【发布时间】：2011-11-25 08:24:40
【问题描述】：
基本上我想要一个按钮在登录后开始一个新的活动。 我发现我无法像以前在登录页面中那样调用 StartActivity()。请指导
这是我成功使用 StartActivity(this,sth.class) 的登录页面
public class Login extends Activity
{
/** Called when the activity is first created. */
Button login;
String name="",pass="";
EditText username,password;
TextView tv;
byte[] data;
HttpPost httppost;
StringBuffer buffer;
HttpResponse response;
HttpClient httpclient;
InputStream inputStream;
SharedPreferences app_preferences ;
List<NameValuePair> nameValuePairs;
CheckBox check;
public void onCreate(Bundle savedInstanceState)
{
super.onCreate(savedInstanceState);
setContentView(R.layout.main);
app_preferences = PreferenceManager.getDefaultSharedPreferences(this);
username = (EditText) findViewById(R.id.username);
password = (EditText) findViewById(R.id.password);
login = (Button) findViewById(R.id.login);
check = (CheckBox) findViewById(R.id.check);
String Str_user = app_preferences.getString("username","0" );
String Str_pass = app_preferences.getString("password", "0");
String Str_check = app_preferences.getString("checked", "no");
if(Str_check.equals("yes"))
{
username.setText(Str_user);
password.setText(Str_pass);
check.setChecked(true);
}
login.setOnClickListener(new View.OnClickListener()
{
public void onClick(View v)
{
name = username.getText().toString();
pass = password.getText().toString();
String Str_check2 = app_preferences.getString("checked", "no");
if(Str_check2.equals("yes"))
{
SharedPreferences.Editor editor = app_preferences.edit();
editor.putString("username", name);
editor.putString("password", pass);
editor.commit();
}
if(name.equals("") || pass.equals(""))
{
Toast.makeText(Login.this, "Blank Field..Please Enter", Toast.LENGTH_LONG).show();
}
else
{
try {
httpclient = new DefaultHttpClient();
httppost = new HttpPost("http://fyptest.comyr.com/main.php");
// Add your data
nameValuePairs = new ArrayList<NameValuePair>(2);
nameValuePairs.add(new BasicNameValuePair("UserEmail", name.trim()));
nameValuePairs.add(new BasicNameValuePair("Password", pass.trim()));
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
// Execute HTTP Post Request
response = httpclient.execute(httppost);
inputStream = response.getEntity().getContent();
data = new byte[256];
buffer = new StringBuffer();
int len = 0;
while (-1 != (len = inputStream.read(data)) )
{
buffer.append(new String(data, 0, len));
}
inputStream.close();
}
catch (Exception e)
{
Toast.makeText(Login.this, "error"+e.toString(), Toast.LENGTH_LONG).show();
}
if(buffer.charAt(0)=='Y')
{
Toast.makeText(Login.this, "login successfull", Toast.LENGTH_LONG).show();
Move_to_next();
}
else
{
Toast.makeText(Login.this, "Invalid Username or password", Toast.LENGTH_LONG).show();
}
}
}
});
check.setOnClickListener(new View.OnClickListener()
{
public void onClick(View v)
{
// Perform action on clicks, depending on whether it's now checked
SharedPreferences.Editor editor = app_preferences.edit();
if (((CheckBox) v).isChecked())
{
editor.putString("checked", "yes");
editor.commit();
}
else
{
editor.putString("checked", "no");
editor.commit();
}
}
});
}
public void Move_to_next()
{
//may perform checking based on ID
startActivity(new Intent(this, MainMenu.class));
}
但是这个，我的 startActivity 是用红色下划线的
public class MainMenu extends Activity{
@Override
protected void onCreate(Bundle savedInstanceState) {
// TODO Auto-generated method stub
super.onCreate(savedInstanceState);
setContentView(R.layout.main_menu);
Button new_folder = (Button)findViewById(R.id.new_folder);
new_folder.setOnClickListener(new View.OnClickListener(){
@Override
public void onClick(View v) {
// show another class
startActivity(new Intent(this,Folder_Details.class));
}
});
}
}
它显示“构造函数 Intent(new View.OnClickListener(){}, Class) 未定义”和“删除参数以匹配 Intent()”选项
我已将<Activity></Activity> 包含在清单中。而上面显示的代码，导入被切断了
【问题讨论】：
标签： android