索引0在SQLITE DB中请求大小为0的错误

Question

数据库中有数据(准确地说是2行),每个数据都有信息.

以下是DBADAPTER中的代码(重要的,不是全部):

// Field Names:
    public static final String KEY_ROWID = "_id";
    public static final String KEY_DEVICE = "device";
    public static final String KEY_TYPE = "type";
    public static final String KEY_DEVID = "devid";

    public static final String[] ALL_KEYS = new String[] {KEY_ROWID, KEY_DEVICE, KEY_TYPE, KEY_DEVID};
    public static final String[] DEVID_KEY = new String[] {KEY_DEVID};

    // Column Numbers for each Field Name:
    public static final int COL_ROWID = 0;
    public static final int COL_DEVICE = 1;
    public static final int COL_TYPE = 2;
    public static final int COL_DEVID = 3;

    // DataBase info:
    public static final String DATABASE_NAME = "dbDevices";
    public static final String DATABASE_TABLE = "mainDevices";
    public static final int DATABASE_VERSION = 10; // The version number must be incremented each time a change to DB structure occurs.

    //SQL statement to create database
    private static final String DATABASE_CREATE_SQL = 
            "CREATE TABLE " + DATABASE_TABLE 
            + " (" + KEY_ROWID + " INTEGER PRIMARY KEY AUTOINCREMENT, "
            + KEY_DEVICE + " TEXT NOT NULL, "
            + KEY_TYPE + " TEXT, "
            + KEY_DEVID + " TEXT NOT NULL "
            + ");";

    private final Context context;
    private DatabaseHelper myDBHelper;
    private SQLiteDatabase db;

    // Get a specific row (by devid)
        public String getDevName(String devid) {
            String name;
            String where = KEY_DEVID + " like '%" + devid + "%'";
            Cursor c =  db.query(true, DATABASE_TABLE, ALL_KEYS, 
                            where, null, null, null, null, null);
                c.moveToFirst();
                name = c.getString(COL_DEVICE);
            return name;
        }

这是调用它的代码:

String incomingMessage = " ";
incomingMessage = in.readLine() + System.getProperty("line.separator");
devName = devDB.getDevName(incomingMessage);

我知道这是获取信息,因为它记录:
03-16 15:39:47.072: V/String(18073): 5122

每当我运行这个时,我都会收到错误:

03-16 15:39:47.122: W/System.err(18073): android.database.CursorIndexOutOfBoundsException: Index 0 requested, with a size of 0
03-16 15:39:47.122: W/System.err(18073):    at android.database.AbstractCursor.checkPosition(AbstractCursor.java:426)
03-16 15:39:47.122: W/System.err(18073):    at android.database.AbstractWindowedCursor.checkPosition(AbstractWindowedCursor.java:136)
03-16 15:39:47.132: W/System.err(18073):    at android.database.AbstractWindowedCursor.getString(AbstractWindowedCursor.java:50)
03-16 15:39:47.132: W/System.err(18073):    at com.ti.cc3x.android.DBAdapter.getDevName(DBAdapter.java:121)
03-16 15:39:47.132: W/System.err(18073):    at com.ti.cc3x.android.broadcastListener$1.run(broadcastListener.java:112)
03-16 15:39:47.132: W/System.err(18073):    at java.lang.Thread.run(Thread.java:841)

我试过检查非空游标.所有这一切都告诉我什么.我知道有数据我要求它看起来我只是不知道为什么它不会显示它.谁能告诉我哪里出错了？谢谢!

Answer 1

你在用SQLiteOpenHelper吗？如果是这样,您需要在尝试读取或写入db数据库之前将数据库变量设置为数据库,如下所示:

db = this.getWritableDatabase();

...你的新getDevName(..)方法应该是这样的:

public String getDevName(String devid) {
    db = this.getWritableDatabase();

    String name;
    String where = KEY_DEVID + " like '%" + devid + "%'";
    Cursor c =  db.query(true, DATABASE_TABLE, ALL_KEYS, 
                    where, null, null, null, null, null);
    c.moveToFirst();
    name = c.getString(COL_DEVICE);

    db.close(); // close your db connection when you're done with it

    return name;
}

从Android Developer文档:

public SQLiteDatabase getWritableDatabase() - 第一次调用它时,将打开数据库并调用onCreate(SQLiteDatabase),onUpgrade(SQLiteDatabase,int,int)和/或onOpen(SQLiteDatabase).成功打开后,数据库将被缓存,因此您可以在每次需要写入数据库时​​调用此方法.(确保在不再需要数据库时调用close().)错误权限或完整磁盘等错误可能导致此方法失败,但如果问题得到解决,将来的尝试可能会成功.

Answer 2

扩展@bwegs关于呼叫的答案getWritableDatabase().

尝试使用selectionArgs它,因为它更干净.检查返回值moveToFirst(),并确保关闭光标以避免内存泄漏.

 public String getDevName(String devid) {
    db = myDBHelper.getWritableDatabase();  //get db if not done so already, you may have already done this in code not shown

    String name = null;
    String where = KEY_DEVID + " like ?";
    String[] selectionArgs =  new String[] {"%"+ devid + "%" };
    Cursor c =  db.query(true, DATABASE_TABLE, ALL_KEYS, 
                    where, selectionArgs , null, null, null, null);

    //Only use cursor if moveToFirst() succeeds
    //this will make it so that you don't get the CursorIndexOutOfBoundsException
    if (c.moveToFirst()){
        name = c.getString(COL_DEVICE);
    }

    c.close(); //close your cursor to avoid memory leak!

    db.close(); //close here if you call getWritableDatabase() in beginning of this function

    //This will return null if no results from query
    return name;
}