I have a relatively complex regex that I need to run in Swift. Originally was:
"typedef\W+struct\W+{([^}]*)}\W+(\w+);"
You can see the pattern working in JS here.
To make it compile in Swift I escaped the backslashes to:
"typedef\\W+struct\\W+{([^}]*)}\\W+(\\w+);"
On runtime the expression fails to compile with 2048 error. I tried escaping other characters too and tried also escapedPatternForString but without luck. Is there a script to convert JS regexs to Swift? Thanks!
You need to escape both { and } that are outside of a character class:
let rx = "typedef\\W+struct\\W+\\{([^}]*)\\}\\W+(\\w+);"
A quick demo:
let rx = "typedef\\W+struct\\W+\\{([^}]*)\\}\\W+(\\w+);"
let str = "typedef: struct { something } text;"
print(str.range(of: rx, options: .regularExpression) != nil)
// => true
When the { and } are inside a character class they may stay unescaped (as in [^}]).
Using this code (answer by Confused Vorlon), you may get the first match with all capturing groups:
extension NSTextCheckingResult {
func groups(testedString:String) -> [String] {
var groups = [String]()
for i in 0 ..< self.numberOfRanges
{
let group = String(testedString[Range(self.range(at: i), in: testedString)!])
groups.append(group)
}
return groups
}
}
let str = "typedef: struct { something } text;"
let rx = "typedef\\W+struct\\W+\\{([^}]*)\\}\\W+(\\w+);"
let MyRegex = try! NSRegularExpression(pattern: rx)
if let match = MyRegex.firstMatch(in: str, range: NSMakeRange(0, str.count)) {
let groups = match.groups(testedString: str)
print(groups)
}
// => ["typedef: struct { something } text;", " something ", "text"]
To search for a string included in a struct I use:
let results = myArray.filter( {$0.model.localizedCaseInsensitiveContains("bu")} )
But say the struct has several properties that I'd like to search - or maybe I'd even like to search all of them at one time. I can only filter primitive types so leaving 'model' out won't work.
Solution -------------------------
While I really liked the idea of using key paths as Matt suggested below, I ended up adding a function to my struct that made my view controller code much cleaner:
struct QuoteItem {
var itemIdentifier: UUID
var quoteNumber: String
var customerName: String
var address1: String
func quoteItemContains(_ searchString: String) -> Bool {
if self.address1.localizedCaseInsensitiveContains(searchString) ||
self.customerName.localizedCaseInsensitiveContains(searchString) ||
self.quoteNumber.localizedCaseInsensitiveContains(searchString)
{
return true
}
return false
}
Then, in my controller, quotes is an array of QuoteItem that I can search by simply writing:
searchQuoteArray = quotes.filter({ $0.quoteItemContains(searchString) })
This sounds like a job for Swift key paths. Just supply the key paths for the String properties you want to search.
struct MyStruct {
let manny = "Hi"
let moe = "Hey"
let jack = "Howdy"
}
let paths = [\MyStruct.manny, \MyStruct.moe, \MyStruct.jack]
let s = MyStruct()
let target = "y"
let results = paths.map { s[keyPath:$0].localizedCaseInsensitiveContains(target) }
// [false, true, true]
I hope i understood you correct. I think with this piece of code you can achieve what you want:
struct ExampleStruct {
let firstSearchString: String
let secondSearchString: String
}
let exampleOne = ExampleStruct(firstSearchString: "Hello", secondSearchString: "Dude")
let exampleTwo = ExampleStruct(firstSearchString: "Bye", secondSearchString: "Boy")
let exampleArray = [exampleOne, exampleTwo]
let searchString = "Hello"
let filteredArray = exampleArray.filter { (example) -> Bool in
// check here the properties you want to check
if (example.firstSearchString.localizedCaseInsensitiveContains(searchString) || example.secondSearchString.localizedCaseInsensitiveContains(searchString)) {
return true
}
return false
}
for example in filteredArray {
print(example)
}
This prints the following in Playgrounds:
ExampleStruct(firstSearchString: "Hello", secondSearchString: "Dude")
Let me know if it helps.
