ArrayAdapter.createFromResource问题 [英] ArrayAdapter.createFromResource issue

问题描述

我正在尝试在Spinner上选择选择一个。我看到了有关该主题的所有答案，但仍然有一些问题。制作自定义微调器的常用方法是：

I am trying to make "Select one" on Spinner. I saw all answer regarding this subject but I am still having some issues. Usual way of making custom spinner is:

ArrayAdapter<CharSequence> dataAdapter1 =  ArrayAdapter.createFromResource(this, R.array.entries,
                                        android.R.layout.simple_spinner_item);
        dataAdapter1.setDropDownViewResource(android.R.layout.simple_spinner_dropdown_item);
        spinner1.setAdapter(dataAdapter1);

        spinner1.setAdapter(
                  new NothingSelectedSpinnerAdapter(
                        dataAdapter1,
                        R.layout.contact_spinner_row_nothing_selected,
                                                    this));

在此代码中，我必须在Strings.xml中定义R.array.entries，但我的应用程序是从MySQL填充Spinner，我有一个列表 grad [i] = json.getString（ Grad）; 。如何使用该列表而不是Strings.xml中定义的条目创建此ArrayAdapter.createFromResource？ Tnx

In this code I have to define R.array.entries in Strings.xml, but my app is populating spinner from MySQL and I am having a list grad[i]=json.getString("Grad");. How can I create this ArrayAdapter.createFromResource with that list instead of Entries that are defined in Strings.xml? Tnx

推荐答案

查询数据，将其放入 List 或数组中并使用 Array Adapter

Query the data, put it in a List or Array and use this constructor of Array Adapter

ArrayAdapter<CharSequence> dataAdapter1 =  new ArrayAdapter(this, 
                                    android.R.layout.simple_spinner_item, yourArrayOrList);

此处更多： http://developer.android.com/reference/android /widget/ArrayAdapter.html#ArrayAdapter(android.content.Context，int，java.util.List）

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