这次要和别人开发的服务器端程序接口对接,之前实现的上传文件的方法对方接口无法接收。他们的接口需要类似这样的form提交数据:

<form enctype="multipart/form-data" action=http://url/path  method="POST"> 

<input name=uid type=text value="110">
<input name="filename” type="file" /> 
<input type="submit" value="test" /> 
</form>

解决办法是引入commons-httpclient.jar,下载地址http://hc.apache.org/downloads.cgi,模拟form表单提交数据,代码如下:

File f = new File("device1.png"); 
PostMethod filePost = new PostMethod( 
        "http://url/path"); 
Part[] parts = { new FilePart("filename", f) ,new StringPart("source", "695132533"), new StringPart("status", URLEncoder.encode(status, "utf-8"))}; 
filePost.setRequestEntity(new MultipartRequestEntity(parts, filePost 
        .getParams())); 
HttpClient clients = new HttpClient();

int status = clients.executeMethod(filePost); 
try { 
    BufferedReader rd = new BufferedReader(new InputStreamReader( 
            filePost.getResponseBodyAsStream(), "UTF-8")); 
    StringBuffer stringBuffer = new StringBuffer(); 
    String line; 
    while ((line = rd.readLine()) != null) { 
        stringBuffer .append(line); 
    } 
    rd.close(); 
    System.out.println("接受到的流是:" + stringBuffer + "—-" + status); 
} catch (Exception e) { 
    throw new RuntimeException("error”,e);

}

解释一下

filePost.setRequestEntity(new MultipartRequestEntity(parts, filePost 
        .getParams()));

设置多媒体参数,作用类似form表单中的enctype="multipart/form-data" ,

Part[] parts = { new FilePart("filename", f) };

设定参数名称和值,类似form表单中的<input name="filename” type="file" /> try catch块主要是打印服务器端回传的数据。