【发布时间】:2015-09-21 03:02:02
【问题描述】:
早安,我在 Android Studio 上创建数据库时遇到问题。代码没有问题,但应用程序说“不幸的是停止”。我该如何解决这个问题?
package com.example.maria.database;
import android.app.Activity;
import android.os.Bundle;
import android.view.Menu;
import android.view.MenuItem;
public class MainActivity extends Activity {
DatabaseHelper myDB;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
myDB = new DatabaseHelper(this);
}
@Override
public boolean onCreateOptionsMenu(Menu menu) {
getMenuInflater().inflate(R.menu.menu_main, menu);
return true;
}
@Override
public boolean onOptionsItemSelected(MenuItem item) {
// Handle action bar item clicks here. The action bar will
// automatically handle clicks on the Home/Up button, so long
// as you specify a parent activity in AndroidManifest.xml.
int id = item.getItemId();
//noinspection SimplifiableIfStatement
if (id == R.id.action_settings) {
return true;
}
return super.onOptionsItemSelected(item);
}
}
还有其他活动
package com.example.maria.database;
import android.content.Context;
import android.database.sqlite.SQLiteDatabase;
import android.database.sqlite.SQLiteOpenHelper;
public class DatabaseHelper extends SQLiteOpenHelper{
public static final String DATABASE_NAME = "student.db";
public static final String TABLE_NAME = "student_table";
public static final String COL_1 = "ID";
public static final String COL_2= "NAME";
public static final String COL_3 = "SURNAME";
public static final String COL_4 = "MARKS";
public DatabaseHelper(Context context) {
super(context, DATABASE_NAME, null, 1);
SQLiteDatabase db = this.getWritableDatabase();
}
@Override
public void onCreate(SQLiteDatabase db) {
db.execSQL("create table" + TABLE_NAME + "(ID INTEGER PRIMARY KEY AUTOINCREMENT, NAME TEXT, SURNAME TEXT, MARKS INTEGER)");
}
@Override
public void onUpgrade(SQLiteDatabase db, int oldVersion, intnewVersion) {
db.execSQL("DROP TABLE IF EXISTS "+TABLE_NAME);
onCreate(db);
}
}
【问题讨论】:
-
请附上日志报告,以便我们了解问题。
-
“代码没有问题,但应用程序说“不幸停止””所以代码可能有问题。
-
使用
"create table "代替"create table"。由于table和TABLE_NAME之间没有空格,可能会出现问题