CODEWITHSUNDEEP
iosswift3contacts
user8541344
user8541344

Reputation:

Find duplicate contacts in Contacts Framework

In Swift 3, I use the new Contact Framework to manipulate contacts, but I don't have any solution for fetching duplicate contacts.

Any idea how to achieve this?

Upvotes: 3

Views: 1918

Answers (2)

Rob
Rob

Reputation: 437552

I'd build a dictionary keyed by name, and then filter down to just those with more than one occurrence of the name:

let keys = [CNContactIdentifierKey as CNKeyDescriptor, CNContactFormatter.descriptorForRequiredKeys(for: .fullName)]
let request = CNContactFetchRequest(keysToFetch: keys)
var contactsByName = [String: [CNContact]]()
try! self.store.enumerateContacts(with: request) { contact, stop in
    guard let name = CNContactFormatter.string(from: contact, style: .fullName) else { return }
    contactsByName[name] = (contactsByName[name] ?? []) + [contact]   // or in Swift 4, `contactsByName[name, default: []].append(contact)`
}
let duplicates = contactsByName.filter { $1.count > 1 }

Upvotes: 2

Satish Babariya
Satish Babariya

Reputation: 3096

You can do something like this:

/// Find Duplicates Contacts In Given Contacts Array
func findDuplicateContacts(Contacts contacts : [CNContact], completionHandler : @escaping (_ result : [Array<CNContact>]) -> ()){
    let arrfullNames : [String?] = contacts.map{CNContactFormatter.string(from: $0, style: .fullName)}
    var contactGroupedByDuplicated : [Array<CNContact>] = [Array<CNContact>]()
    if let fullNames : [String] = arrfullNames as? [String]{
        let uniqueArray = Array(Set(fullNames))
        var contactGroupedByUnique = [Array<CNContact>]()
        for fullName in uniqueArray {
            let group = contacts.filter {
                CNContactFormatter.string(from: $0, style: .fullName) == fullName
            }
            contactGroupedByUnique.append(group)
        }
        for items in contactGroupedByUnique{
            if items.count > 1 {
                contactGroupedByDuplicated.append(items)
            }
        }
    }
    completionHandler(contactGroupedByDuplicated)
}

Upvotes: 2

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