从XML在Android中列出反序列化/解组泛型列表序列化、列表、XML、Android

由网友(姑娘别输给爱)分享简介：我做了Java中的web服务与返回一个字符串（泛型列表XML格式）的方法。我使用来自Android的此WebService，我得到这个字符串，但几次尝试后Android模拟器尝试反序列化字符串时只是崩溃。这是一个字符串，我得到一个例子：＆LT;？XML版本=1.0编码=UTF-8独立=YES＆GT？;＆LT; p...

我做了Java中的web服务与返回一个字符串（泛型列表XML格式）的方法。我使用来自Android的此WebService，我得到这个字符串，但几次尝试后Android模拟器尝试反序列化字符串时只是崩溃。这是一个字符串，我得到一个例子：

 ＆LT;？XML版本=1.0编码=UTF-8独立=YES＆GT？;＆LT; peliculas＆GT;    ＆LT;＆PELICULA GT;        ＆LT;＆ID GT; 18329＆LT; / ID＆GT;        ＆LT;＆海报GT; HTTP：//cache-cmx.netmx.mx/image/muestras/5368.rrr.jpg< /海报＆GT;        ＆LT;＆TITULO GT; 007 Operaci＆放大器;放大器; oacute; N 007：大破天幕杀机＆LT; / TITULO＆GT;    ＆LT; / PELICULA＆GT;...＆LT; / peliculas＆GT;

这是在web服务类：

  @XmlRootElement公共类Peliculas {    @XmlElement（NAME =PELICULA）    保护列表与LT; PELICULA＆GT; peliculas;    公共Peliculas（）{peliculas =新的ArrayList＆LT;＆PELICULA GT;（）;}    公共Peliculas（列表＆LT;＆PELICULA GT; PE）{        peliculas = PE;    }    公开名单＆LT;＆PELICULA GT;的GetList（）{        返回peliculas;    }    公共无效添加（PELICULA PELICULA）{        peliculas.add（PELICULA）;    }}

________EDIT______________

好像你不能与Android使用JAXB，并有更好/更轻库这一点。所以我想简单的XML。这是方法：

 公共Peliculas解组（XML字符串）抛出异常{    Peliculas peliculas =新Peliculas（）;    串行串行=新的持留（）;    StringBuffer的xmlStr =新的StringBuffer（XML）;    peliculas = serializer.read（Peliculas.class，（新StringReader（xmlStr.toString（））））;    返回peliculas;}

但我得到这个异常，好像它不能保存对象数据：

  20 11-12：30：10.898：I /错误（1058）：元素PELICULA'没有在课堂上app.cinemexservice.Pelicula比赛在第3行

解决方案

我用SAX解析该文件，然后手动将其转换为一个对象。这是code：

 公开名单＆LT;＆PELICULA GT;解组（XML字符串）抛出异常{        清单＆LT;＆PELICULA GT; peliculas =新的ArrayList＆LT;＆PELICULA GT;（）;        InputStream为=新ByteArrayInputStream进行（xml.getBytes（UTF-8））;        XmlPullParser解析器= Xml.newPullParser（）;        的char []℃;        串ID =，TITULO =，海报=，atributo =;        INT DATOS = 0;        尝试{            parser.setInput（是，UTF-8）;            INT事件= parser.next（）;        而（事件！= XmlPullParser.END_DOCUMENT）{            如果（事件== XmlPullParser.START_TAG）{                Log.d（TAG，＆下;+ parser.getName（）+＆gt;中）;                atributo = parser.getName（）;                的for（int i = 0; I＆LT; parser.getAttributeCount（）;我++）{                    Log.d（TAG， t的+ parser.getAttributeName（I）+=+ parser.getAttributeValue（ⅰ））;                }            }            如果（事件== XmlPullParser.TEXT＆放大器;＆放大器;！parser.getText（）修剪（）长度（）= 0）            {                Log.d（TAG， t  t+ parser.getText（））;                如果（atributo ==ID）{ID = parser.getText（）; DATOS ++;}                否则，如果（atributo ==TITULO）{TITULO = parser.getText（）; DATOS ++;}                否则，如果（atributo ==海报）{海报= parser.getText（）; DATOS ++;}                如果（DATOS == 3）{peliculas.add（新PELICULA（ID，TITULO，海报））; DATOS = 0;}            }                如果（事件== XmlPullParser.END_TAG）                    Log.d（TAG，＆所述; /+ parser.getName（）+＆gt;中）;                事件= parser.next（）;            is.close（）;        }        }赶上（例外五）{Toast.makeText（这一点，e.getMessage（），Toast.LENGTH_LONG）.show（）; }        对于（PELICULA号码：peliculas）{            Log.d（PELICULA恩LISTA：p.titulo）;        }        返回peliculas;    }

