我正在尝试在 android 上制作一个应用程序,该应用程序将联系人姓名作为字符串输入,如果电话簿中存在该联系人,则返回他的电话号码......
我尝试四处搜索,但没有明确的教程说明如何做到这一点
输入:联系人姓名输出:电话号码
请帮忙
Cursor cursor = context.getContentResolver().query(ContactsContract.Contacts.CONTENT_URI,null, null, null, null);
while (cursor.moveToNext()) {
String name = cursor.getString(cursor.getColumnIndex(ContactsContract.Contacts.DISPLAY_NAME));
if(name.equalsIgnoreCase(token3)) {
try{ ContentResolver cr = context.getContentResolver();
Cursor pCur = cr.query(ContactsContract.CommonDataKinds.Phone.CONTENT_URI, null, ContactsContract.CommonDataKinds.Phone.CONTACT_ID + " = ?", new String[] { ContactsContract.CommonDataKinds.Phone._ID}, null);
String lname = pCur.getString(pCur.getColumnIndex(ContactsContract.CommonDataKinds.Phone.NUMBER));
Toast.makeText(context, "number is "+lname, Toast.LENGTH_LONG).show();
}catch (Exception e) {
// TODO: handle exception
}
}
}
这就是我目前所拥有的。 try catch 块中的代码段总是崩溃。
我最终写了这个方法来解决我的问题
public String get_Number(String name,Context context)
{String number="";
Uri uri = ContactsContract.CommonDataKinds.Phone.CONTENT_URI;
String[] projection = new String[] {ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME,
ContactsContract.CommonDataKinds.Phone.NUMBER};
Cursor people = context.getContentResolver().query(uri, projection, null, null, null);
int indexName = people.getColumnIndex(ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME);
int indexNumber = people.getColumnIndex(ContactsContract.CommonDataKinds.Phone.NUMBER);
people.moveToFirst();
do {
String Name = people.getString(indexName);
String Number = people.getString(indexNumber);
if(Name.equalsIgnoreCase(name)){return Number.replace("-", "");}
// Do work...
} while (people.moveToNext());
if(!number.equalsIgnoreCase("")){return number.replace("-", "");}
else return number;
}
它可能不是很有效,但嘿,它有效
试试这个方式...
Cursor cur = cr.query(ContactsContract.Contacts.CONTENT_URI, null,);
String name = cur.getString(cur.getColumnIndex(ContactsContract.Contacts.DISPLAY_NAME));
if(name.equals(Your_String)) {
Cursor pCur = cr.query(ContactsContract.CommonDataKinds.Phone.CONTENT_URI, null, ContactsContract.CommonDataKinds.Phone.CONTACT_ID + " = ?", new String[] { id }, null);
String lname = pCur.getString(pCur.getColumnIndex(ContactsContract.CommonDataKinds.Phone.NUMBER));
}
试试这种方式。
Uri uri = ContactsContract.CommonDataKinds.Phone.CONTENT_URI;
String[] projection = new String[] {ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME,
ContactsContract.CommonDataKinds.Phone.NUMBER};
Cursor cursor = getContentResolver().query(uri, projection, null, null, null);
int idxName = cursor.getColumnIndex(ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME);
int idxNumber = cursor.getColumnIndex(ContactsContract.CommonDataKinds.Phone.NUMBER);
if(cursor.moveToFirst()) {
do {
contactname = cursor.getString(idxName);
contactNumber = cursor.getString(idxNumber);
if (contactname.equals("YOUR CONTACT NAME")){
Log.d(LOG_TAG,"Contact Name -> "+ contactname +" Contact Number -> "+ contactNumber);
}
} while (cursor.moveToNext());
}
cursor.close();