try! NSDataDetector(types: NSTextCheckingResult.CheckingType.link.rawValue)
let matches = detector.matches(in: text, range: NSRange(location: 0, length: text.utf16.count))
for match in matches {
if let range = Range(match.range, in: text), range.contains(index) {
print(text[range])
}
}
Different approach:
You have the string and the Int index
let string = "jim#domain.com, bill#domain.com, chad#domain.com, tom#domain.com"
let characterIndex = 39
Get the String.Index from the Int
let stringIndex = string.index(string.startIndex, offsetBy: characterIndex)
Convert the string into an array of addresses
let addresses = string.components(separatedBy: ", ")
Map the addresses to their ranges (Range<String.Index>) in the string
let ranges = addresses.map{string.range(of: $0)!}
Get the (Int) index of the range which contains stringIndex
if let index = ranges.index(where: {$0.contains(stringIndex)}) {
Get the corresponding address
let address = addresses[index] }
One approach could be to split the original string on the “,” and then using simple math to find in what element of the array the given position (39) exist and from there get the right string or indexes for the previous space and next comma depending on what your end goal is.
Objective-C code:
NSString *str = #"hi";
NSString *strDigit = #"1934"; (or #"193" may be a 3 digit or 4 digit value)
[dayText appendFormat:#"%#%4s,str,[strDigit UTF8String]];
The Objective-C code handles the output string with current alignment when it appears with 3 or 4 digits as output. It is correctly aligning to left and doesn't matter how much digits it is. Any one know how to handle this in Swift?
In Swift I tried with below code and the string is not adjusting the alignment according to the number of digits.
textForTrip += "\(str) \(String(format:"%4s", (strDigit.utf8))"
The %s format expects a pointer to a (NULL-terminated) C string
as argument, that can be obtained with the withCString method.
This would produce the same output as your Objective-C code:
let str = "Hi"
let strDigit = "193"
let text = strDigit.withCString {
String(format: "%#%4s", str, $0)
}
print(text)
It becomes easier if you store the number as integer instead of a
string:
let str = "Hi"
let number = 934
let text = String(format: "%#%4d", str, number)
print(text)
Try this below approach, that might help you
let strDigit = "\("1934".utf8)" //(or #"193" may be a 3 digit or 4 digit value)
var dayText = "Hello, good morning."
dayText += "\(strDigit.prefix(3))"
Sorry,I'm new of swift. I want to calculate the target char in string.But I don't know how to do.Have any good suggestion to me?Thanks.
let string = "hello\nNice to meet you.\nMy name is Leo.\n" //I want to get 3
If you simply want a count of newline characters then you can use a filter on the string's characters:
let string = "hello\nNice to meet you.\nMy name is Leo.\n"
let count = string.characters.filter { $0 == "\n" }.count
print(count)
This outputs 3 as expected.
An alternative is to split the lines with the components(separatedBy method:
let string = "hello\nNice to meet you.\nMy name is Leo.\n"
let lineCounter = string.components(separatedBy: "\n").count - 1
or more versatile to consider all kinds of newline characters
let lineCounter = string.components(separatedBy: CharacterSet.newlines).count - 1
Due to the trailing newline character the result is 4. To ignore a trailing new line you have to decrement the result.
My Swift app involves searching through text in a UITextView. The user can search for a certain substring within that text view, then jump to any instance of that string in the text view (say, the third instance). I need to find out the integer value of which character they are on.
For example:
Example 1: The user searches for "hello" and the text view reads "hey hi hello, hey hi hello", then the user presses down arrow to view second instance. I need to know the integer value of the first h in the second hello (i.e. which # character that h in hello is within the text view). The integer value should be 22.
Example 2: The user searches for "abc" while the text view reads "abcd" and they are looking for the first instance of abc, so the integer value should be 1 (which is the integer value of that a since it's the first character of the instance they're searching for).
How can I get the index of the character the user is searching for?
Xcode 11 • Swift 5 or later
let sentence = "hey hi hello, hey hi hello"
let query = "hello"
var searchRange = sentence.startIndex..<sentence.endIndex
var indices: [String.Index] = []
while let range = sentence.range(of: query, options: .caseInsensitive, range: searchRange) {
searchRange = range.upperBound..<searchRange.upperBound
indices.append(range.lowerBound)
}
print(indices) // "[7, 21]\n"
Another approach is NSRegularExpression which is designed to easily iterate through matches in an string. And if you use the .ignoreMetacharacters option, it will not apply any sophisticated wildcard/regex logic, but will just look for the string in question. So consider:
let string = "hey hi hello, hey hi hello" // string to search within
let searchString = "hello" // string to search for
let matchToFind = 2 // grab the second occurrence
let regex = try! NSRegularExpression(pattern: searchString, options: [.caseInsensitive, .ignoreMetacharacters])
You could use enumerateMatches:
var count = 0
let range = NSRange(string.startIndex ..< string.endIndex, in: string)
regex.enumerateMatches(in: string, range: range) { result, _, stop in
count += 1
if count == matchToFind {
print(result!.range.location)
stop.pointee = true
}
}
Or you can just find all of them with matches(in:range:) and then grab the n'th one:
let matches = regex.matches(in: string, range: range)
if matches.count >= matchToFind {
print(matches[matchToFind - 1].range.location)
}
Obviously, if you were so inclined, you could omit the .ignoreMetacharacters option and allow the user to perform regex searches, too (e.g. wildcards, whole word searches, start of word, etc.).
For Swift 2, see previous revision of this answer.
This question already has answers here:
NSRange to Range<String.Index>
(16 answers)
Closed 7 years ago.
I ran into a very strange problem today with Swift 2.
I have this simple method to extract a substring based on NSRange:
func substringWithRange(string: String, range: NSRange) -> String {
let startIndex = string.startIndex.advancedBy(range.location)
let endIndex = startIndex.advancedBy(range.length)
let substringRange = Range<String.Index>(start: startIndex, end: endIndex)
return string.substringWithRange(substringRange)
}
With ordinary strings or strings containing unicode characters everything works fine. But one string contains the newline characters "\r\n" and suddenly
let startIndex = string.startIndex.advancedBy(range.location)
is always 1 greater than it should be.
let string = "<html>\r\n var info={};</html>"
let range = NSMakeRange(9, 12)
let substring = substringWithRange(string, range: range)
//Expected: var info={};
//Actual: ar info={};<
//string.startIndex = 0
//range.location = 9
//startIndex after advancedBy = 10
Does anyone know why advancedBy is acting that way and how I can solve this problem?
The reason is that Swift treats \r\n as one character
let cr = "\r"
cr.characters.count // 1
let lf = "\n"
lf.characters.count // 1
let crlf = "\r\n"
crlf.characters.count // 1