How to get the first name and last name from Android contacts?

问题

How to get the following fields from Android contacts? I used Android 2.2.

Name prefix
First name
Middle name
Last name
Name prefix
Phonetic given name
Phonetic middle name
Phonetic family name

回答1:

Look at ContactsContract.CommonDataKinds.StructuredName class. You can find there all columns you are looking for. Try sth like this:

    String whereName = ContactsContract.Data.MIMETYPE + " = ?";
    String[] whereNameParams = new String[] { ContactsContract.CommonDataKinds.StructuredName.CONTENT_ITEM_TYPE };
    Cursor nameCur = contentResolver.query(ContactsContract.Data.CONTENT_URI, null, whereName, whereNameParams, ContactsContract.CommonDataKinds.StructuredName.GIVEN_NAME);
    while (nameCur.moveToNext()) {
        String given = nameCur.getString(nameCur.getColumnIndex(ContactsContract.CommonDataKinds.StructuredName.GIVEN_NAME));
        String family = nameCur.getString(nameCur.getColumnIndex(ContactsContract.CommonDataKinds.StructuredName.FAMILY_NAME));
        String display = nameCur.getString(nameCur.getColumnIndex(ContactsContract.CommonDataKinds.StructuredName.DISPLAY_NAME));
    }
    nameCur.close();

It returns all names in contacts. To be more precise you can add a contact id as an additional parameter to the query - you will get address for particular contact.

回答2:

For a specified contact you can do this :

String whereName = ContactsContract.Data.MIMETYPE + " = ? AND " + ContactsContract.CommonDataKinds.StructuredName.CONTACT_ID + " = ?";
String[] whereNameParams = new String[] { ContactsContract.CommonDataKinds.StructuredName.CONTENT_ITEM_TYPE, contact_ID };
Cursor nameCur = contentResolver.query(ContactsContract.Data.CONTENT_URI, null, whereName, whereNameParams, ContactsContract.CommonDataKinds.StructuredName.GIVEN_NAME);
while (nameCur.moveToNext()) {
    String given = nameCur.getString(nameCur.getColumnIndex(ContactsContract.CommonDataKinds.StructuredName.GIVEN_NAME));
    String family = nameCur.getString(nameCur.getColumnIndex(ContactsContract.CommonDataKinds.StructuredName.FAMILY_NAME));
    String display = nameCur.getString(nameCur.getColumnIndex(ContactsContract.CommonDataKinds.StructuredName.DISPLAY_NAME));
}
nameCur.close();

回答3:

try to use this code to get the required information about the contact,the code is here-

import android.provider.ContactsContract.Contacts;
import android.database.Cursor;

// Form an array specifying which columns to return, you can add more.
String[] projection = new String[] {
                         ContactsContract.Contacts.DISPLAY_NAME,
                         ContactsContract.CommonDataKinds.Phone
                         ContactsContract.CommonDataKinds.Email
                      };

Uri contacts =  ContactsContract.Contacts.CONTENT_LOOKUP_URI;
// id of the Contact to return.
long id = 3;

// Make the query. 
Cursor managedCursor = managedQuery(contacts,
                     projection, // Which columns to return 
                     null,       // Which rows to return (all rows)
                                 // Selection arguments (with a given ID)
                     ContactsContract.Contacts._ID = "id", 
                                 // Put the results in ascending order by name
                     ContactsContract.Contacts.DISPLAY_NAME + " ASC");

回答4:

As the other example (just for fun) but for fetching contact name of a single user:

// A contact ID is fetched from ContactList
Uri resultUri = data.getData(); 
Cursor cont = getContentResolver().query(resultUri, null, null, null, null);
if (!cont.moveToNext()) {   
    Toast.makeText(this, "Cursor contains no data", Toast.LENGTH_LONG).show(); 
                return;
}
int columnIndexForId = cont.getColumnIndex(ContactsContract.Contacts._ID);
String contactId = cont.getString(columnIndexForId);

