本文介绍了进不去的Andr​​oid 2.3.3电话号码的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我写了一个简单的应用程序取得联系的电话号码。然而,在电话号码返回空。

下面是我的code:

 私人无效queryContactPhoneNumber(){
    // TODO自动生成方法存根
    的String [] COLS =新的String [] {People.NAME,People.NUMBER};
    乌里myContacts = People.CONTENT_URI;
    光标mqCur = managedQuery(myContacts,COLS,NULL,NULL,NULL);
    如果(mqCur.moveToFirst())
    {
        字符串MYNAME = NULL;
        字符串为mynumber = NULL;
        做
        {
            MYNAME = mqCur.getString(mqCur.getColumnIndex(People.NAME));
            为mynumber = mqCur.getString(mqCur.getColumnIndex(People.NUMBER));
            Toast.makeText(这一点,MYNAME ++为mynumber,Toast.LENGTH_SHORT).show();
        }
        而(mqCur.moveToNext());
    }
}
 

解决方案

试试这个,

 乌里myContacts = ContactsContract.CommonDataKinds.Phone.CONTENT_URI; // People.CONTENT_URI;
        光标mqCur = managedQuery(myContacts,NULL,NULL,NULL,NULL);
        如果(mqCur.moveToFirst())
        {
            字符串MYNAME = NULL;
            字符串为mynumber = NULL;
            做
            {
                MYNAME = mqCur.getString(mqCur.getColumnIndexOrThrow(ContactsContract.Contacts.DISPLAY_NAME));
                为mynumber = mqCur.getString(mqCur
                        .getColumnIndexOrThrow(ContactsContract.CommonDataKinds.Phone.NUMBER));
                Toast.makeText(这一点,MYNAME ++为mynumber,Toast.LENGTH_SHORT).show();
            }
            而(mqCur.moveToNext());
        }
 

我想这会帮助你。

I write a simple application to get phone number in Contacts. However, the phone number return "null".

Here is my code:

private void queryContactPhoneNumber() {
    // TODO Auto-generated method stub
    String[] cols = new String[] {People.NAME, People.NUMBER};
    Uri myContacts = People.CONTENT_URI;
    Cursor mqCur =  managedQuery(myContacts, cols, null, null, null);
    if(mqCur.moveToFirst())
    {
        String myname = null;
        String mynumber = null;
        do
        {
            myname = mqCur.getString(mqCur.getColumnIndex(People.NAME));
            mynumber = mqCur.getString(mqCur.getColumnIndex(People.NUMBER));
            Toast.makeText(this, myname + " " + mynumber, Toast.LENGTH_SHORT).show();
        }
        while(mqCur.moveToNext());
    }
}
解决方案

Try this,

Uri myContacts = ContactsContract.CommonDataKinds.Phone.CONTENT_URI ;//People.CONTENT_URI;
        Cursor mqCur =  managedQuery(myContacts, null, null, null, null);
        if(mqCur.moveToFirst())
        {
            String myname = null;
            String mynumber = null;
            do
            {
                myname = mqCur.getString(mqCur.getColumnIndexOrThrow(ContactsContract.Contacts.DISPLAY_NAME));
                mynumber = mqCur.getString(mqCur
                        .getColumnIndexOrThrow(ContactsContract.CommonDataKinds.Phone.NUMBER));
                Toast.makeText(this, myname + " " + mynumber, Toast.LENGTH_SHORT).show();
            }
            while(mqCur.moveToNext());
        }

I think this will help you.

这篇关于进不去的Andr​​oid 2.3.3电话号码的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!

10-27 03:39