问题描述
我已经建立了一个Android应用程序。它会得到默认地址簿(我的手机)的所有联系人(1000接触)。然后,我显示在列表视图(我的应用程序)所有。但是,我花费约13S加载和显示在列表视图。在我的下面code,我用了3个命令来查询:产品名称,每个联系人的电话号码和公司。我想这就是为什么我的应用程序,它花太多时间来加载和列表视图显示数据的原因。
我有2个问题:
- 如何只用1 SQLite的命令来获得一个机器人接触的所有信息?
- 有没有加载任何方式和显示数据的列表视图上快???
这是code查询:姓名,电话没有,公司联系。查询的结果将被存储到一个光标。然后我用光标设置listAdapter。
公共无效addNewContacts(字符串名称,字符串电话,弦乐COM){
字符串显示名称=名称;
字符串移动电话号码=手机;
字符串公司= com的;
字符串JOBTITLE =工程师; ArrayList的< ContentProviderOperation> OPS =新的ArrayList< ContentProviderOperation> (); ops.add(ContentProviderOperation.newInsert(
ContactsContract.RawContacts.CONTENT_URI)
.withValue(ContactsContract.RawContacts.ACCOUNT_TYPE,NULL)
.withValue(ContactsContract.RawContacts.ACCOUNT_NAME,NULL)
。建立()); // ------------------------------------------------ ------名称
如果(显示名称!= NULL){
ops.add(ContentProviderOperation.newInsert(
ContactsContract.Data.CONTENT_URI)
.withValueBackReference(ContactsContract.Data.RAW_CONTACT_ID,0)
.withValue(ContactsContract.Data.MIMETYPE,
ContactsContract.CommonDataKinds.StructuredName.CONTENT_ITEM_TYPE)
.withValue(
ContactsContract.CommonDataKinds.StructuredName.DISPLAY_NAME,
显示名称).build());
} // ------------------------------------------------ - - - 手机号
如果(移动电话号码!= NULL){
ops.add(ContentProviderOperation。
newInsert(ContactsContract.Data.CONTENT_URI)
.withValueBackReference(ContactsContract.Data.RAW_CONTACT_ID,0)
.withValue(ContactsContract.Data.MIMETYPE,
ContactsContract.CommonDataKinds.Phone.CONTENT_ITEM_TYPE)
.withValue(ContactsContract.CommonDataKinds.Phone.NUMBER,移动电话号码)
.withValue(ContactsContract.CommonDataKinds.Phone.TYPE,
ContactsContract.CommonDataKinds.Phone.TYPE_MOBILE)
。建立());
} // ------------------------------------------------ ------组织
如果(company.equals()及与放大器;!jobTitle.equals(!)){
ops.add(ContentProviderOperation.newInsert(ContactsContract.Data.CONTENT_URI)
.withValueBackReference(ContactsContract.Data.RAW_CONTACT_ID,0)
.withValue(ContactsContract.Data.MIMETYPE,
ContactsContract.CommonDataKinds.Organization.CONTENT_ITEM_TYPE)
.withValue(ContactsContract.CommonDataKinds.Organization.COMPANY,公司)
.withValue(ContactsContract.CommonDataKinds.Organization.TYPE,ContactsContract.CommonDataKinds.Organization.TYPE_WORK)
.withValue(ContactsContract.CommonDataKinds.Organization.TITLE,JOBTITLE)
.withValue(ContactsContract.CommonDataKinds.Organization.TYPE,ContactsContract.CommonDataKinds.Organization.TYPE_WORK)
。建立());
} //问联系人提供商创建新的联系人
尝试{
。getContentResolver()applyBatch(ContactsContract.AUTHORITY,OPS);
}赶上(例外五){
Log.e(错误:,e.getMessage());
}
}
我不这么认为你可以,也应该是复杂的。
在读取的任何信息,我需要验证,如果名称中至少存在。
下面是code我用:
公共类ContactListControl {/ **确保final字段等于input_display_name * /
私人最终静态字符串NAME =名;
私人最终静态字符串GMAIL =Gmail的;
私人最终静态字符串PHONE =手机;私人ContactListControl(){}公共静态无效onGetContactInfo(意向数据,上下文的背景下,列表与LT; InputView> inputViewList){ 地图<字符串,字符串> contactDataMap =新的HashMap<字符串,字符串>(); ContentResolver的CR = context.getContentResolver(); 乌里联络资料= data.getData();
//光标指针= managedQuery(联络资料,NULL,NULL,NULL,NULL);
光标光标= cr.query(联络资料,NULL,NULL,NULL,NULL);
cursor.moveToFirst();
字符串名称= cursor.getString(cursor.getColumnIndexOrThrow(ContactsContract.Contacts.DISPLAY_NAME));
字符串ID = cursor.getString(cursor.getColumnIndexOrThrow(ContactsContract.Contacts._ID)); contactDataMap.put(NAME(名称= NULL)名称:!?);
如果(的Integer.parseInt(cursor.getString(
cursor.getColumnIndex(ContactsContract.Contacts.HAS_PHONE_NUMBER)))> 0){
光标pCur = cr.query(
ContactsContract.CommonDataKinds.Phone.CONTENT_URI,
空值,
ContactsContract.CommonDataKinds.Phone.CONTACT_ID +=?,
新的String [] {ID},
