I need to filter the array of the Dictionary based on the search string. In Dictionary, I have a key "name", I need to compare the entered string in textfield to the name key in Dictionary.
func textChanged(textField: UITextField) {
if textField.text == "" {
self.filteredArrayStoreList = self.arrayStoreList
} else {
self.filteredArrayStoreList.removeAll()
print("Text: \(textField.text ?? "")")
let filteredArray = self.arrayStoreList.filter { ($0.name?.contains((textField.text ?? "")))!}
self.filteredArrayStoreList = filteredArray
self.noLocationFoundView.isHidden = true //filteredArrayStoreList.count == 0
}
self.searchLocationTabelView.reloadData()
}
In Dictionary, the name key value is " Vimal's ".
But in textfield, I will get the text as " Vimal‘s ". How to search for the single quotes.
' (ascii-39) is the normal quote, and ‘ (ascii-8216) is called "smart" or "curly" quote. this quote is the appear on iOS device keyboard.User can on/off this feature from settings
Settings -> General -> Keyboard-> Smart punctuation
That is the reason for not matching these 2 strings.You can do 2 things to solve your problem. Because we don't know app user's device on/off that feature
simply you can enter dictionary names without special characters
you can find and replace that smart/curl quote with normal quote before check
this is the example
let name:String = "vimal‘s"
print(name.replacingOccurrences(of: "‘", with: "'"))
//result "vimal's"
Related
I saw a similar post to this but none of the answers there helped me. I am looking to check if a string does not contain any special characters using swift regular expressions. I want to allow other iPhone keyboard letters though like Chinese, Hindi, Arabic, Korean, etc.. I only do not want to allow special characters like +, &, #, $, % though.
I have tried using:
func usernameTest(testStr:String) -> Bool {
return testStr.range(of: "^[ !\"#$%&'()*+,-./:;<=>?#\\[\\\\\\]^_`{|}~]+", options: .regularExpression) != nil
}
This did not work though. I thought it would check to see if any special characters were used, but when I tested phrases like "Tom###." and "%Will!!" it returned false. I would have expected this to return true since the strings that I passed in contained one or more of the special characters in the range.
Any help would be appreciated as we want our users to be able to create their usernames in any language but we still do not want to allow spaces or special characters.
you can try this way
var charSet = CharacterSet.init(charactersIn: "##$%+_)(")
var string2 = "test#3"
if let strvalue = string2.rangeOfCharacter(from: charSet)
{
print("true")
}
in the characterset init you can give the special characters which you want to check the occurance.
You must remove character "^" in the regex. So your function should look like this:
func usernameTest(testStr:String) -> Bool {
return testStr.range(of: "[ !\"#$%&'()*+,-./:;<=>?#\\[\\\\\\]^_`{|}~]+", options: .regularExpression) != nil }
func usernameTest(testStr:String) -> Bool {
let letters = CharacterSet.punctuationCharacters
let range = testStr.rangeOfCharacter(from: letters)
// range will be nil if no letters is found
if range != nil {
return true
}
else {
return false
}
}
try this one
I'm using Swift 3 to make an app. On the register screen I want the user to type in their first and last name and then a random alphanumeric string will be generated. This string should automatically be filled in to the username text field, and the user should not be able to change it.
I already have the code for an alphanumeric string:
func randomString(length:Int) -> String {
let charSet = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"
var c = charSet.characters.map { String($0) }
var s:String = ""
for _ in (1...length) {
s.append(c[Int(arc4random()) % c.count])
}
return s
}
I just want this string to be filled in automatically by the program itself, and the user should not be able to change it.
So first set your string to your texitField
textField.text = randmString(length: 10)
Then, disable editing of it
textField.isUserInteractionEnabled = false
I'm using Regex to search for a word in a textView. I implemented a textField and two switch as options (Whole words and Match case). All work fine when you enter a plain word in the search filed but I get an error when I enter a special character like \ or *.
The error I get is like this one:
Error Domain=NSCocoaErrorDomain Code=2048 "The value “*” is invalid." UserInfo={NSInvalidValue=*}
Is there a way to avoid this problem and have the code handle all the text like plain text?
Because I would like to search also for special characters, I would like to prefer to not interdict to enter them. At the beginning I though to programmatically add an escape backslash to all special character before to perform a search, but maybe there are some more smart approaches?
