基本上，我想在登录后有一个按钮来启动新活动。
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));
}
但这，我的startActivity用红色下划线
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“未定义构造函数Intent（new View.OnClickListener（）{} {}，Class）”和“删除参数以匹配Intent（）”选项
我在清单中包括了
<Activity></Activity>。以及上面显示的代码，进口被切断了 最佳答案
是的，这是因为，您正在尝试使用“ this”来引用您的按钮而不是您的活动。您必须像这样替换它，
代替startActivity(new Intent(this, Folder_Details.class));
做这个，
startActivity(new Intent(MainMenu.this, Folder_Details.class));
关于android - StartActivity()红色突出显示，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/8266355/