Question

我正在开发一个SQLite程序，并显示一条错误消息

    int i = 0;
    int j = 0;

    for (i = 0; i < lungDR; i++) { valDreapta[i] = input[i + mijloc + 1]; }
    for (i = 0; i < lunST; i++) { valStanga[i] = input[i + stanga]; }      // fix

    i = 0;
    j = 0;

    for (int k = stanga; k <= dreapta; k++)                                // fix
    {
        if (valStanga[i] <= valDreapta[j])

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

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.

Answer 1

您可以更改代码

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”，并且可以通过不对名称进行硬编码来解决此问题，因为您可以在变量COLUMN_PRODUCTNAME中使用它。
同样，您的代码也会丢失行，因为仅在需要moveToFirst()时您错误地使用了moveToNext()，isAfterLast()和moveToNext()：

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();
}

除非您以后想要将其更改为有意义的内容，否则不需要sql语句中的WHERE 1。

从SQLite中的游标访问数据时出错

2 个答案: