如何从Android联系人中获取以下字段？我使用的是Android 2.2。
答案 0 :(得分：55)
查看ContactsContract.CommonDataKinds.StructuredName课程。您可以在那里找到您要查找的所有列。试试这样：
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();
返回联系人中的所有姓名。更准确地说，您可以将联系人ID添加为查询的附加参数 - 您将获得特定联系人的地址。
答案 1 :(得分：16)
对于指定的联系人，您可以这样做：
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();
答案 2 :(得分：6)
尝试使用此代码获取有关联系人的所需信息，代码在此处 -
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");
答案 3 :(得分：6)
作为另一个例子（仅仅是为了好玩），但是用于获取单个用户的联系人姓名：
// 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();
答案 4 :(得分：3)
除了Raunak的建议外，还有一些链接可以帮助您入门：
答案 5 :(得分：1)
2015年底在棉花糖上尝试ContactsContract.Data.CONTENT_URI。我无法获得GIVEN_NAME或类似字段。我认为后来的apis已经弃用了这些。运行以下代码以打印出手机上的列
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");
}
}
答案 6 :(得分：0)
试试这个，
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());
}
}
答案 7 :(得分：0)
在这里结合各种解决方案，并看到结果中有重复的记录（由于多个帐户），我决定制作一个函数，将常见帐户类型优先于其他帐户类型。在这个示例中，我也忽略了完全空/空名称的记录（如果全部都是这样），但如果您愿意，您可以更改它：
int用法：
@RequiresPermission(
allOf = [Manifest.permission.READ_CONTACTS])
@WorkerThread
fun getContactIdToContactNameMap(context: Context): LongSparseArray<ContactObject> {
val contactIdToContactObjectMap = LongSparseArray<ContactObject>()
val contentResolver = context.contentResolver
contentResolver.query(ContactsContract.Data.CONTENT_URI,
arrayOf(
ContactsContract.CommonDataKinds.StructuredName.CONTACT_ID,
ContactsContract.CommonDataKinds.StructuredName.GIVEN_NAME,
ContactsContract.CommonDataKinds.StructuredName.FAMILY_NAME,
ContactsContract.CommonDataKinds.StructuredName.MIDDLE_NAME,
ContactsContract.RawContacts.ACCOUNT_TYPE),
ContactsContract.Data.MIMETYPE + " = ? AND " + ContactsContract.Data.IN_VISIBLE_GROUP + " = ?",
arrayOf(ContactsContract.CommonDataKinds.StructuredName.CONTENT_ITEM_TYPE, "1"),
null)?.use { cursor ->
// Log.d("AppLog", "got ${cursor.count} records for names")
val colContactId = cursor.getColumnIndex(
ContactsContract.CommonDataKinds.StructuredName.CONTACT_ID)
val colFirstName = cursor.getColumnIndex(
ContactsContract.CommonDataKinds.StructuredName.GIVEN_NAME)
val colFamilyName = cursor.getColumnIndex(
ContactsContract.CommonDataKinds.StructuredName.FAMILY_NAME)
val colMiddleName = cursor.getColumnIndex(
ContactsContract.CommonDataKinds.StructuredName.MIDDLE_NAME)
val colAccountType =
cursor.getColumnIndex(ContactsContract.RawContacts.ACCOUNT_TYPE)
val googleAccount = "com.google"
//https://stackoverflow.com/a/44802016/878126
val prioritizedAccountTypes =
hashSetOf("vnd.sec.contact.phone", "com.htc.android.pcsc",
"com.sonyericsson.localcontacts", "com.lge.sync", "com.lge.phone",
"vnd.tmobileus.contact.phone", "com.android.huawei.phone",
"Local Phone Account",
"")
val contactIdToAccountTypeMap = LongSparseArray<String>()
while (cursor.moveToNext()) {
val contactId = cursor.getLong(colContactId)
val accountType = cursor.getString(colAccountType).orEmpty()
val existingContact = contactIdToContactObjectMap.get(contactId)
if (existingContact != null) {
//this can occur, as we go over all of the items, including duplicate ones made by various sources
// https://stackoverflow.com/a/4599474/878126
val previousAccountType = contactIdToAccountTypeMap.get(contactId)
//google account is most prioritized, so we skip current one if previous was of it
if (previousAccountType == googleAccount)
continue
if (accountType != googleAccount && previousAccountType != null && prioritizedAccountTypes.contains(
previousAccountType))
//we got now a name of an account that isn't prioritized, but we already had a prioritized one, so ignore
continue
}
contactIdToAccountTypeMap.put(contactId, accountType)
val firstName = cursor.getString(colFirstName)?.trim()
val lastName = cursor.getString(colFamilyName)?.trim()
val middleName = cursor.getString(colMiddleName)?.trim()
if (firstName.isNullOrBlank() && lastName.isNullOrBlank() && middleName.isNullOrBlank())
continue
val contactObject = existingContact ?: ContactObject()
contactObject.firstName = firstName
contactObject.lastName = lastName
contactObject.middleName = middleName
contactIdToContactObjectMap.put(contactId, contactObject)
}
}
return contactIdToContactObjectMap
}
class ContactObject {
var firstName: String? = null
var middleName: String? = null
var lastName: String? = null
}
答案 8 :(得分：-1)