如何为ArrayAdapter

时间:2018-06-10 16:28:35

标签: android

我正在尝试获取我的Object的位置但它返回-1,即使我知道它存在。我是否必须定义自定义ArrayAdapter,而不是ArrayAdapter使用对象toString方法?

spinner_formula = findViewById(R.id.formula_spinner);
    List<Formula> formulas = formulaViewModel.getAllFormulas_();
    formulaArrayAdapter = new ArrayAdapter<>(this, android.R.layout.simple_spinner_item, formulas);
    formulaArrayAdapter.setDropDownViewResource(android.R.layout.simple_spinner_dropdown_item);
    spinner_formula.setAdapter(formulaArrayAdapter);
    if (toClass.equals("BottleCurrent")) {
        Log.v("test", "Ok");
        Bottle bottle = bottleViewModel.getBottle(id);
        Log.v("test", bottle.getId());
        Log.v("test", formulaViewModel.getFormula(bottle.getFormulaID()).getName());
        Log.v("test", ""+formulaArrayAdapter.getPosition(formulaViewModel.getFormula(bottle.getFormulaID())));
        ArrayAdapter<Formula> x = (ArrayAdapter)spinner_formula.getAdapter();
        spinner_formula.setSelection(formulaArrayAdapter.getPosition(formulaViewModel.getFormula(bottle.getFormulaID())), true);
    }

0 个答案:

没有答案