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android - StartActivity() 红色突出显示

转载 作者:行者123 更新时间:2023-11-29 14:41:32 24 4
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基本上我想要一个按钮来在登录后开始一个新的 Activity 。我发现我无法像之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 被加了红色下划线

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));

}

});
}

}

它显示了“The constructor Intent(new View.OnClickListener(){}, Class) is undefined”和“Remove arguments to match Intent()”选项

我已经包含了 <Activity></Activity>在 list 中。上面的代码显示,导入被切断

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 Activity 。你必须像这样替换它,

代替 startActivity(new Intent(this, Folder_Details.class));

这样做,

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/

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