空值); 而(pCur.moveToNext()){
串号= pCur.getString(pCur.getColumnIndexOrThrow(ContactsContract.CommonDataKinds.Phone.NUMBER)); contactDataMap.put(电话,(数量= NULL)号:!?);
打破; //?我们只想要1值
}
pCur.close();
} 光标emailCur = cr.query(
ContactsContract.CommonDataKinds.Email.CONTENT_URI,
空值,
ContactsContract.CommonDataKinds.Email.CONTACT_ID +=?,
新的String [] {ID},NULL); 而(emailCur.moveToNext()){
//这将让你得到一些电子邮件地址
//如果电子邮件地址被存储在一个阵列
字符串email = emailCur.getString(
emailCur.getColumnIndex(ContactsContract.CommonDataKinds.Email.DATA)); contactDataMap.put(GMAIL,(电子邮件= NULL)电子邮件:!?);
打破; //?我们只想要1值
}
emailCur.close(); cursor.close();
Log.d(活动结果,的onActivityResult - 有联系:+ contactDataMap.get(NAME)+,+ contactDataMap.get(GMAIL)+,+ contactDataMap.get(电话));
InputView outputViewName = getInputputViewByDisplayName(姓名,inputViewList);
字符串tempName = onGetContactInfoAppend(outputViewName,contactDataMap.get(名));
outputViewName.setValue(tempName); InputView outputViewGmail = getInputputViewByDisplayName(GMAIL,inputViewList);
串tempGmail = onGetContactInfoAppend(outputViewGmail,contactDataMap.get(GMAIL));
outputViewGmail.setValue(tempGmail);
InputView outputViewPhone = getInputputViewByDisplayName(电话,inputViewList);
字符串tempPhone = onGetContactInfoAppend(outputViewPhone,contactDataMap.get(电话));
outputViewPhone.setValue(tempPhone);
}私人静态字符串onGetContactInfoAppend(InputView outputView,串联络资料){ 如果(联络资料== NULL){
联络资料=;
} 字符串TEMP = outputView.getValue(); //如果(!。等于(联络资料)){
如果(!。等于(TEMP)){
TEMP =温度+|;
} TEMP =温度+联络资料;
//} 返回温度;
}私有静态InputView getInputputViewByDisplayName(字符串显示名,名单,LT; InputView> inputViewList){
对于(InputView inputView:inputViewList){
如果(inputView.getDisplayName()。等于(显示名)){
返回inputView;
}
} 返回null;
}
}
I have built an android app. It will get all contacts (1000 contact) from default address book (of my phone). And then, I show all of them on list view (of my app). BUT, I spend about 13s to load and display on list view. In my code below, I used 3 commands to query: Name, Phone Number and Company of each contact. I think this is the reason why my app which spend too much time to load and display data on listview.
I have 2 questions:
- How to get all information of an contact in android by using only 1 SQLite command?
- Is there any ways to load and display data onto listview faster???
This is code to query: name, phone no, company of a contact. The result of the query will be stored into a cursor. And then I use the cursor to set listAdapter.
public void addNewContacts(String name, String phone, String com){
String DisplayName = name;
String MobileNumber = phone;
String company = com;
String jobTitle = "Engineer";
ArrayList < ContentProviderOperation > ops = new ArrayList < ContentProviderOperation > ();
ops.add(ContentProviderOperation.newInsert(
ContactsContract.RawContacts.CONTENT_URI)
.withValue(ContactsContract.RawContacts.ACCOUNT_TYPE, null)
.withValue(ContactsContract.RawContacts.ACCOUNT_NAME, null)
.build());
//------------------------------------------------------ Names
if (DisplayName != null) {
ops.add(ContentProviderOperation.newInsert(
ContactsContract.Data.CONTENT_URI)
.withValueBackReference(ContactsContract.Data.RAW_CONTACT_ID, 0)
.withValue(ContactsContract.Data.MIMETYPE,
ContactsContract.CommonDataKinds.StructuredName.CONTENT_ITEM_TYPE)
.withValue(
ContactsContract.CommonDataKinds.StructuredName.DISPLAY_NAME,
DisplayName).build());
}
//------------------------------------------------------ Mobile Number
if (MobileNumber != null) {
ops.add(ContentProviderOperation.
