Android数据类型不匹配(代码20)

时间:2017-04-28 15:37:48

标签: android database sqlite

我遇到有关数据库服务的问题。 PUT方法正在运行,但POST无效。我弄错了#39;数据类型不匹配(代码20)'

public class DatabaseService
{
    public static final DatabaseService instance = new DatabaseService();
    public static DatabaseService getInstance() { return instance; };

    public DatabaseService() {

    }

    public Cursor GET() {
        try {
            Cursor cursor = DatabaseConnectionService
                    .getInstance()
                    .getDatabaseInstance()
                    .rawQuery(VehicleTableModel.getInstance().GetSelectSQL(""), null);
            return cursor;
        }
        catch (Exception ex)
        {
            Logs.error(ex);
            return null;
        }
    }

    public boolean POST(ContentValues parametersAndValues) {
        try {
            DatabaseConnectionService
                    .getInstance()
                    .getDatabaseInstance()
                    .insertOrThrow(VehicleTableModel.getInstance().TableName, null, parametersAndValues);
        }
        catch (Exception ex)
        {
            Logs.error(ex);
            return false;
        }
        finally {
            return true;
        }
 public boolean PUT(ContentValues parametersAndValues ,String id) {
    try {
        VehicleTableModel vehicleTableModel = VehicleTableModel.getInstance();
        DatabaseConnectionService
                .getInstance()
                .getDatabaseInstance()
                .update(vehicleTableModel.TableName, parametersAndValues, vehicleTableModel.Columns.Id  + " = ?", new String[] { String.valueOf(id) });
    }
    catch (Exception ex)
    {
        Logs.error(ex);
        return false;
    }
    finally {
        return true;
    }

这是Table Helper模型类;

public class BaseTableModel {
    public String DatabaseName = "";
    public int VersionNumber = 0;
    public String TableName = "";
}

关于表;

public class VehicleTableModel extends BaseTableModel {

private static final VehicleTableModel instance = new VehicleTableModel();
public static VehicleTableModel getInstance() {
    return instance;
}
public VehicleTableModel() {
    this.DatabaseName = "vehicle";
    this.VersionNumber = 1;
    this.TableName = "vehicles";
}

我该怎么办?我研究过,但我找不到任何东西。 感谢。

0 个答案:

没有答案
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