// Fetch contact name with a specific ID
String whereName = ContactsContract.Data.MIMETYPE + " = ? AND " + ContactsContract.CommonDataKinds.StructuredName.CONTACT_ID + " = " + contactId; 
String[] whereNameParams = new String[] { ContactsContract.CommonDataKinds.StructuredName.CONTENT_ITEM_TYPE };
Cursor nameCur = getContentResolver().query(ContactsContract.Data.CONTENT_URI, null, whereName, whereNameParams, ContactsContract.CommonDataKinds.StructuredName.GIVEN_NAME);
while (nameCur.moveToNext()) {
    String given = nameCur.getString(nameCur.getColumnIndex(ContactsContract.CommonDataKinds.StructuredName.GIVEN_NAME));
    String family = nameCur.getString(nameCur.getColumnIndex(ContactsContract.CommonDataKinds.StructuredName.FAMILY_NAME));
    String display = nameCur.getString(nameCur.getColumnIndex(ContactsContract.CommonDataKinds.StructuredName.DISPLAY_NAME));
    Toast.makeText(this, "Name: " + given + " Family: " +  family + " Displayname: "  + display, Toast.LENGTH_LONG).show();
}
nameCur.close();
cont.close();

回答5:

Some links to get you started, in addition to the suggestions from Raunak:

http://www.higherpass.com/Android/Tutorials/Working-With-Android-Contacts/
How to obtain all details of a contact in Android
http://developer.android.com/resources/samples/ContactManager/index.html

回答6:

Experimenting with the ContactsContract.Data.CONTENT_URI in late 2015 on marshmallow. I'm unable to get the GIVEN_NAME or similar fields. I think the later apis have deprecated these. Run the following code to print out the columns you have on your phone

Uri uri = ContactsContract.Data.CONTENT_URI;
String selection = ContactsContract.Data.MIMETYPE + " = ?";
String[] selectionArgs = new String[] { ContactsContract.CommonDataKinds.StructuredName.CONTENT_ITEM_TYPE};
Cursor cursor = contentResolver.query(
            uri,       // URI representing the table/resource to be queried
            null,      // projection - the list of columns to return.  Null means "all"
            selection, // selection - Which rows to return (condition rows must match)
            selectionArgs,      // selection args - can be provided separately and subbed into selection.
            null);   // string specifying sort order

if (cursor.getCount() == 0) {
  return;
}
Log.i("Count:", Integer.toString(cursor.getCount())); // returns number of names on phone

while (cursor.moveToNext()) {
  // Behold, the firehose!
  Log.d(TAG, "-------------------new record\n");
  for(String column : cursor.getColumnNames()) {
    Log.d(TAG, column + ": " + cursor.getString(cursor.getColumnIndex(column)) + "\n");
  }
}

回答7:

try this,

public void onActivityResult(int reqCode, int resultCode, Intent data) { super.onActivityResult(reqCode, resultCode, data);

    try {
        if (resultCode == Activity.RESULT_OK) {
            Uri contactData = data.getData();
            Cursor cur = managedQuery(contactData, null, null, null, null);
            ContentResolver contect_resolver = getContentResolver();

            if (cur.moveToFirst()) {
                String id = cur.getString(cur.getColumnIndexOrThrow(ContactsContract.Contacts._ID));
                String name = "";
                String no = "";

                Cursor phoneCur = contect_resolver.query(ContactsContract.CommonDataKinds.Phone.CONTENT_URI, null,
                        ContactsContract.CommonDataKinds.Phone.CONTACT_ID + " = ?", new String[]{id}, null);

                if (phoneCur.moveToFirst()) {
                    name = phoneCur.getString(phoneCur.getColumnIndex(ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME));
                    no = phoneCur.getString(phoneCur.getColumnIndex(ContactsContract.CommonDataKinds.Phone.NUMBER));
                }

                Log.e("Phone no & name :***: ", name + " : " + no);
                txt.append(name + " : " + no + "\n");

                id = null;
                name = null;
                no = null;
                phoneCur = null;
            }
            contect_resolver = null;
            cur = null;
            //                      populateContacts();
        }
    } catch (IllegalArgumentException e) {
        e.printStackTrace();
        Log.e("IllegalArgumentException::", e.toString());
    } catch (Exception e) {
        e.printStackTrace();
        Log.e("Error :: ", e.toString());
    }
}

回答8:

Check here there is example code exactly for that: http://developer.android.com/guide/topics/ui/layout/listview.html

来源：https://stackoverflow.com/questions/4301064/how-to-get-the-first-name-and-last-name-from-android-contacts

标签

android

android-contacts