在 SQLite 中从游标访问数据时出错

Question

我正在处理 SQLite 程序并收到一条错误消息

2019-07-14 21:07:37.465 13538-13538/? E/CursorWindow: Failed to read row 0, column -1 from a CursorWindow which has 1 rows, 2 columns.
2019-07-14 21:07:37.466 13538-13538/? D/AndroidRuntime: Shutting down VM
2019-07-14 21:07:37.467 13538-13538/? E/AndroidRuntime: FATAL EXCEPTION: main
    Process: com.example.xyz.sqlitelist, PID: 13538
    java.lang.RuntimeException: Unable to start activity ComponentInfo{com.example.xyz.sqlitelist/com.example.xyz.sqlitelist.MainActivity}: java.lang.IllegalStateException:

 Couldn't read row 0, col -1 from CursorWindow.  Make sure the Cursor is initialized correctly before accessing data from it.
.
.

     Caused by: java.lang.IllegalStateException: Couldn't read row 0, col -1 from CursorWindow.  Make sure the Cursor is initialized correctly before accessing data from it.

W/ActivityManager:强制完成activity com.example.xyz.sqlitelist/.MainActivity

String databaseToString(){
    StringBuilder dbString= new StringBuilder();
    SQLiteDatabase db = getWritableDatabase();
    String query=" SELECT * FROM " + TABLE_PRODUCTS + " WHERE 1";
    Cursor c=db.rawQuery(query,null);
    c.moveToFirst();
    while (!c.isAfterLast()){
        if(c.getString(c.getColumnIndex("productname")) != null) {
            c.moveToNext();
            dbString.append(c.getString(c.getColumnIndex("productname")));
            dbString.append("\n");
    }
    c.close();
    db.close();
    return dbString.toString();
}

Answer 1

您可以更改密码

String sql = " SELECT * FROM " + TABLE_PRODUCTS + " WHERE 1";
        Cursor c = getWritableDatabase().rawQuery(sql, null);        
        while (c.moveToNext()) {
           if(c.getString(c.getColumnIndex("productname")) != null) 
            {

             dbString.append(c.getString(c.getColumnIndex("productname")));

             dbString.append("\n");
            }
        }
c.close();
db.close();
return dbString.toString();

Answer 2

您的列名称是“_productname”而不是 "productname"，这个问题可以通过不硬编码这个名称来解决，因为您在变量 COLUMN_PRODUCTNAME.
中有它此外，您的代码将丢失行，因为您在只需要 moveToNext():

时错误地使用了 moveToFirst()、moveToNext() 和 isAfterLast()

String databaseToString() {
    StringBuilder dbString = new StringBuilder();
    SQLiteDatabase db = getWritableDatabase();
    String query = "SELECT * FROM " + TABLE_PRODUCTS + " WHERE 1";
    Cursor c = db.rawQuery(query, null);
    int index = c.getColumnIndex(COLUMN_PRODUCTNAME);
    while (c.moveToNext()) {
        dbString.append(c.getString(index)).append("\n");
    }
    c.close();
    db.close();
    return dbString.toString();
}

此外，sql 语句中的 WHERE 1 也不是必需的，除非您想稍后将其更改为有意义的内容。

在 SQLite 中从游标访问数据时出错

Error in accessing data from cursor in SQLite

android

debugging

cursor