Here is the code I'm using (based on this tutorial: NSRegularExpression Tutorial: Getting Started)
struct SearchOptions {
let searchString: String
var replacementString: String
let matchCase: Bool
let wholeWords: Bool
}
extension NSRegularExpression {
convenience init?(options: SearchOptions) {
let searchString = options.searchString
let isCaseSensitive = options.matchCase
let isWholeWords = options.wholeWords
// handle case sensitive option
var regexOption: NSRegularExpressionOptions = .CaseInsensitive
if isCaseSensitive { // if it is match case remove case sensitive option
regexOption = []
}
// put the search string in the pattern
var pattern = searchString
// if it's whole word put the string between word boundary \b
if isWholeWords {
pattern = "\\b\(searchString)\\b" // the second \ is used as escape
}
do {
try self.init(pattern: pattern, options: regexOption)
} catch {
print(error)
}
}
}
You may use NSRegularExpression.escapedPatternForString:
Returns a string by adding backslash escapes as necessary to protect any characters that would match as pattern metacharacters.
Thus, you need
var pattern = NSRegularExpression.escapedPatternForString(searchString)
Also, note that this piece:
if isWholeWords {
pattern = "\\b\(searchString)\\b"
might fail if a user inputs (text) and wishes to search for it as a whole word. The best way to match whole words is by means of lookarounds disallowing word chars on both ends of the search word:
if isWholeWords {
pattern = "(?<!\\w)" + NSRegularExpression.escapedPatternForString(searchString) + "(?!\\w)"
Is there a function to capitalize each word in a string or is this a manual process?
For e.g. "bob is tall"
And I would like "Bob Is Tall"
Surely there is something and none of the Swift IOS answers I have found seemed to cover this.
Are you looking for capitalizedString
Discussion
A string with the first character in each word changed to its corresponding uppercase value, and all remaining characters set to their corresponding lowercase values.
and/or capitalizedStringWithLocale(_:)
Returns a capitalized representation of the receiver using the specified locale.
For strings presented to users, pass the current locale ([NSLocale currentLocale]). To use the system locale, pass nil.
Swift 3:
var lowercased = "hello there"
var stringCapitalized = lowercased.capitalized
//prints: "Hello There"
Since iOS 9 a localised capitalization function is available as capitalised letters may differ in languages.
if #available(iOS 9.0, *) {
"istanbul".localizedCapitalizedString
// In Turkish: "İstanbul"
}
An example of the answer provided above.
var sentenceToCap = "this is a sentence."
println(sentenceToCap.capitalizedStringWithLocale(NSLocale.currentLocale()) )
End result is a string "This Is A Sentence"
For Swift 3 it has been changed to capitalized .
Discussion
This property performs the canonical (non-localized) mapping. It is suitable for programming operations that require stable results not depending on the current locale.
A capitalized string is a string with the first character in each word changed to its corresponding uppercase value, and all remaining characters set to their corresponding lowercase values. A “word” is any sequence of characters delimited by spaces, tabs, or line terminators (listed under getLineStart(_:end:contentsEnd:for:)). Some common word delimiting punctuation isn’t considered, so this property may not generally produce the desired results for multiword strings.
Case transformations aren’t guaranteed to be symmetrical or to produce strings of the same lengths as the originals. See lowercased for an example.
There is a built in function for that
nameOfString.capitalizedString
This will capitalize every word of string. To capitalize only the first letter you can use:
nameOfString.replaceRange(nameOfString.startIndex...nameOfString.startIndex, with: String(nameOfString[nameOfString.startIndex]).capitalizedString)
Older Thread
Here is what I came up with that seems to work but I am open to anything that is better.
func firstCharacterUpperCase(sentenceToCap:String) -> String {
//break it into an array by delimiting the sentence using a space
var breakupSentence = sentenceToCap.componentsSeparatedByString(" ")
var newSentence = ""
//Loop the array and concatinate the capitalized word into a variable.
for wordInSentence in breakupSentence {
newSentence = "\(newSentence) \(wordInSentence.capitalizedString)"
}
// send it back up.
return newSentence
}
or if I want to use this as an extension of the string class.
extension String {
var capitalizeEachWord:String {
//break it into an array by delimiting the sentence using a space
var breakupSentence = self.componentsSeparatedByString(" ")
var newSentence = ""
//Loop the array and concatinate the capitalized word into a variable.
for wordInSentence in breakupSentence {
newSentence = "\(newSentence) \(wordInSentence.capitalizedString)"
}
// send it back up.
return newSentence
}
}
Again, anything better is welcome.
Swift 5 version of Christopher Wade's answer
let str = "my string"
let result = str.capitalized(with: NSLocale.current)
print(result) // prints My String
I'm new to Swift and taking a class to learn iOS programming. I find myself stumped on how to search in an array of dictionaries for a string value and dump the string value into an array. This is taken from my Xcode playground.
