我想在两行ListView
中列出联系人详细信息(人员姓名、人员号码)(第一行:人员姓名,第二行:电话号码)。我成功获取了联系人的所有必要信息,但在 ListView
中列出它们时遇到问题。 ListView
中仅显示第一个联系人。
在onCreate()
中,我创建了ContentResolver
对象,调用方法get_list()
来获取姓名和电话号码并尝试显示ListView
:
ContentResolver cr = getContentResolver();
Cursor cur = cr.query(ContactsContract.Contacts.CONTENT_URI, null, null, null, null);
int anzahl = cur.getCount();
Log.d("anzahl09", "anzahl09 " + anzahl);
telefonname = new String[anzahl];
telefonnummer = new String[anzahl];
zusammengesetzt = new String[anzahl];
kontakte_laden = new String[anzahl][anzahl];
String[] telefonname_telefonnummer = new String[anzahl];
telefonname_telefonnummer = get_list(cur, cr);
int i =0;
do {
Log.d("schleife09", "schleife09 " + i + "|" + telefonname_telefonnummer[i] + "|" + telefonname[i] + "|" + telefonnummer[i] + "|" + kontakte_laden[i][i]);
kontakte_laden[i][0] = telefonname[i];
kontakte_laden[i][1] = telefonnummer[i];
i++;
} while (i < anzahl);
ListView listview = ( ListView ) findViewById ( android.R.id.list );
listview.setPadding ( 20, 20, 20, 20 );
@SuppressWarnings("Convert2Diamond") ArrayList< HashMap< String, String > > list = new ArrayList< HashMap< String, String > > ();
for (String[] anAussehen : kontakte_laden) {
//noinspection Convert2Diamond
item = new HashMap<String, String>();
item.put("line1", anAussehen[0]);
item.put("line2", anAussehen[1]);
list.add(item);
}
SimpleAdapter sa = new SimpleAdapter(this, list,
android.R.layout.two_line_list_item,
new String[]{"line1", "line2"},
new int[]{android.R.id.text1, android.R.id.text2});
listview.setAdapter(sa);
listview.setOnItemClickListener(listViewOnItemClickListener);
读取联系信息的方法get_list()
如下所示:
private String[] get_list(Cursor cur, ContentResolver cr){
if (cur.getCount() > 0) {
while (cur.moveToNext()) {
int i=0;
String id = cur.getString(cur.getColumnIndex(ContactsContract.Contacts._ID));
telefonname[i] = cur.getString(cur.getColumnIndex(ContactsContract.Contacts.DISPLAY_NAME));
Log.d("telefonname09", "telefonname09 " +telefonname[i]);
if (Integer.parseInt(cur.getString(cur.getColumnIndex(ContactsContract.Contacts.HAS_PHONE_NUMBER))) > 0) {
if (Integer.parseInt(cur.getString(cur.getColumnIndex(ContactsContract.Contacts.HAS_PHONE_NUMBER))) > 0) {
Cursor pCur = cr.query(ContactsContract.CommonDataKinds.Phone.CONTENT_URI, null, ContactsContract.CommonDataKinds.Phone.CONTACT_ID +" = ?", new String[]{id}, null);
while (pCur.moveToNext()) {
telefonnummer[i] = pCur.getString( pCur.getColumnIndex( ContactsContract.CommonDataKinds.Phone.NUMBER ) );
Log.d("telefonnummer09", "telefonnummer09 " + telefonnummer[i]); // correct values
}
pCur.close();
}
}
zusammengesetzt[i] = telefonname[i] + "|" + telefonnummer[i];
}
}
// only one value
for(int i=0; i < cur.getCount(); i++){
Log.d("telefonnummer07", "telefonnummer07 " + telefonname[i] + "|" + telefonnummer[i]);
}
cur.close();
return zusammengesetzt;
}
更新
问题是,人名数组 (telefonname[]
) 和电话号码 (telefonnummer[]
) 只进入了 while 循环get_list()
方法获取正确的内容。在 while 循环之外,数组仅获得最后一个值。请参阅上面的方法get_list()
。
解决方案 我更改了以下几点:
- 我之前直接将编号和名称定义到全局字符串数组中。因此,在
get_list()
方法中,我必须定义两个新的字符串数组 - 我将姓名和电话号码引用到这些新的本地字符串数组
- 将这些新的本地字符串数组引用到全局字符串数组
这是工作代码:
onCreate()
方法:
ContentResolver cr = getContentResolver();
Cursor cur = cr.query(ContactsContract.Contacts.CONTENT_URI, null, null, null, null);
int anzahl = cur.getCount();
Log.d("anzahl09", "anzahl09 " + anzahl);
telefonname_global = new String[anzahl];
telefonnummer_global = new String[anzahl];
kontakte_laden = new String[anzahl][anzahl];
get_list(cur, cr);
int i =0;
do {
Log.d("schleife09", "schleife09 " + i + "|" /*+ telefonname_telefonnummer[i] + "|" */ + telefonname_global[i] + "|" + telefonnummer_global[i] + "|" + kontakte_laden[i][i]);
kontakte_laden[i][0] = telefonname_global[i];
