Android Spinner - onItemSelected/setOnItemSelectedListener未触发

Question

这让我疯了,因为这是我以前做过的事情,但是无法弄清楚为什么它现在不起作用......

我有一个菜单按钮,以通常的方式通过menu.xml文件和onOptionsItemSelected带有开关的方法实现,创建并显示微调器.

我添加了setOnItemSelectedListener,但它似乎永远不会触发.在微调的出现,我选择一个选项或背出来,既不onItemSelected或onNothingSelected调用.

这是菜单按钮处理开关语句的"case"和"return true"之间的所有代码.(topThis是一个引用活动上下文的变量 - 适用于应用程序中的所有其他toast)

String[] widgetModes = {"Mode 1", "Mode2"};
ArrayAdapter<String> widgetModeAdapter = new ArrayAdapter<String> (this, android.R.layout.simple_spinner_item, widgetModes);
widgetModeAdapter.setDropDownViewResource(android.R.layout.simple_spinner_dropdown_item);

Spinner widgetModeSpinner = new Spinner(this);
widgetModeSpinner.setAdapter(widgetModeAdapter);
widgetModeSpinner.setPrompt("Choose Widget Mode");

widgetModeSpinner.setOnItemSelectedListener(new OnItemSelectedListener() 
{
    @Override
    public void onItemSelected(AdapterView<?> parentView, View selectedItemView, int position, long id) 
    {
        Toast.makeText(topThis, "derp", Toast.LENGTH_LONG).show();
    }

    @Override
    public void onNothingSelected(AdapterView<?> parentView) 
    {
        Toast.makeText(topThis, "herf", Toast.LENGTH_LONG).show();
    }
});

widgetModeSpinner.performClick();

有任何想法吗？我模糊地怀疑我正在以Spinner编程方式创建的事实是问题......

Answer 1

当我实现一个微调器时,我遇到了类似的问题,我通过获取父视图并设置适配器来解决它

spinner1 =(Spinner)findViewById(R.id.spinner1);
spinner1.setAdapter(BindSpinner("ProgramMaster",cols,null,true,""));
spinner1.setOnItemSelectedListener(new OnItemSelectedListener() 
{
protected Adapter initializedAdapter=null;
    @Override
    public void onItemSelected(AdapterView<?> parentView, View selectedItemView, int position, long id) 
    {
        if(initializedAdapter !=parentView.getAdapter() ) {
            initializedAdapter = parentView.getAdapter();
           return;
        }

        String selected = parentView.getItemAtPosition(position).toString();

        if(abc.equals("Select") && !selected.equals("Select"))
        {
            do something
        }

        else 
        {
            Do something
        }
        textQualification=selected;
        SearchUpdated("Qualification");
    }

    @Override
    public void onNothingSelected(AdapterView<?> parentView) {
        // your code here
    }
});

Answer 2

请记住,如果不添加一些自定义代码来处理微调器选择,则无法重新选择相同的微调器项,它始终将第一个项设置为选中.

对于Toast没有显示,我建议在侦听器界面中创建Toast时始终使用"MyActivity.this"作为上下文:

spinner.setOnItemSelectedListener(new AdapterView.OnItemSelectedListener() {
        /**
         * Called when a new item is selected (in the Spinner)
         */
         public void onItemSelected(AdapterView<?> parent, View view, 
                    int pos, long id) {
                // An spinnerItem was selected. You can retrieve the selected item using
                // parent.getItemAtPosition(pos)

                Toast.makeText(MyActivity.this, "Hello Toast",Toast.LENGTH_SHORT).show();

            }

            public void onNothingSelected(AdapterView<?> parent) {
                // Do nothing, just another required interface callback
            }

    }); // (optional)

最后的.show()有时会很容易忘记;)

Answer 3

实际上,如果您的微调器可见性设置为已消失,那么当您调用performclick()方法时它将触发它的单击,但它不会触发其setOnItemSelectedListener, 因此您需要更改可见性然后它将起作用