这是太长了我的口味，但我就是无法找出简单的XML，以配合我的课。

I made a webservice in java with a method that returns a string (a generic list in XML format). I consume this webservice from Android, and I get this string, but after several tries the Android emulator just crashes when trying to deserialize the string. This is an example for the string I get:

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<peliculas>
    <pelicula>
        <id>18329</id>
        <poster>http://cache-cmx.netmx.mx/image/muestras/5368.rrr.jpg</poster>
        <titulo>007 Operaci&amp;oacute;n Skyfall</titulo>
    </pelicula>
...
</peliculas>

This is the class in the webservice:

@XmlRootElement
public class Peliculas{

    @XmlElement(name="pelicula")
    protected List<Pelicula> peliculas;
    public Peliculas(){ peliculas = new ArrayList<Pelicula>();}

    public Peliculas(List<Pelicula> pe){
        peliculas = pe;
    }


    public List<Pelicula> getList(){
        return peliculas;       
    }

    public void add(Pelicula pelicula) {
        peliculas.add(pelicula);
    }
}

________EDIT______________

Seems like you can't use JAXB with Android, and there's better/lighter libraries for that. so I tried Simple XML. This is the method:

public Peliculas unmarshal(String xml) throws Exception{            
    Peliculas peliculas = new Peliculas();  
    Serializer serializer = new Persister();
    StringBuffer xmlStr = new StringBuffer( xml );
    peliculas = serializer.read(Peliculas.class, ( new StringReader( xmlStr.toString() ) )  );
    return peliculas;
}

BUT I get this exception, seems like it can't save data in object:

11-12 20:30:10.898: I/Error(1058): Element 'Pelicula' does not have a match in class app.cinemexservice.Pelicula at line 3

解决方案

I used SAX to parse the file, and then convert it manually to an object. This is the code:

public List<Pelicula> unmarshal(String xml) throws Exception{           
        List<Pelicula> peliculas = new ArrayList<Pelicula>();       
        InputStream is = new ByteArrayInputStream(xml.getBytes("UTF-8"));
        XmlPullParser parser = Xml.newPullParser(); 
        char[] c;
        String id="", titulo="", poster="", atributo="";
        int datos =0;
        try{ 
            parser.setInput(is, "UTF-8"); 
            int event = parser.next();  
        while(event != XmlPullParser.END_DOCUMENT) { 
            if(event == XmlPullParser.START_TAG) { 
                Log.d(TAG, "<"+ parser.getName() + ">"); 
                atributo = parser.getName();
                for(int i = 0; i < parser.getAttributeCount(); i++) { 
                    Log.d(TAG, "t"+ parser.getAttributeName(i) + " = "+ parser.getAttributeValue(i)); 
                } 
            } 
            if(event == XmlPullParser.TEXT&& parser.getText().trim().length() != 0) 
            {
                Log.d(TAG, "tt"+ parser.getText());
                if (atributo=="id"){id=parser.getText(); datos++;}
                else if(atributo=="titulo"){titulo=parser.getText(); datos++;}
                else if(atributo=="poster"){poster=parser.getText(); datos++;}
                if(datos==3){peliculas.add(new Pelicula(id, titulo, poster)); datos=0;} 
            }
                if(event == XmlPullParser.END_TAG) 
                    Log.d(TAG, "</"+ parser.getName() + ">");               
                event = parser.next(); 

            is.close();
        }
        } catch(Exception e) { Toast.makeText(this, e.getMessage(), Toast.LENGTH_LONG).show(); }        
        for (Pelicula p : peliculas){
            Log.d("Película en lista: ", p.titulo);
        }           
        return peliculas;
    }

It's way too long for my taste, but I just couldn't figure out Simple XML to match my classes.

阅读全文

相关专题：序列化；列表；xml ；Android ；解组泛型；发布时间：2023-09-07 16:25:50