I'm trying to figure out how to:
1) search through an array of dictionaries
2) dump the results of the search into an array (which I've created)
These are the character dictionaries.
let worf = [
"name": "Worf",
"rank": "lieutenant",
"information": "son of Mogh, slayer of Gowron",
"favorite drink": "prune juice",
"quote" : "Today is a good day to die."]
let picard = [
"name": "Jean-Luc Picard",
"rank": "captain",
"information": "Captain of the USS Enterprise",
"favorite drink": "tea, Earl Grey, hot"]
This is an array of the character dictionaries listed above.
let characters = [worf, picard]
This is the function I'm trying to write.
func favoriteDrinksArrayForCharacters(characters:Array<Dictionary<String, String>>) -> Array<String> {
// create an array of Strings to dump in favorite drink strings
var favoriteDrinkArray = [String]()
for character in characters {
// look up favorite drink
// add favorite drink to favoriteDrinkArray
}
return favoriteDrinkArray
}
let favoriteDrinks = favoriteDrinksArrayForCharacters(characters)
favoriteDrinks
I would be grateful for any assistance on how to move forward on this. I've dug around for examples, but I'm coming up short finding one that's applicable to what I'm trying to do here.
Inside the loop, you need to fetch the "favorite drink" entry from the dictionary, and append it to the array:
for character in characters {
if let drink = character["favorite drink"] {
favoriteDrinkArray.append(drink)
}
}
Note, the if let drink = guards against the possibility there is no such entry in the array – if there isn't, you get a nil back, and the if is checking for that, only adding the entry if it's not nil.
You might sometimes see people skip the if let part and instead just write let drink = character["favorite drink"]!, with an exclamation mark on the end. Do not do this. This is known as "force unwrapping" an optional, and if there is ever not a valid value returned from the dictionary, your program will crash.
The behavior with the first example is, if there is no drink you don't add it to the array. But this might not be what you want since you may be expecting a 1-to-1 correspondence between entries in the character array and entries in the drinks array.
If that's the case, and you perhaps want an empty string, you could do this instead:
func favoriteDrinksArrayForCharacters(characters: [[String:String]]) -> [String] {
return characters.map { character in
character["favorite drink"] ?? ""
}
}
The .map means: run through every entry in characters, and put the result of running this expression in a new array (which you then return).
The ?? means: if you get back a nil from the left-hand side, replace it with the value on the right-hand side.
Airspeed Velocity's answer is very comprehensive and provides a solution that works. A more compact way of achieving the same result is using the filter and map methods of swift arrays:
func favoriteDrinksArrayForCharacters(characters:Array<Dictionary<String, String>>) -> Array<String> {
// create an array of Strings to dump in favorite drink strings
return characters.filter { $0["favorite drink"] != nil }.map { $0["favorite drink"]! }
}
The filter takes a closure returning a boolean, which states whether an element must be included or not - in our case, it checks for the existence of an element for key "favorite drink". This method returns the array of dictionaries satisfying that condition.
The second step uses the map method to transform each dictionary into the value corresponding to the "favorite drink" key - taking into account that a dictionary lookup always returns an optional (to account for missing key), and that the filter has already excluded all dictionaries not having a value for that key, it's safe to apply the forced unwrapping operator ! to return a non optional string.
The combined result is an array of strings - copied from my playground:
["prune juice", "tea, Earl Grey, hot"]
let drinks = characters.map({$0["favorite drink"]}) // [Optional("prune juice"), Optional("tea, Earl Grey, hot")]
or
let drinks = characters.filter({$0["favorite drink"] != nil}).map({$0["favorite drink"]!}) // [prune juice, tea, Earl Grey, hot]
It may help you
var customerNameDict = ["firstName":"karthi","LastName":"alagu","MiddleName":"prabhu"];
var clientNameDict = ["firstName":"Selva","LastName":"kumar","MiddleName":"m"];
var employeeNameDict = ["firstName":"karthi","LastName":"prabhu","MiddleName":"kp"];
var attributeValue = "karthi";
var arrNames:Array = [customerNameDict,clientNameDict,employeeNameDict];
var namePredicate = NSPredicate(format: "firstName like %#",attributeValue);
let filteredArray = arrNames.filter { namePredicate.evaluateWithObject($0) };
println("names = ,\(filteredArray)");
Use the following code to search from NSArray of dictionaries whose keys are ID and Name.
var txtVal:NSString
let path = NSBundle.mainBundle().pathForResource(plistName, ofType: "plist")
var list = NSArray(contentsOfFile: path!) as [[String:String]]
var namePredicate = NSPredicate(format: "ID like %#", String(forId));
let filteredArray = list.filter { namePredicate!.evaluateWithObject($0) };
if filteredArray.count != 0
{
let value = filteredArray[0] as NSDictionary
txtVal = value.objectForKey("Name") as String
}
i have array of customer ,each customer having name,phone number and other stubs .so i used the below code to search by phone number in the array of dictionary in search bar
for index in self.customerArray
{
var string = index.valueForKey("phone")
if let phoneNumber = index.valueForKey("phone") as? String {
string = phoneNumber
}
else
{
string = ""
}
if string!.localizedCaseInsensitiveContainsString(searchText) {
filtered.addObject(index)
searchActive = true;
}
}