如何利用改造和SimpleXML一起下载，并从现场解析XML文件？

问题描述

我刚开始与改造工作。我正在使用的SimpleXML项目。有人可以给我，其中一个从网站如取一个XML的例子 http://www.w3schools.com/xml/simple.xml 中记载其出？

I just started working with Retrofit. I am working on a project that uses SimpleXML. Can somebody provide me an example in which one fetches an XML from a site e.g. http://www.w3schools.com/xml/simple.xml" and reads it out?

推荐答案

您将在项目中创建一个界面，一个新的类：

You will create an interface as a new class in your project:

public interface ApiService
{

    @GET("/xml/simple.xml")
    YourObject getUser();
}

然后在你的活动，你会调用以下内容：

Then in your activity you will call the following:

RestAdapter restAdapter = new RestAdapter.Builder()
                    .setEndpoint("http://www.w3schools.com")
                    .setConverter(new SimpleXmlConverter())
                    .build();

ApiService apiService = restAdapter.create(ApiService.class);
YourObject object = apiService.getXML();

要正确地得到您的图书馆，在build.gradle文件，你需要做到以下几点：

To get your libraries correctly, in your build.gradle file you need to do the following:

configurations {
compile.exclude group: 'stax'
compile.exclude group: 'xpp3'
}

dependencies {
compile fileTree(dir: 'libs', include: ['*.jar'])
compile 'com.squareup.retrofit:retrofit:1.6.1'
compile 'com.mobprofs:retrofit-simplexmlconverter:1.1'
compile 'org.simpleframework:simple-xml:2.7.1'
compile 'com.google.code.gson:gson:2.2.4'
}

然后，你需要指定YourObject并根据XML文件的结构添加注释它

Then you need to specify YourObject and add annotations to it according to the structure of the xml file

@Root(name = "breakfast_menu")
public class BreakFastMenu
{
@ElementList(inline = true)
List<Food> foodList;
}

@Root(name="food")
public class Food
{
@Element(name = "name")
String name;

@Element(name = "price")
String price;

@Element(name = "description")
String description;

@Element(name = "calories")
String calories;
}

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