Question

我正在尝试创建自定义联系人应用，该应用仅显示联系人号的联系人。首先，有没有自动化的方法呢？假设没有，那么我试图通过其名称搜索联系人，例如的罗汉

以下是代码： -

Cursor photoCursor = getContentResolver().query(
            android.provider.ContactsContract.Contacts.CONTENT_URI,
            new String[] { ContactsContract.Contacts.PHOTO_ID,
                    ContactsContract.Contacts.DISPLAY_NAME },
            ContactsContract.Contacts.DISPLAY_NAME + " = ?",
            new String[]{"Rohan"}, null);
    photoCursor.moveToFirst();
    while (photoCursor.moveToNext()) {
        Log.d("Photo Thumbnail", "" + photoCursor.getString(1));
    }

虽然联系人存在但我没有收到任何记录，如果我删除选择＆amp; 选择参数我在日志中看到 Rohan 。我做错了什么？

Answer 1

试试这个：

    Cursor contactLookupCursor =
            getContentResolver().query(
                    Uri.withAppendedPath(PhoneLookup.CONTENT_FILTER_URI,
                            Uri.encode("Rohan")),
                    new String[] {PhoneLookup.DISPLAY_NAME, PhoneLookup.NUMBER},
                    null,
                    null,
                    null);

    try {
        while (contactLookupCursor.moveToNext()) {
            contactName = contactLookupCursor.getString(contactLookupCursor.getColumnIndexOrThrow(PhoneLookup.DISPLAY_NAME));
            contactNumber = contactLookupCursor.getString(contactLookupCursor.getColumnIndexOrThrow(PhoneLookup.NUMBER));
        }
    } finally {
        contactLookupCursor.close();
    }

Answer 2

我是通过使用以下代码

完成的

Cursor cursor = getContentResolver().query(
            android.provider.ContactsContract.Contacts.CONTENT_URI,
            new String[] { ContactsContract.Contacts.PHOTO_ID,
                    ContactsContract.Contacts.DISPLAY_NAME,
                    ContactsContract.Contacts._ID },
            ContactsContract.Contacts.HAS_PHONE_NUMBER, null,
            ContactsContract.Contacts.DISPLAY_NAME);

此cursor为所有拥有任意电话号码的联系人提供，然后我将ID中的唯一ArrayList保存为

cursor.moveToFirst();

    while (cursor.moveToNext()) {
        contactsID.add(cursor.getString(2));
    }

然后在选择联系人时，我会使用此

找到联系号码

Cursor cursor = getContentResolver()
                    .query(android.provider.ContactsContract.CommonDataKinds.Phone.CONTENT_URI,
                            new String[] {
                                    ContactsContract.CommonDataKinds.Phone.NUMBER,
                                    ContactsContract.CommonDataKinds.Phone.CONTACT_ID,
                                    ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME },
                            ContactsContract.CommonDataKinds.Phone.CONTACT_ID
                                    + " = ?",
                            new String[] { contactsID.get(position) }, null);
            contactNumbers = new ArrayList<String>();
            while (cursor.moveToNext()) {
                contactNumbers.add(cursor.getString(0));
                Log.d("number", cursor.getString(0));
            }

Answer 3

搜索部分显示名称的简单解决方案。

ContentResolver contentResolver = getCurrentActivity().getContentResolver();

String   whereString = "display_name LIKE ?";
String[] whereParams = new String[]{ "%" + searchText + "%" };

Cursor contactCursor = contentResolver.query(
        ContactsContract.Data.CONTENT_URI,
        null,
        whereString,
        whereParams,
        null );

while( contactCursor.moveToNext() ) {

    int contactId = getIntFromCursor( contactCursor, ContactsContract.Data.CONTACT_ID );

    Log.d( "Contact ID", contactId)

}

contactCursor.close();

Answer 4

看起来您正在尝试实现一个屏幕，允许用户选择联系人，然后选择该联系人的电话号码。

如果是这种情况，您可以改用电话选择器意图：

Intent intent = Intent(Intent.ACTION_PICK);
intent.setType(CommonDataKinds.Phone.CONTENT_TYPE);
startActivityForResult(intent, REQUEST_SELECT_PHONE_NUMBER);

这将打开本机通讯录应用，并允许用户选择联系人和电话号码。然后，您将在您的应用中收到如下结果：

@Override
protected void onActivityResult(int requestCode, int resultCode, Intent data) {
    if (requestCode == REQUEST_SELECT_PHONE_NUMBER && resultCode == RESULT_OK) {
        // Get the URI and query the content provider for the phone number
        Uri contactUri = data.getData();
        String[] projection = new String[]{CommonDataKinds.Phone.NUMBER};
        Cursor cursor = getContentResolver().query(contactUri, projection,
                null, null, null);
        // If the cursor returned is valid, get the phone number
        if (cursor != null && cursor.moveToFirst()) {
            int numberIndex = cursor.getColumnIndex(CommonDataKinds.Phone.NUMBER);
            String number = cursor.getString(numberIndex);
            // Do something with the phone number
            ...
        }
    }
}

按名称搜索联系人

4 个答案:

搜索部分显示名称的简单解决方案。