Android SQL查询和强制关闭挑战
在这里,我试图从数据库中获取用户名和密码,如果用户已经有一个帐户,或者通过输入他的数据注册他 这是第一个通过单击第一个按钮登录数据库而强制关闭的活动Android SQL查询和强制关闭挑战,android,nullpointerexception,android-sqlite,Android,Nullpointerexception,Android Sqlite,在这里,我试图从数据库中获取用户名和密码,如果用户已经有一个帐户,或者通过输入他的数据注册他 这是第一个通过单击第一个按钮登录数据库而强制关闭的活动 public class SelesMeter2Activity extends Activity implements OnClickListener { EditText ed1; EditText ed2; Button b1; Button b2; SQLiteDatabase sql; Cu
public class SelesMeter2Activity extends Activity implements OnClickListener {
EditText ed1;
EditText ed2;
Button b1;
Button b2;
SQLiteDatabase sql;
Cursor c;
Intent in;
/** Called when the activity is first created. */
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.main);
ed1 = (EditText) findViewById(R.id.ed1);
ed2 = (EditText) findViewById(R.id.ed2);
b1 = (Button) findViewById(R.id.bt1);
b2 = (Button) findViewById(R.id.bt2);
b1.setOnClickListener(this);
b2.setOnClickListener(this);
sql = openOrCreateDatabase("db", 0, null);
sql.execSQL("CREATE TABLE if not exists "
+ "Employee2 (password integer NOT NULL PRIMARY KEY,name text NOT NULL)");
}
@Override
public void onClick(View arg0) {
// log in
if (arg0.getId() == R.id.bt1) {
int p = 0;
String name = ed1.getText().toString();
String sp = ed2.getText().toString();
try {
// Attempt to parse the number as an integer
p = Integer.parseInt(sp);
} catch (NumberFormatException nfe) {
// parseInt failed, so tell the user it's not a number
Toast.makeText(this,
"Sorry, " + sp + " is not a number. Please try again.",
Toast.LENGTH_LONG).show();
}
if (c.getCount() != 0) {
c = sql.rawQuery("select * from Employee", null);
while (c.moveToNext()) {
if (name.equals("c.getString(1)") && p == c.getInt(0)) {
in = new Intent(this, secondview.class);
startActivity(in);
break;
}
}
}
else {
Toast.makeText(this,
"please sign up first or enter " + "correct data", 2000)
.show();
}
} else if (arg0.getId() == R.id.bt2) {
// sign up
Intent in2 = new Intent(this, signup.class);
startActivity(in2);
}
}
}
输入未按预期工作的新用户的第二个类,
祝酒词不起作用:
public class signup extends Activity implements OnClickListener {
EditText e1;
EditText e2;
SQLiteDatabase sql;
Button b;
@Override
protected void onCreate(Bundle savedInstanceState) {
// TODO Auto-generated method stub
super.onCreate(savedInstanceState);
setContentView(R.layout.singupx);
Intent in = getIntent();
e1 = (EditText) findViewById(R.id.ed1s);
e2 = (EditText) findViewById(R.id.ed2s);
b = (Button) findViewById(R.id.bt1s);
b.setOnClickListener(this);
}
@Override
public void onClick(View v) {
String n = e1.getText().toString();
String sp = e2.getText().toString();
try {
// Attempt to parse the number as an integer
int p = Integer.parseInt(sp);
// This insertion will *only* execute if the parseInt was successful
// sql.execSQL("insert into Employee2(password,name)values('"+n+"',"+p+")");
ContentValues values = new ContentValues();
values.put("password", p);
values.put("name", n);
sql.insert("Employee2", null, values);
Toast.makeText(this, "Data inserted", Toast.LENGTH_LONG).show();
Intent in2 = new Intent(this, secondview.class);
startActivity(in2);
} catch (NumberFormatException nfe) {
// parseInt failed, so tell the user it's not a number
Toast.makeText(this,
"Sorry, " + sp + " is not a number. Please try again.",
Toast.LENGTH_LONG).show();
}
}
}
这就是为什么会出现错误
// you are calling the `c.getCount();` before you are assigning
// It will throw null pointer exception
if (c.getCount() != 0) {
c = sql.rawQuery("select * from Employee", null);
while (c.moveToNext()) {
if (name.equals(c.getString(1)) && p == c.getInt(0)) {
in = new Intent(this, secondview.class);
startActivity(in);
break;
}
}
}
像这样改变逻辑
c = sql.rawQuery("select * from Employee", null);
c.moveToFirst();
if(!c.isAfterLast()) {
do {
if (name.equals(c.getString(1)) && p == c.getInt(0)) {
in = new Intent(this, secondview.class);
startActivity(in);
break;
}
} while (c.moveToNext());
}
和name.equals(“c.getString(1)”)
应该是name.equals(c.getString(1))
编辑
插入法示例
ContentValues values = new ContentValues();
values.put("password", n);
values.put("name", p);
database.insert("Employee2", null, values);
在
SelesMeter2Activity
中,您将在以下行出现NullPointerException
:
if (c.getCount() != 0) {
因为您没有在该行之前初始化光标
。将查询移动到上一行之前:
c = sql.rawQuery("select * from Employee", null);
if (c.getCount() != 0) {
// ...
您应该发布从logcat获得的异常
另外,关于您的
注册
活动,请不要实例化从中访问字段的第一个活动。在第二个活动中再次打开数据库并插入值。您可以发布日志跟踪吗?您需要具有相同的sql=openOrCreateDatabase(“db”,0,null)第二个活动中的代码>语句。是的,您是对的:)错误已消失,但注册活动未将数据输入数据库!祝酒辞没有什么意义work@SarahAabed你指的是哪个祝酒词?注册活动是否运行?是的,它已运行,但sql查询后的toast不起作用,因此我无法输入数据@LuKsprong@SarahAabed您是否修改了注册以不实例化SelesMeter2Activity
活动,而是直接打开数据库?您是否已检查logcat以了解有关插入的任何警告。另外,请考虑使用一个代替<代码> Excel SQL >代码来实际查看插入是否正确。@ SalaHabeBe您解决了这个问题吗?您没有看到任何祝酒
,因为您没有在注册
活动中为按钮
设置OnClickListener
(在布局中找到按钮(就像您为编辑文本
所做的那样)并分配侦听器按钮。setOnClickListener(此);
)。另外,我建议您使用SQLiteOpenHelper
来管理数据库。是的,没错:)注册活动如何?它不接受数据并将其输入数据中base@SarahAabed同一个包中的这两个类?@SarahAabed ok,正如Luksprog所说,更改实现。我确信,sql
仍然没有为不同的上下文保留数据库的引用,因为您正在实例化新对象,而不是启动它onCreate()
这是每个活动的入口点,您在这里进行数据库初始化。我已经更改了逻辑,我将尝试使用插入方法,但您的意思是我必须将其放在onCreate中吗?再次感谢:)