I am consuming an API that gives me the next page in the Header inside a field called Link. (For example Github does the same, so it isn't weird.Github Doc)
The service that I am consuming retrieve me the pagination data in the following way:
As we can see in the "Link" gives me the next page,
With $0.response?.allHeaderFields["Link"]: I get </api/games?page=1&size=20>; rel="next",</api/games?page=25&size=20>; rel="last",</api/games?page=0&size=20>; rel="first".
I have found the following code to read the page, but it is very dirty... And I would like if anyone has dealt with the same problem or if there is a standard way of face with it. (I have also searched if alamofire supports any kind of feature for this but I haven't found it)
// MARK: - Pagination
private func getNextPageFromHeaders(response: NSHTTPURLResponse?) -> String? {
if let linkHeader = response?.allHeaderFields["Link"] as? String {
/* looks like:
<https://api.github.com/user/20267/gists?page=2>; rel="next", <https://api.github.com/user/20267/gists?page=6>; rel="last"
*/
// so split on "," the on ";"
let components = linkHeader.characters.split {$0 == ","}.map { String($0) }
// now we have 2 lines like '<https://api.github.com/user/20267/gists?page=2>; rel="next"'
// So let's get the URL out of there:
for item in components {
// see if it's "next"
let rangeOfNext = item.rangeOfString("rel=\"next\"", options: [])
if rangeOfNext != nil {
let rangeOfPaddedURL = item.rangeOfString("<(.*)>;", options: .RegularExpressionSearch)
if let range = rangeOfPaddedURL {
let nextURL = item.substringWithRange(range)
// strip off the < and >;
let startIndex = nextURL.startIndex.advancedBy(1) //advance as much as you like
let endIndex = nextURL.endIndex.advancedBy(-2)
let urlRange = startIndex..<endIndex
return nextURL.substringWithRange(urlRange)
}
}
}
}
return nil
}
I think that the forEach() could have a better solution, but here is what I got:
let linkHeader = "</api/games?page=1&size=20>; rel=\"next\",</api/games?page=25&size=20>; rel=\"last\",</api/games?page=0&size=20>; rel=\"first\""
let links = linkHeader.components(separatedBy: ",")
var dictionary: [String: String] = [:]
links.forEach({
let components = $0.components(separatedBy:"; ")
let cleanPath = components[0].trimmingCharacters(in: CharacterSet(charactersIn: "<>"))
dictionary[components[1]] = cleanPath
})
if let nextPagePath = dictionary["rel=\"next\""] {
print("nextPagePath: \(nextPagePath)")
}
//Bonus
if let lastPagePath = dictionary["rel=\"last\""] {
print("lastPagePath: \(lastPagePath)")
}
if let firstPagePath = dictionary["rel=\"first\""] {
print("firstPagePath: \(firstPagePath)")
}
Console output:
$> nextPagePath: /api/games?page=1&size=20
$> lastPagePath: /api/games?page=25&size=20
$> firstPagePath: /api/games?page=0&size=20
I used components(separatedBy:) instead of split() to avoid the String() conversion at the end.
I created a Dictionary for the values to hold and removed the < and > with a trim.
I'm trying to implement search inside my app that I'm making. I have an array that I'm trying to search and I find this code online:
func filterContentForSearchText(searchText: String) {
filteredCandies = candies.filter({( candy : Candies) -> Bool in
if candy.name.lowercaseString.containsString(searchText.lowercaseString) == true {
return true
} else {
return false
}
})
tableView.reloadData()
}
The issue is that the database that I'm trying to implement search on has text that is all scrambled because it was supposed to shortened. How can I make it so that the search will check if all the letters are there instead of searching exactly the right name. Example of object from database (USDA): CRAB, DUNGINESS, RAW
If you have an answer, please make it fast enough for live searching. Non live searching makes searching terrible (at least for me)!
