【发布时间】:2018-01-17 16:24:32
【问题描述】:
我必须创建什么类来接收一个 XML 元素?
我正在从 API 接收 XML:
<string xmlns="http://schemas.microsoft.com/2003/10/Serialization/">cat</string>
我有课:
@Root(strict = false)
public class Translation {
@Element(name = "string")
private String string;
public String getString(){
return string;
}
public Translation() {
}
}
并发现错误:
org.simpleframework.xml.core.ValueRequiredException: Unable to satisfy @org.simpleframework.xml.Element(data=false, name=string, required=true, type=void) on field 'string' private java.lang.String com.antonioleiva.mvpexample.app.main.Utils.Network.Translation.Translation.string for class com.antonioleiva.mvpexample.app.main.Utils.Network.Translation.Translation at line 1
【问题讨论】:
-
Xml 来自 SOAP 服务?
-
尝试使用 ksoap2 代替改造肥皂