navigator.contacts.find無法正常工作（cordova / phonegap）

Question

我正在嘗試訪問手機上的聯系人。 我使用下面的代碼來執行此操作，但navigator.contacts.find無法正常工作。 它不會返回錯誤或成功消息。 如果我在該行代碼之后放置任何類型的警報，它將不會出現。

function read_contacts(){

var options = new ContactFindOptions( );

options.filter = "";  //leaving this empty will find return all contacts

options.multiple = true;  //return multiple results

var filter = ["displayName"];    //an array of fields to compare against the options.filter 

navigator.contacts.find(filter, successFunc, errFunc, options); //breaking the code

function successFunc( matches ){
    alert("reading contacts...");
  for( var i=0; i<matches.length; i++){
    alert( matches[i].displayName );
  }

function errFunc(){
    alert("Error finding contacts");
    }
}

}

Answer 1

Phonegap Docs

試試這個 -

function onDeviceReady() {
    var options = new ContactFindOptions();
    options.filter = "";          // empty search string returns all contacts
    options.multiple = true;      // return multiple results
    filter = ["displayName", "name"];   // return contact.displayName 
    navigator.contacts.find(filter, onSuccess, onError, options);
}

// onSuccess: Get a snapshot of the current contacts

function onSuccess(contacts) {
       for (var i = 0; i < contacts.length; i++) {
            console.log("Display Name = " + contacts[i].displayName);
        }
}

// onError: Failed to get the contacts

function onError(contactError) {
     alert('onError!');
}

嘗試使用此代碼查找具有顯示名稱的所有聯系人。

Answer 2

HTML

<ol id="contact"></ol>

JAVASCRIPT

function read_contacts(){
   var options = new ContactFindOptions();
   options.filter="";
   options.filter="";
   options.multiple=true;
   var fields = ["*"];  //"*" will return all contact fields
   navigator.contacts.find(fields, onSuccess, onError, options);
}

// display the address information for all contacts
function onSuccess(contacts) {
  //console.log(JSON.stringify(contacts))
   var li = '';
   $.each(contacts, function(key, value) {
        if(value.name){
            $.each(value.name, function(key, value) {
               if(key == 'formatted'){
                   name = value;
               }                      
            });
        }
        if(value.phoneNumbers){
            $.each(value.phoneNumbers, function(key, value) {
                phone = value.value;
            });
        }                    
        li += '<li style="text-decoration:none;">'+name+' '+phone+'</li>';
   }); 

   $("#contact").html(li);   
}

function onError(contactError) {
   alert('onError!');
}

Answer 3

你不要求所需的字段：

options.desiredFields = [navigator.contacts.fieldType.id, navigator.contacts.fieldType.formatted, navigator.contacts.fieldType.name, navigator.contacts.fieldType.phoneNumbers];