我刚刚开始使用Retrofit。我正在使用SimpleXML的项目中。有人可以给我提供一个例子吗,其中有人从某个网站获取XML。 http://www.w3schools.com/xml/simple.xml“并读出来吗?

最佳答案

您将在项目中创建一个作为新类的接口(interface):

public interface ApiService {
    @GET("/xml/simple.xml")
    YourObject getUser();
}

然后在您的 Activity 中,您将调用以下内容:
RestAdapter restAdapter = new RestAdapter.Builder()
                    .setEndpoint("http://www.w3schools.com")
                    .setConverter(new SimpleXmlConverter())
                    .build();

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

为了正确获取库,需要在build.gradle文件中执行以下操作:
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,您需要指定YourObject并根据xml文件的结构向其添加注释
@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;
}

09-09 21:52
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