newInsert(ContactsContract.Data.CONTENT_URI)
.withValueBackReference(ContactsContract.Data.RAW_CONTACT_ID, 0)
.withValue(ContactsContract.Data.MIMETYPE,
ContactsContract.CommonDataKinds.Phone.CONTENT_ITEM_TYPE)
.withValue(ContactsContract.CommonDataKinds.Phone.NUMBER, MobileNumber)
.withValue(ContactsContract.CommonDataKinds.Phone.TYPE,
ContactsContract.CommonDataKinds.Phone.TYPE_MOBILE)
.build());
}
//------------------------------------------------------ Organization
if (!company.equals("") && !jobTitle.equals("")) {
ops.add(ContentProviderOperation.newInsert(ContactsContract.Data.CONTENT_URI)
.withValueBackReference(ContactsContract.Data.RAW_CONTACT_ID, 0)
.withValue(ContactsContract.Data.MIMETYPE,
ContactsContract.CommonDataKinds.Organization.CONTENT_ITEM_TYPE)
.withValue(ContactsContract.CommonDataKinds.Organization.COMPANY, company)
.withValue(ContactsContract.CommonDataKinds.Organization.TYPE, ContactsContract.CommonDataKinds.Organization.TYPE_WORK)
.withValue(ContactsContract.CommonDataKinds.Organization.TITLE, jobTitle)
.withValue(ContactsContract.CommonDataKinds.Organization.TYPE, ContactsContract.CommonDataKinds.Organization.TYPE_WORK)
.build());
}
// Asking the Contact provider to create a new contact
try {
getContentResolver().applyBatch(ContactsContract.AUTHORITY, ops);
} catch (Exception e) {
Log.e("ERROR: ", e.getMessage());
}
}
I don't think so you can, or it should be something complicated.
Before fetch any info i need validate if Name exists at least.
Here is code I used:
public class ContactListControl {
/** be sure that final fields equal to input_display_name */
private final static String NAME = "name";
private final static String GMAIL = "gmail";
private final static String PHONE = "phone";
private ContactListControl(){}
public static void onGetContactInfo(Intent data, Context context, List<InputView> inputViewList) {
Map<String, String> contactDataMap = new HashMap<String, String>();
ContentResolver cr = context.getContentResolver();
Uri contactData = data.getData();
//Cursor cursor = managedQuery(contactData, null, null, null, null);
Cursor cursor = cr.query(contactData, null, null, null, null);
cursor.moveToFirst();
String name = cursor.getString(cursor.getColumnIndexOrThrow(ContactsContract.Contacts.DISPLAY_NAME));
String id = cursor.getString(cursor.getColumnIndexOrThrow(ContactsContract.Contacts._ID));
contactDataMap.put(NAME, (name != null)?name:"");
if (Integer.parseInt(cursor.getString(
cursor.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()) {
String number = pCur.getString(pCur.getColumnIndexOrThrow(ContactsContract.CommonDataKinds.Phone.NUMBER));
contactDataMap.put(PHONE, (number != null)?number:"");
break; // ? we want only 1 value
}
pCur.close();
}
Cursor emailCur = cr.query(
ContactsContract.CommonDataKinds.Email.CONTENT_URI,
null,
ContactsContract.CommonDataKinds.Email.CONTACT_ID + " = ?",
new String[]{id}, null);
while (emailCur.moveToNext()) {
// This would allow you get several email addresses
// if the email addresses were stored in an array
String email = emailCur.getString(
emailCur.getColumnIndex(ContactsContract.CommonDataKinds.Email.DATA));
contactDataMap.put(GMAIL, (email != null)?email:"");
break;// ? we want only 1 value
}
emailCur.close();
cursor.close();
Log.d("activity result","onActivityResult - got contact: "+contactDataMap.get(NAME) + "; " + contactDataMap.get(GMAIL) + "; " + contactDataMap.get(PHONE));
InputView outputViewName = getInputputViewByDisplayName(NAME, inputViewList);
String tempName = onGetContactInfoAppend(outputViewName, contactDataMap.get(NAME));
outputViewName.setValue(tempName);
InputView outputViewGmail = getInputputViewByDisplayName(GMAIL, inputViewList);
String tempGmail = onGetContactInfoAppend(outputViewGmail, contactDataMap.get(GMAIL));
outputViewGmail.setValue(tempGmail);
InputView outputViewPhone = getInputputViewByDisplayName(PHONE, inputViewList);
String tempPhone = onGetContactInfoAppend(outputViewPhone, contactDataMap.get(PHONE));
outputViewPhone.setValue(tempPhone);
}
private static String onGetContactInfoAppend(InputView outputView, String contactData) {
if(contactData == null){
contactData = "";
}
String temp = outputView.getValue();
//if(!"".equals(contactData)){
if(!"".equals(temp)){
temp = temp + " | ";
}
temp = temp + contactData;
//}
return temp;
}
private static InputView getInputputViewByDisplayName(String displayName, List<InputView> inputViewList) {
for (InputView inputView : inputViewList){
if (inputView.getDisplayName().equals(displayName)){
return inputView;
}
}
return null;
}
}
这篇关于如何只用1 SQLite的命令来获得一个机器人接触的所有信息?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!