如何挑选到我的应用程序从电话簿的Android的联系方式吗? [英] How to pick contact number from phone-book of android in to my application?
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问题描述
所有 我想利用人数从机器人的电话簿在我的应用程序数据库.. 我曾尝试用以下code,但在这里我得到的人名,而不是我想要的号码从电话簿中,并希望将其存储在我的database..how实现这一????任何一个可以指导我..
@覆盖
公共无效onActivityResult(INT REQ code,INT结果code,意图数据){
super.onActivityResult(REQ code,因此code,数据);
开关(REQ code){
案例(PICK_CONTACT):
如果(结果code == Activity.RESULT_OK){
乌里联系数据= data.getData();
光标C = managedQuery(联络人资料,NULL,NULL,NULL,NULL);
如果(c.moveToFirst()){
//其他的数据可用于联系。我已决定
//只拿到的联系人的名字。
字符串名称= c.getString(c.getColumnIndexOrThrow(ContactsContract.Contacts.CONTENT_TYPE));
Toast.makeText(getApplicationContext(),姓名,Toast.LENGTH_SHORT).show();
}
}
}
在此先感谢 -
解决方案
试试这个code,
@覆盖 公共无效onActivityResult(INT REQ code,INT结果code,意图 数据){ super.onActivityResult(REQ code,因此code,数据); 开关(REQ code){ 案例(PICK_CONTACT): 如果(结果code == Activity.RESULT_OK) { 乌里联系数据= data.getData(); 光标C = managedQuery(联络人资料,NULL,NULL,NULL,NULL); 如果(c.moveToFirst()){ 字符串ID = c.getString(c.getColumnIndexOrThrow(ContactsContract.Contacts._ID)); 字符串hasPhone = c.getString(c.getColumnIndex(ContactsContract.Contacts.HAS_PHONE_NUMBER)); 如果(hasPhone.equalsIgnoreCase(1)){ 光标手机= getContentResolver()查询( ContactsContract.CommonDataKinds.Phone.CONTENT_URI,空, ContactsContract.CommonDataKinds.Phone.CONTACT_ID +=+的id, NULL,NULL); phones.moveToFirst(); 串cNumber = phones.getString(phones.getColumnIndex(数据1)); } } } }
all i want to take number from phonebook of android in my application database.. i have tried it with below code but here am getting name of person instead i want number from phonebook and want to store it in my database..how to achieve this????can any one guide me..
@Override
public void onActivityResult(int reqCode, int resultCode, Intent data){
super.onActivityResult(reqCode, resultCode, data);
switch(reqCode){
case (PICK_CONTACT):
if (resultCode == Activity.RESULT_OK){
Uri contactData = data.getData();
Cursor c = managedQuery(contactData, null, null, null, null);
if (c.moveToFirst()){
// other data is available for the Contact. I have decided
// to only get the name of the Contact.
String name = c.getString(c.getColumnIndexOrThrow(ContactsContract.Contacts.CONTENT_TYPE));
Toast.makeText(getApplicationContext(), name, Toast.LENGTH_SHORT).show();
}
}
}
Thanks in Advance--
解决方案
try this code,
@Override public void onActivityResult(int reqCode, int resultCode, Intent data){ super.onActivityResult(reqCode, resultCode, data); switch(reqCode){ case (PICK_CONTACT): if (resultCode == Activity.RESULT_OK) { Uri contactData = data.getData(); Cursor c = managedQuery(contactData, null, null, null, null); if (c.moveToFirst()) { String id = c.getString(c.getColumnIndexOrThrow(ContactsContract.Contacts._ID)); String hasPhone = c.getString(c.getColumnIndex(ContactsContract.Contacts.HAS_PHONE_NUMBER)); if (hasPhone.equalsIgnoreCase("1")) { Cursor phones = getContentResolver().query( ContactsContract.CommonDataKinds.Phone.CONTENT_URI,null, ContactsContract.CommonDataKinds.Phone.CONTACT_ID +" = "+ id, null, null); phones.moveToFirst(); String cNumber = phones.getString(phones.getColumnIndex("data1")); } } } }
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