Question

我正在尝试使用SQLITE实现一个简单的数据库示例。这是我的代码

package com.example.dbex;

import java.sql.SQLOutput;
import java.sql.SQLPermission;

import android.content.ContentValues;
import android.content.Context;
import android.database.Cursor;
import android.database.sqlite.SQLiteDatabase;
import android.database.sqlite.SQLiteDatabase.CursorFactory;
import android.database.sqlite.SQLiteOpenHelper;

public class DBHandler extends SQLiteOpenHelper {

    public static final int DATABASE_VERSION=1;
    public static final String DATABASE_NAME="contactsmanager";
    public static final String TABLE_CONTACTS="contacts";
    public static final String KEY_ID="id";
    public static final String KEY_NAME="name";
    public static final String KEY_NUMBER="number";
    public DBHandler(Context context) {
        super(context,DATABASE_NAME, null, DATABASE_VERSION);
        // TODO Auto-generated constructor stub
    }

    @Override
    public void onCreate(SQLiteDatabase db) {
        // TODO Auto-generated method stub
        String CREATE_TABLE_CONTACTS="CREATE TABLE "+TABLE_CONTACTS+"("+KEY_ID+" INTEGER PRIMARY KEY,"+KEY_NAME+" TEXT,"+KEY_NUMBER+" TEXT"+")";
        db.execSQL(CREATE_TABLE_CONTACTS);
    }

    @Override
    public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion) {
        // TODO Auto-generated method stub
        db.execSQL("DROP TABLE IF EXISTS "+TABLE_CONTACTS);
        onCreate(db);
    }

    public void addContact(Contacts contacts)
    {
        SQLiteDatabase db=this.getWritableDatabase();
        ContentValues values=new ContentValues();
        values.put(KEY_NAME, contacts.getname());
        values.put(KEY_NUMBER, contacts.getnumber());
        db.insert(TABLE_CONTACTS,null,values);
        db.close();
    }

    public Contacts getContact(int id)
    {
        SQLiteDatabase db=this.getReadableDatabase();
        Cursor c=db.query(TABLE_CONTACTS, new String[]{KEY_ID,KEY_NAME,KEY_NUMBER},KEY_ID+" =?",new String[]{String.valueOf(id)}, null, null, null, null);
        if(c!=null)
        {
            c.moveToFirst();
        }
        Contacts contacts=new Contacts(Integer.parseInt(c.getString(0)),c.getString(1),c.getString(2));
        return contacts;
    }

    public void deleteContact(Contacts contacts)
    {
        SQLiteDatabase db=this.getWritableDatabase();
        db.delete(TABLE_CONTACTS, KEY_ID+" =?", new String[]{String.valueOf(contacts.getID())});
        db.close();
    }


}

`

我在行

上收到光标超出范围异常

Contacts contacts=new Contacts(Integer.parseInt(c.getString(0)),c.getString(1),c.getString(2));

我哪里错了？

Answer 1

你的光标可能是空的。

它可以不为null但仍然没有结果（返回0行）。只有通过将逻辑更改为：

，才能确保访问它

Contact contacts = null;
if (c!=null && c.moveToFirst()) {
    contacts = new Contacts(Integer.parseInt(c.getString(0)),c.getString(1),c.getString(2));
}
return contacts;

Answer 2

当光标内部没有那么多行时，你试图访问光标位置时，通常会出现光标超出范围异常.Print日志有关游标的行数，如果不是0，则只尝试访问更多行。 / p>

Answer 3

尝试以下

    Contacts getContact(int id) {
    SQLiteDatabase db = this.getReadableDatabase();

    Cursor cursor = db.query(TABLE_CONTACTS, new String[] { KEY_ID,
            KEY_NAME, KEY_NUMBER }, KEY_ID + "=?",
            new String[] { String.valueOf(id) }, null, null, null, null);
    Contacts contacts; 
    if (cursor != null)
    {
    cursor.moveToFirst();
    contacts = new Contacts(Integer.parseInt(cursor.getString(0)),
            cursor.getString(1), cursor.getString(2));
    }
    return contacts

}

android SQLITE中的游标索引异常

3 个答案: