Question

EDIT-现在，我了解到我应该对toString方法使用Override以实际获取正确的文本，但是，在这种情况下，我仍然对如何实现它感到困惑。有人有主意吗？

我有一个活动中的微调器，用于显示构建器的名称。这些名称是从SQLite数据库中的表中检索的。按下按钮后，微调器的选定选项随后存储在另一个表中，该表稍后将用于设置TextView的文本。

问题是，当从新表中检索名称时，它不能正确显示名称。

我尝试创建一个新表以从中检索数据，但是得到的结果相同。

将列出的构建器从第一张表加载到微调器中的方法

public void loadListedBuilders() {
    String sql = "SELECT * FROM buildersTable ORDER BY name ASC"; //fetching all values from the database called "builders"

    Cursor cursor = SQLiteBuilderDatabase.rawQuery(sql, null);  //cursor holds all the data in the builders table

    if (cursor.moveToFirst()) {  //if the cursor can move to the first record, it has data and it can be fetched
        do {
            builderList.add(new Builder(
                    cursor.getInt(0), //ID
                    cursor.getString(1) //Name
            ));
        } while (cursor.moveToNext()); //run the loop until the cursor has exhausted all the records

        BuilderSpinnerAdapter adapter = new BuilderSpinnerAdapter(this, R.layout.newproject_builderspinner_layout, builderList, SQLiteBuilderDatabase);
        editProjectBuilder.setAdapter(adapter);

    }
}

将微调器中选定的构建器添加到第二个表的方法

String sql = "INSERT INTO activeProjectsTable(name, builderName, projectRate)" +
            "VALUES (?, ?, ?)";
SQLiteBuilderDatabase.execSQL(sql, new String[]{projectName, projectBuilder, projectRate});

从第二个表中检索数据的方法

private void loadDailyData() {
    String sql = "SELECT * FROM activeProjectsTable WHERE ProjectId = (SELECT MAX(ProjectId) FROM activeProjectsTable)"; //fetching all values from the database called "builders", and ordering the name column in alphabetical order

    Cursor cursor = AusWideDatabase.rawQuery(sql,null);  //cursor holds all the data in the builders table


    if (cursor.moveToFirst()) {  //if the cursor can move to the first record, it has data and it can be fetched
        do {
            projectList.add(new Project(
                    cursor.getInt(0), //ProjectID
                    cursor.getString(1),  //Project Name
                    cursor.getString(2),  //Builder Name
                    cursor.getDouble(3),  //Project Rate
                    cursor.getDouble(4)   //Bricks Quota
            ));
        }
        while (cursor.moveToNext()); //run the loop until the cursor has exhausted all the records

        DailyDataAdapter adapter = new DailyDataAdapter(this, R.layout.layout_daily_data, projectList);
        listView.setAdapter(adapter);

    }
}

适配器代码，使用第二个表中的数据设置TextViews的值

public View getView(int position, @Nullable View convertView, @NonNull ViewGroup parent) {
    LayoutInflater inflater = LayoutInflater.from(Ctx);

    View view = inflater.inflate(layoutRes, null);

    TextView dataProjectName = view.findViewById((R.id.textView_dailyData_ProjectName));
    TextView dataBuilderName = view.findViewById(R.id.textView_dailyData_BuilderName);
    TextView dataBuilderRate = view.findViewById(R.id.textView_dailyData_ProjectRate);

    Project project = projectList.get(position);

    dataProjectName.setText(project.getName());
    dataBuilderName.setText(project.getBuilderName());
    dataBuilderRate.setText(String.valueOf(project.getProjectRate()));

    return view;
}

表1

   String sql = "CREATE TABLE IF NOT EXISTS buildersTable (\n" +
            "   BuilderId INTEGER CONSTRAINT builders_pk PRIMARY KEY AUTOINCREMENT, \n" +
            "   name varchar(200) \n" +
            ");";

表2

   String sql = "CREATE TABLE IF NOT EXISTS activeProjectsTable (\n" +
            "    ProjectId INTEGER CONSTRAINT activeProjectsTable_pk PRIMARY KEY AUTOINCREMENT,\n" +
            "    name varchar(200),\n" +
            "    builderName varchar(200),\n" +
            "    projectRate double, \n" +
            "    averageBricks double \n" +
            ");";

它应该做的是显示要提取的项目的字符串值，例如生成器1。但是，目前，它显示为某种值，例如com.auswide.auswideapp.Builder@9444717代替。不知道问题出在哪里或为什么会发生。

无法从SQLite数据库正确传递字符串

0 个答案: