Android - 从资源中检索字符串数组

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时间:2020-08-20 04:04:26  来源:igfitidea点击:

Android - retrieve string array from resources

androidstringresources

提问by Yuen Tong

Below is the code that i made to retrieve the string array item:

下面是我用来检索字符串数组项的代码:

String[] menuArray;

@Override
public void onCreate(Bundle savedInstanceState) 
{       
    super.onCreate(savedInstanceState);
    setContentView(R.layout.main);

    ScrollView sv = new ScrollView(this);
    LinearLayout ll = new LinearLayout(this);
    ll.setOrientation(LinearLayout.VERTICAL);
    sv.addView(ll);

 // Create an ArrayAdapter that will contain all list items
    ArrayAdapter<String> adapter;

    menuArray = getResources().getStringArray(R.array.menu); 


    for(int i = 0; i < menuArray.length; i++) 
    {
        Button b = new Button(this);
        b.setText(menuArray[i]);
        ll.addView(b);
    }

    this.setContentView(sv);
 }

This is the strings.xml file:

这是 strings.xml 文件:

 <string-array name="menu">
        <item>1</item>
        <item>2</item>
        <item>3</item>
        </string-array>

However, the R.array.menuhaving this issue to compile: As of ADT 14, resource fields cannot be used as switch cases. Invoke this fix to get more information.

然而,R.array.menu有这个问题需要编译:从 ADT 14 开始,资源字段不能用作 switch case。调用此修复程序以获取更多信息。

采纳答案by c2dm

for(int i = 0;i<menuArray.length; i++) 
{
    Button b = new Button(this);
    b.setText(menuArray[i]);
    ll.addView(b);
}

Delete the below statement

删除下面的语句

 try {
        x = count();
    } catch (IOException e1) {
        // TODO Auto-generated catch block
        e1.printStackTrace();
    }

and try to give layout height and width to all Layouts..

并尝试为所有布局提供布局高度和宽度。

回答by Suragch

How to retrieve a string array from resources:

如何从资源中检索字符串数组:

array.xml

数组.xml

<?xml version="1.0" encoding="utf-8"?>
<resources>

    <string-array name="my_string_array">
        <item>one</item>
        <item>two</item>
        <item>three</item>
    </string-array>

</resources>

Code

代码

String[] stringArray = getResources().getStringArray(R.array.my_string_array);

(The OP already had their question answered but based on the question title, other people may come here looking for this answer.)

(OP 已经回答了他们的问题,但根据问题标题,其他人可能会来这里寻找这个答案。)