在Restlet中带有描述的文件上载
在Restlet中带有描述的文件上载
提问于 2015-02-03 06:14:43
我需要使用restlet上传一个带有其他数据的文件。所以我创建了一个示例html页面，如下所示。
<html>
<body>
<h1>*****Upload File with RESTFul WebService*****</h1>
<form action="http://localhost:8080/test/api/streams/sample.json" method="post" enctype="multipart/form-data">
<fieldset>
<legend>Upload File</legend>
<input type="file" name="fileToUpload"/><br />
<br /><br />
Party ID<input type="text" name="mybody" /><br />
<input type="submit" name="Upload" id="Upload" value="Upload" />
</fieldset>
</form>
</body>
我需要从输入字段和文件data.Now中读取值，可以在同一个data.Now调用中读取文件content.how，可以从输入框中获得值。
@Post
public Representation accept(Representation entity) throws Exception {
Representation result = null;
if (entity != null) {
if (MediaType.MULTIPART_FORM_DATA.equals(entity.getMediaType(), true)) {
// 1/ Create a factory for disk-based file items
DiskFileItemFactory factory = new DiskFileItemFactory();
factory.setSizeThreshold(1000240);
// 2/ Create a new file upload handler based on the Restlet
// FileUpload extension that will parse Restlet requests and
// generates FileItems.
RestletFileUpload upload = new RestletFileUpload(factory);
// 3/ Request is parsed by the handler which generates a
// list of FileItems
FileItemIterator fileIterator = upload.getItemIterator(entity);
// Process only the uploaded item called "fileToUpload"
// and return back
boolean found = false;
while (fileIterator.hasNext() && !found) {
FileItemStream fi = fileIterator.next();
Extractor extractor = new Extractor(getContext());
if (fi.getFieldName().equals("fileToUpload")) {
found = true;
// consume the stream immediately, otherwise the stream
// will be closed.
StringBuilder sb = new StringBuilder("media type: ");
sb.append(fi.getContentType()).append("\n");
sb.append("file name : ");
sb.append(fi.getName()).append("\n");
BufferedReader br = new BufferedReader(new InputStreamReader(fi.openStream()));
String line = null;
while ((line = br.readLine()) != null) {
sb.append(line);
}
sb.append("\n");
result = new StringRepresentation(sb.toString(), MediaType.TEXT_PLAIN);
}
}
} else {
setStatus(Status.CLIENT_ERROR_BAD_REQUEST);
}
System.out.println("result==" + result);
}
return result;}
回答 1
Stack Overflow用户
回答已采纳
发布于 2015-02-03 08:55:04
首先，您需要理解请求的内容。如果您查看发送的请求(firebug或其他东西)，您将看到以下内容：
-----------------------------2003194375274723921294130757
Content-Disposition: form-data; name="fileToUpload";
filename="mysql.sql" Content-Type: application/sql
<<YOUR FILE CONTENT>>
-----------------------------2003194375274723921294130757
Content-Disposition: form-data; name="mybody"
my value
-----------------------------2003194375274723921294130757
Content-Disposition: form-data; name="Upload"
Upload
-----------------------------2003194375274723921294130757--如您所见，在您的请求中有几个部分。这意味着您可以在服务器资源中迭代这些部分。在您提供的代码中，您只查找名称为fileToUpload的条目，当找到时，就会中断循环。
您可以更新代码以执行完整的循环，并检查输入字段的值mybody，如下所述：
while (fileIterator.hasNext()) {
FileItemStream fi = fileIterator.next();
if ("mybody".equals(fi.getFieldName())) {
BufferedReader br = new BufferedReader(
new InputStreamReader(fi.openStream()));
String fieldValue = null;
if ((line = br.readLine()) != null) {
fieldValue = line;
}
} else if ("fileToUpload".equals(fi.getFieldName())) {
(...)
}
(...)
}希望能帮上忙，蒂埃里
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/28292523
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