1
這是我的主要活動的.java[Android]如何僅對名字和姓氏進行搜索？
package com.example.search;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.Iterator;
import java.util.List;
import android.app.Activity;
import android.content.Context;
import android.os.Bundle;
import android.support.v4.app.FragmentActivity;
import android.text.Editable;
import android.text.TextWatcher;
import android.widget.ArrayAdapter;
import android.widget.EditText;
import android.widget.Filter;
import android.widget.ListView;
public User getUserByName(String firstName, String lastName) {
synchronized (userList) {
Iterator<User> iterator = userList.iterator();
while (iterator.hasNext()) {
User user = (User) iterator.next();
if (user.getFirstName().equals(firstName)
&& user.getLastName().equals(lastName)) {
return user;
}
}
// no user found
return null;
}
}
public class MainActivity extends FragmentActivity {
private EditText firstNameEditText;
private EditText lastNameEditText;
private UserListAdapter<User> adapter;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
firstNameEditText = (EditText) findViewById(R.id.firstNameEditText);
lastNameEditText = (EditText) findViewById(R.id.lastNameEditText);
final ListView listView = (ListView) findViewById(R.id.listView);
ArrayList<User> usersList = new ArrayList<User>();
// add items to list
adapter = new UserListAdapter<User>(this,
R.layout.textview_layout, usersList);
listView.setAdapter(adapter);
firstNameEditText.setOnFocusChangeListener(new View.OnFocusChangeListener() {
@Override
public void onFocusChange(View v, boolean hasFocus) {
if(hasFocus) {
adapter.setFilterMode(UserListAdapter.MODE_FILTER_BY_FIRST_NAME);
}
}
});
lastNameEditText.setOnFocusChangeListener(new View.OnFocusChangeListener() {
@Override
public void onFocusChange(View v, boolean hasFocus) {
if(hasFocus) {
adapter.setFilterMode(UserListAdapter.MODE_FILTER_BY_LAST_NAME);
}
}
});
TextWatcher textWatcher = new TextWatcher() {
@Override
public void onTextChanged(CharSequence s, int start, int before, int count) {
}
@Override
public void beforeTextChanged(CharSequence s, int start, int count,
int after) {
}
@Override
public void afterTextChanged(Editable s) {
adapter.getFilter().filter(s.toString());
}
};
firstNameEditText.addTextChangedListener(textWatcher);
lastNameEditText.addTextChangedListener(textWatcher);
}
public static class UserListAdapter<T> extends ArrayAdapter<T> {
public static final int MODE_FILTER_BY_FIRST_NAME = 0xFEED;
public static final int MODE_FILTER_BY_LAST_NAME = 0xDEAF;
private List<T> objects;
private int filterMode;
public UserListAdapter(Context context, int textViewResourceId,
List<T> objects) {
super(context, textViewResourceId, objects);
init(objects);
}
public UserListAdapter(Context context, int resource,
int textViewResourceId, T[] objects) {
super(context, resource, textViewResourceId, objects);
init(Arrays.asList(objects));
}
private void init(List<T> objects) {
this.objects = objects;
}
public UserListAdapter(Context context, int resource,
int textViewResourceId, List<T> objects) {
super(context, resource, textViewResourceId, objects);
init(objects);
}
public UserListAdapter(Context context, int resource,
int textViewResourceId) {
super(context, resource, textViewResourceId);
}
public UserListAdapter(Context context, int resource) {
super(context, resource);
}
@Override
public Filter getFilter() {
return new Filter() {
@Override
protected void publishResults(CharSequence constraint,
FilterResults results) {
UserListAdapter.this.clear();
ArrayList<T> list = (ArrayList<T>) results.values;
UserListAdapter.this.addAll(list);
if (results.count > 0) {
notifyDataSetChanged();
} else {
notifyDataSetInvalidated();
}
}
@Override
protected synchronized FilterResults performFiltering(
CharSequence constraint) {
FilterResults results = new FilterResults();
ArrayList<T> list = new ArrayList<T>();
Iterator<T> iterator = objects.iterator();
while (iterator.hasNext()) {
T t = (T) iterator.next();
User user = (User) t;
String s = null;
if (filterMode == MODE_FILTER_BY_FIRST_NAME) {
s = user.getFirstName();
} else if (filterMode == MODE_FILTER_BY_LAST_NAME) {
s = user.getLastName();
}
if (s.equals(constraint.toString())) {
list.add(t);
}
}
results.values = list;
results.count = list.size();
return results;
}
};
}
public int getFilterMode() {
return filterMode;
}
public void setFilterMode(int filterMode) {
this.filterMode = filterMode;
}
}
public static class User {
// attributes
private String accountID;
private String companyName;
private String designation;
private String firstName;
private String lastName;
private String profileimageBLOB;
// behaviors
public User() // default constructor
{
// attributes will be initialized to their default values
}
public User(String accID, String comName, String design, String fName,
String lName, String image) // parameterized constructor
{
accountID = accID;
companyName = comName;
designation = design;
firstName = fName;
lastName = lName;
profileimageBLOB = image;
}
// set and get methods
public void setAccountID(String accID) {
accountID = accID;
}
public String getAccountID() {
return accountID;
}
public void setCompanyName(String comName) {
companyName = comName;
}
public String getCompanyName() {
return companyName;
}
public void setDesignation(String design) {
designation = design;
}
public String getDesignation() {
return designation;
}
public void setFirstName(String fName) {
firstName = fName;
}
public String getFirstName() {
return firstName;
}
public void setLastName(String lName) {
lastName = lName;
}
public String getLastName() {
return lastName;
}
public void setProfileImage(String image) {
profileimageBLOB = image;
}
public String getProfileImage() {
return firstName;
}
// toString() method
public String toString() {
return accountID + "\t" + companyName + "\t" + designation + "\t\t"
+ firstName + "\t" + lastName + "\t" + profileimageBLOB;
}
}
我做出相應的調整，以@Gopal饒，這是我上面的代碼，但是我面對了很多的錯誤，我無法理解，我在做正確嗎？
請指教，但仍然沒有工作
什麼叫整個字符串是什麼意思？ –
你的搜索邏輯/方法在哪裏？ – TheLostMind
他可能是指「而不是toString（）方法」。至於解決方案，@GopalRao的答案將適用於完全匹配。 – NitroNbg