I'm using Swift 2.2 and Xcode 7
As an improvement to #appzYourLife's solution, you could do this with a native Swift Set, as a counted set isn't necessarily needed in this case. This will save having to map(_:) over the characters of each name and bridging them to Objective-C. You can now just use a set of Characters, as they're Hashable.
For example:
struct Candy {
let name: String
}
let candies = [Candy(name: "CRAB"), Candy(name: "DUNGINESS"), Candy(name: "RAW")]
var filteredCandies = [Candy]()
func filterContentForSearchText(searchText: String) {
let searchCharacters = Set(searchText.lowercaseString.characters)
filteredCandies = candies.filter {Set($0.name.lowercaseString.characters).isSupersetOf(searchCharacters)}
tableView.reloadData()
}
filterContentForSearchText("RA")
print(filteredCandies) // [Candy(name: "CRAB"), Candy(name: "RAW")]
filterContentForSearchText("ED")
print(filteredCandies) // Candy(name: "DUNGINESS")]
Also depending on whether you can identify this as a performance bottleneck (you should do some profiling first) – you could potentially optimise the above further by caching the sets containing the characters of your 'candy' names, saving from having to recreate them at each search (although you'll have to ensure that they're updated if you update your candies data).
When you come to search, you can then use zip(_:_:) and flatMap(_:) in order to filter out the corresponding candies.
let candies = [Candy(name: "CRAB"), Candy(name: "DUNGINESS"), Candy(name: "RAW")]
// cached sets of (lowercased) candy name characters
let candyNameCharacterSets = candies.map {Set($0.name.lowercaseString.characters)}
var filteredCandies = [Candy]()
func filterContentForSearchText(searchText: String) {
let searchCharacters = Set(searchText.lowercaseString.characters)
filteredCandies = zip(candyNameCharacterSets, candies).flatMap {$0.isSupersetOf(searchCharacters) ? $1 : nil}
tableView.reloadData()
}
First of all a block of code like this
if someCondition == true {
return true
} else {
return false
}
can also be written this ways
return someCondition
right? :)
Refactoring
So your original code would look like this
func filterContentForSearchText(searchText: String) {
filteredCandies = candies.filter { $0.name.lowercaseString.containsString(searchText.lowercaseString) }
tableView.reloadData()
}
Scrambled search
Now, given a string A, your want to know if another string B contains all the character of A right?
For this we need CountedSet which is available from Swift 3. Since you are using Swift 2.2 we'll use the old NSCountedSet but some bridging to Objective-C is needed.
Here's the code.
struct Candy {
let name: String
}
let candies = [Candy]()
var filteredCandies = [Candy]()
func filterContentForSearchText(searchText: String) {
let keywordChars = NSCountedSet(array:Array(searchText.lowercaseString.characters).map { String($0) })
filteredCandies = candies.filter {
let candyChars = NSCountedSet(array:Array($0.name.lowercaseString.characters).map { String($0) }) as Set<NSObject>
return keywordChars.isSubsetOfSet(candyChars)
}
tableView.reloadData()
}
Swift 3 code update :
func filterContentForSearchText(searchText: String, scope: String = "All") {
filteredCandies = candies.filter { candy in
return candy.name.localizedLowercase.contains(searchText.lowercased())
}
tableView.reloadData()
}
I have the need to parse some unknown data which should just be a numeric value, but may contain whitespace or other non-alphanumeric characters.
Is there a new way of doing this in Swift? All I can find online seems to be the old C way of doing things.
I am looking at stringByTrimmingCharactersInSet - as I am sure my inputs will only have whitespace/special characters at the start or end of the string. Are there any built in character sets I can use for this? Or do I need to create my own?
I was hoping there would be something like stringFromCharactersInSet() which would allow me to specify only valid characters to keep
I was hoping there would be something like stringFromCharactersInSet() which would allow me to specify only valid characters to keep.