kontakte_laden[i][1] = telefonnummer_global[i];
i++;
} while (i < anzahl);
ListView listview = ( ListView ) findViewById ( android.R.id.list );
listview.setPadding ( 20, 20, 20, 20 );
@SuppressWarnings("Convert2Diamond") ArrayList< HashMap< String, String > > list = new ArrayList< HashMap< String, String > > ();
for (String[] anAussehen : kontakte_laden) {
//noinspection Convert2Diamond
item = new HashMap<String, String>();
item.put("line1", anAussehen[0]);
item.put("line2", anAussehen[1]);
list.add(item);
}
SimpleAdapter sa = new SimpleAdapter(this, list,
android.R.layout.two_line_list_item,
new String[]{"line1", "line2"},
new int[]{android.R.id.text1, android.R.id.text2});
listview.setAdapter(sa);
listview.setOnItemClickListener(listViewOnItemClickListener);
以及 get_list()
方法:
private void get_list(Cursor cur, ContentResolver cr){
int i = 0;
if (cur.getCount() > 0) {
String[] telefonname = new String[cur.getCount()];
String[] telefonnummer = new String[cur.getCount()];
while (cur.moveToNext()) {
String id = cur.getString(cur.getColumnIndex(ContactsContract.Contacts._ID));
telefonname[i] = cur.getString(cur.getColumnIndex(ContactsContract.Contacts.DISPLAY_NAME));
Log.d("telefonname09", "telefonname09 " +telefonname[i]);
if (Integer.parseInt(cur.getString(cur.getColumnIndex(ContactsContract.Contacts.HAS_PHONE_NUMBER))) > 0) {
if (Integer.parseInt(cur.getString(cur.getColumnIndex(ContactsContract.Contacts.HAS_PHONE_NUMBER))) > 0) {
Cursor pCur = cr.query(ContactsContract.CommonDataKinds.Phone.CONTENT_URI, null, ContactsContract.CommonDataKinds.Phone.CONTACT_ID +" = ?", new String[]{id}, null);
while (pCur.moveToNext()) {
telefonnummer[i] = pCur.getString( pCur.getColumnIndex( ContactsContract.CommonDataKinds.Phone.NUMBER ) );
Log.d("telefonnummer09", "telefonnummer09 " + telefonnummer[i]);
}
pCur.close();
}
}
Log.d("telefonnummer08", "telefonnummer08 " + telefonnummer[i]);
telefonname_global[i] = telefonname[i];
telefonnummer_global[i] = telefonnummer[i];
i++;
}
}
// Fehler: Array telefonnummer hat nur einen Wert
for(int j=0; j < cur.getCount(); j++){
Log.d("telefonnummer07", "telefonnummer07 " + telefonname_global[j] + "|" + telefonnummer_global[j]);
}
cur.close();
}
非常感谢您的帮助。
另一个问题是,没有电话号码的电子邮件地址或联系人不应在此列表中。如何实现这一目标?
最佳答案
我不是增量,这就是问题所在
private String[] get_list(Cursor cur, ContentResolver cr){
if (cur.getCount() > 0) {
// inialize outside
int i=0;
while (cur.moveToNext()) {
String id = cur.getString(cur.getColumnIndex(ContactsContract.Contacts._ID));
telefonname[i] = cur.getString(cur.getColumnIndex(ContactsContract.Contacts.DISPLAY_NAME));
Log.d("telefonname09", "telefonname09 " +telefonname[i]);
if (Integer.parseInt(cur.getString(cur.getColumnIndex(ContactsContract.Contacts.HAS_PHONE_NUMBER))) > 0) {
if (Integer.parseInt(cur.getString(cur.getColumnIndex(ContactsContract.Contacts.HAS_PHONE_NUMBER))) > 0) {
Cursor pCur = cr.query(ContactsContract.CommonDataKinds.Phone.CONTENT_URI, null, ContactsContract.CommonDataKinds.Phone.CONTACT_ID +" = ?", new String[]{id}, null);
while (pCur.moveToNext()) {
telefonnummer[i] = pCur.getString( pCur.getColumnIndex( ContactsContract.CommonDataKinds.Phone.NUMBER ) );
Log.d("telefonnummer09", "telefonnummer09 " + telefonnummer[i]); // correct values
}
pCur.close();
}
}
zusammengesetzt[i] = telefonname[i] + "|" + telefonnummer[i];
//i is need to increment
i++;
}
}
仅在电话号码存在时添加
for (String[] anAussehen : kontakte_laden) {
if(anAussehen[1] != null)
{
//noinspection Convert2Diamond
item = new HashMap<String, String>();
item.put("line1", anAussehen[0]);
item.put("line2", anAussehen[1]);
list.add(item);
}
}
关于android - 在两行ListView中列出联系信息,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/32666551/