我有带有网络线程的 Android / Java 项目。它连接到为我提供 GetAddonsTypes 方法的 WCF WebService。
GetAddonsTypes 返回 long、int 和 string 项，因此我使用自己的 Parser (String / regex)，它还在 myDataSource（它是 SQLite 数据库）中创建对象。现在我对称为 GetProducts 的第二种方法有疑问。GetProducts 返回 long、int 和 Image。
我想将 Image 存储为 byte[] 类型。但是如何使用 SoapObjects 处理二进制文件呢？也许我应该将此 anyType{} 转换为二进制文件，但我该怎么做呢？
这就是 GetProducts (.toString()) 的 SoapObject 结果的样子
anyType{DocumentElement=anyType{Tabela=anyType{ID=701; lg=1; ProductImage=anyType{}; };
我的网络主题
Thread networkThread1 = new Thread() {
@Override
public void run() {
try {
final String METHOD_NAME = "GetAddonsTypes";
final String SOAP_ACTION = "http://tempuri.org/IService1/GetAddonsTypes";
SoapObject request = new SoapObject(NAMESPACE, METHOD_NAME);
SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11);
envelope.setOutputSoapObject(request);
envelope.dotNet = true;
HttpTransportSE ht = new HttpTransportSE(URL);
ht.call(SOAP_ACTION, envelope);
final SoapObject result=(SoapObject)envelope.getResponse();
runOnUiThread (new Runnable(){
public void run() {
ParseTable(result.getProperty(1).toString());
}
});
}
catch (Exception e) {
Log.e("WS", e.toString());
}
}
};
public void ParseTable(String input)
{
myDataSource myDatasource;
Pattern p = Pattern.compile("(PID=)(\\d*); (flg=)(\\d*); (Name=)(\\w*);");
Matcher m = p.matcher(input);
myDatasource = new myDatasource(this);
myDatasource.open();
while (m.find()) {
try {
myDatasource.createMyItem(Long.parseLong(m.group(2)), Integer.parseInt(m.group(4)), m.group(6));
}
catch (Exception e) {
Log.e("Parser Error", e.toString());
}
}
myDatasource.close();
}