You can either use trimmingCharacters with the inverted character set to remove characters from the start or the end of the string. In Swift 3 and later:
let result = string.trimmingCharacters(in: CharacterSet(charactersIn: "0123456789.").inverted)
Or, if you want to remove non-numeric characters anywhere in the string (not just the start or end), you can filter the characters, e.g. in Swift 4.2.1:
let result = string.filter("0123456789.".contains)
Or, if you want to remove characters from a CharacterSet from anywhere in the string, use:
let result = String(string.unicodeScalars.filter(CharacterSet.whitespaces.inverted.contains))
Or, if you want to only match valid strings of a certain format (e.g. ####.##), you could use regular expression. For example:
if let range = string.range(of: #"\d+(\.\d*)?"#, options: .regularExpression) {
let result = string[range] // or `String(string[range])` if you need `String`
}
The behavior of these different approaches differ slightly so it just depends on precisely what you're trying to do. Include or exclude the decimal point if you want decimal numbers, or just integers. There are lots of ways to accomplish this.
For older, Swift 2 syntax, see previous revision of this answer.
let result = string.stringByReplacingOccurrencesOfString("[^0-9]", withString: "", options: NSStringCompareOptions.RegularExpressionSearch, range:nil).stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceCharacterSet())
Swift 3
let result = string.replacingOccurrences( of:"[^0-9]", with: "", options: .regularExpression)
You can upvote this answer.
I prefer this solution, because I like extensions, and it seems a bit cleaner to me. Solution reproduced here:
extension String {
var digits: String {
return components(separatedBy: CharacterSet.decimalDigits.inverted)
.joined()
}
}
You can filter the UnicodeScalarView of the string using the pattern matching operator for ranges, pass a UnicodeScalar ClosedRange from 0 to 9 and initialise a new String with the resulting UnicodeScalarView:
extension String {
private static var digits = UnicodeScalar("0")..."9"
var digits: String {
return String(unicodeScalars.filter(String.digits.contains))
}
}
"abc12345".digits // "12345"
edit/update:
Swift 4.2
extension RangeReplaceableCollection where Self: StringProtocol {
var digits: Self {
return filter(("0"..."9").contains)
}
}
or as a mutating method
extension RangeReplaceableCollection where Self: StringProtocol {
mutating func removeAllNonNumeric() {
removeAll { !("0"..."9" ~= $0) }
}
}
Swift 5.2 • Xcode 11.4 or later
In Swift5 we can use a new Character property called isWholeNumber:
extension RangeReplaceableCollection where Self: StringProtocol {
var digits: Self { filter(\.isWholeNumber) }
}
extension RangeReplaceableCollection where Self: StringProtocol {
mutating func removeAllNonNumeric() {
removeAll { !$0.isWholeNumber }
}
}
To allow a period as well we can extend Character and create a computed property:
extension Character {
var isDecimalOrPeriod: Bool { "0"..."9" ~= self || self == "." }
}
extension RangeReplaceableCollection where Self: StringProtocol {
var digitsAndPeriods: Self { filter(\.isDecimalOrPeriod) }
}
Playground testing:
"abc12345".digits // "12345"
var str = "123abc0"
str.removeAllNonNumeric()
print(str) //"1230"
"Testing0123456789.".digitsAndPeriods // "0123456789."
Swift 4
I found a decent way to get only alpha numeric characters set from a string.
For instance:-
func getAlphaNumericValue() {
var yourString = "123456789!##$%^&*()AnyThingYouWant"
let unsafeChars = CharacterSet.alphanumerics.inverted // Remove the .inverted to get the opposite result.
let cleanChars = yourString.components(separatedBy: unsafeChars).joined(separator: "")
print(cleanChars) // 123456789AnyThingYouWant
}
A solution using the filter function and rangeOfCharacterFromSet
let string = "sld [f]34é7*˜µ"
let alphaNumericCharacterSet = NSCharacterSet.alphanumericCharacterSet()
let filteredCharacters = string.characters.filter {
return String($0).rangeOfCharacterFromSet(alphaNumericCharacterSet) != nil
}
let filteredString = String(filteredCharacters) // -> sldf34é7µ
To filter for only numeric characters use
let string = "sld [f]34é7*˜µ"
let numericSet = "0123456789"
let filteredCharacters = string.characters.filter {
return numericSet.containsString(String($0))
}
let filteredString = String(filteredCharacters) // -> 347
or
let numericSet : [Character] = ["0", "1", "2", "3", "4", "5", "6", "7", "8", "9"]
let filteredCharacters = string.characters.filter {
return numericSet.contains($0)
}
let filteredString = String(filteredCharacters) // -> 347
Swift 4
But without extensions or componentsSeparatedByCharactersInSet which doesn't read as well.
let allowedCharSet = NSCharacterSet.letters.union(.whitespaces)
let filteredText = String(sourceText.unicodeScalars.filter(allowedCharSet.contains))
let string = "+1*(234) fds567#-8/90-"
let onlyNumbers = string.components(separatedBy: CharacterSet.decimalDigits.inverted).joined()
print(onlyNumbers) // "1234567890"
or
extension String {
func removeNonNumeric() -> String {
return self.components(separatedBy: CharacterSet.decimalDigits.inverted).joined()
}
}
let onlyNumbers = "+1*(234) fds567#-8/90-".removeNonNumeric()
print(onlyNumbers)// "1234567890"
Swift 3, filters all except numbers
let myString = "dasdf3453453fsdf23455sf.2234"
let result = String(myString.characters.filter { String($0).rangeOfCharacter(from: CharacterSet(charactersIn: "0123456789")) != nil })
print(result)
Swift 4.2
let numericString = string.filter { (char) -> Bool in
return char.isNumber
}
You can do something like this...
let string = "[,myString1. \"" // string : [,myString1. "
let characterSet = NSCharacterSet(charactersInString: "[,. \"")
let finalString = (string.componentsSeparatedByCharactersInSet(characterSet) as NSArray).componentsJoinedByString("")
print(finalString)
//finalString will be "myString1"
The issue with Rob's first solution is stringByTrimmingCharactersInSet only filters the ends of the string rather than throughout, as stated in Apple's documentation:
Returns a new string made by removing from both ends of the receiver characters contained in a given character set.
Instead use componentsSeparatedByCharactersInSet to first isolate all non-occurrences of the character set into arrays and subsequently join them with an empty string separator:
"$$1234%^56()78*9££".componentsSeparatedByCharactersInSet(NSCharacterSet(charactersInString: "0123456789").invertedSet)).joinWithSeparator("")
Which returns 123456789
Swift 3
extension String {
var keepNumericsOnly: String {
return self.components(separatedBy: CharacterSet(charactersIn: "0123456789").inverted).joined(separator: "")
}
}
Swift 4.0 version
extension String {
var numbers: String {
return String(describing: filter { String($0).rangeOfCharacter(from: CharacterSet(charactersIn: "0123456789")) != nil })
}
}
Swift 4
String.swift
import Foundation
extension String {
func removeCharacters(from forbiddenChars: CharacterSet) -> String {
let passed = self.unicodeScalars.filter { !forbiddenChars.contains($0) }
return String(String.UnicodeScalarView(passed))
}
func removeCharacters(from: String) -> String {
return removeCharacters(from: CharacterSet(charactersIn: from))
}
}
ViewController.swift
let character = "1Vi234s56a78l9"
let alphaNumericSet = character.removeCharacters(from: CharacterSet.decimalDigits.inverted)
print(alphaNumericSet) // will print: 123456789
let alphaNumericCharacterSet = character.removeCharacters(from: "0123456789")
print("no digits",alphaNumericCharacterSet) // will print: Vishal
Swift 4.2
let digitChars = yourString.components(separatedBy:
CharacterSet.decimalDigits.inverted).joined(separator: "")
Swift 3 Version
extension String
{
func trimmingCharactersNot(in charSet: CharacterSet) -> String
{
var s:String = ""
for unicodeScalar in self.unicodeScalars
{
if charSet.contains(unicodeScalar)
{
s.append(String(unicodeScalar))
}
}
return s
}
}