JAX-RS之上传文件

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Spring REST

今天学习的是jax-rs中的上传文件.
1 首先要包含的是resteasy-multipart-provider.jar这个文件

2) 之后是简单的HTML FORM
   <html>
<body>
<h1>JAX-RS Upload Form</h1>

<form action="rest/file/upload" method="post" enctype="multipart/form-data">


Select a file : <input type="file" name="uploadedFile" size="50" />


   <input type="submit" value="Upload It" />
</form>

</body>
</html>

3 代码如下,先列出,再讲解:

 import java.io.File;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.util.List;
import java.util.Map;
import javax.ws.rs.Consumes;
import javax.ws.rs.POST;
import javax.ws.rs.Path;
import javax.ws.rs.core.MultivaluedMap;
import javax.ws.rs.core.Response;
import org.apache.commons.io.IOUtils;
import org.jboss.resteasy.plugins.providers.multipart.InputPart;
import org.jboss.resteasy.plugins.providers.multipart.MultipartFormDataInput;
 
@Path("/file")
public class UploadFileService {
 
	private final String UPLOADED_FILE_PATH = "d:\\";
 
	@POST
	@Path("/upload")
	@Consumes("multipart/form-data")
	public Response uploadFile(MultipartFormDataInput input) {
 
		String fileName = "";
 
		Map<String, List<InputPart>> uploadForm = input.getFormDataMap();
		List<InputPart> inputParts = uploadForm.get("uploadedFile");
 
		for (InputPart inputPart : inputParts) {
 
		 try {
 
			MultivaluedMap<String, String> header = inputPart.getHeaders();
			fileName = getFileName(header);
 
			//convert the uploaded file to inputstream
			InputStream inputStream = inputPart.getBody(InputStream.class,null);
 
			byte [] bytes = IOUtils.toByteArray(inputStream);
 
			//constructs upload file path
			fileName = UPLOADED_FILE_PATH + fileName;
 
			writeFile(bytes,fileName);
 
			System.out.println("Done");
 
		  } catch (IOException e) {
			e.printStackTrace();
		  }
 
		}
 
		return Response.status(200)
		    .entity("uploadFile is called, Uploaded file name : " + fileName).build();
 
	}
 
	/**
	 * header sample
	 * {
	 * 	Content-Type=[image/png], 
	 * 	Content-Disposition=[form-data; name="file"; filename="filename.extension"]
	 * }
	 **/
	//get uploaded filename, is there a easy way in RESTEasy?
	private String getFileName(MultivaluedMap<String, String> header) {
 
		String[] contentDisposition = header.getFirst("Content-Disposition").split(";");
 
		for (String filename : contentDisposition) {
			if ((filename.trim().startsWith("filename"))) {
 
				String[] name = filename.split("=");
 
				String finalFileName = name[1].trim().replaceAll("\"", "");
				return finalFileName;
			}
		}
		return "unknown";
	}
 
	//save to somewhere
	private void writeFile(byte[] content, String filename) throws IOException {
 
		File file = new File(filename);
 
		if (!file.exists()) {
			file.createNewFile();
		}
 
		FileOutputStream fop = new FileOutputStream(file);
 
		fop.write(content);
		fop.flush();
		fop.close();
 
	}
}

   这里,用户选择了文件上传后,会URL根据REST的特性,自动map
到uploadFile方法中,然后通过:
   Map<String, List<InputPart>> uploadForm = input.getFormDataMap();

List<InputPart> inputParts = uploadForm.get("uploadedFile");
   找出所有的上传文件框(可以是多个),然后进行循环工作:
首先是获得每个文件头的HEADER,用这个
MultivaluedMap<String, String> header = inputPart.getHeaders();
然后在header中取出文件名,这里使用的方法是getFileName,另外写了个方法:

   private String getFileName(MultivaluedMap<String, String> header) {
 
		String[] contentDisposition = header.getFirst("Content-Disposition").split(";");
 
		for (String filename : contentDisposition) {
			if ((filename.trim().startsWith("filename"))) {
 
				String[] name = filename.split("=");
 
				String finalFileName = name[1].trim().replaceAll("\"", "");
				return finalFileName;
			}
		}
		return "unknown";
	}

这里,比较麻烦,要获得header,比如header是如下形式的,然后要再提取其中的文件名:
Content-Disposition=[form-data; name="file"; filename="filename.extension"]
最后用writeFile写入磁盘.真麻烦呀,还是用spring mvc好.

2
MultipartForm 的例子, MultipartForm 中,将上传的文件中的属性配置到
POJO中,例子为:FileUploadForm 类,这个POJO类对应上传的文件类.

 import javax.ws.rs.FormParam;
import org.jboss.resteasy.annotations.providers.multipart.PartType;
 
public class FileUploadForm {
 
	public FileUploadForm() {
	}
 
	private byte[] data;
 
	public byte[] getData() {
		return data;
	}
 
	@FormParam("uploadedFile")
	@PartType("application/octet-stream")
	public void setData(byte[] data) {
		this.data = data;
	}
 
}

处理部分就简单多了,可以这样:

Path("/file")
public class UploadFileService {
 
	@POST
	@Path("/upload")
	@Consumes("multipart/form-data")
	public Response uploadFile(@MultipartForm FileUploadForm form) {
 
		String fileName = "d:\\anything";
 
		try {
			writeFile(form.getData(), fileName);
		} catch (IOException e) {
 
			e.printStackTrace();
		}
 
		System.out.println("Done");
 
		return Response.status(200)
		    .entity("uploadFile is called, Uploaded file name : " + fileName).build();
 
	}

即可.

但总的感觉,REST有点蛋疼,上传个文件,用SPIRNG MVC或者其他方法都可以了,还用
REST这个方法?

2
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3
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4 楼笛安小六 2016-09-05

您好，我按照您的方法写完，测试的时候总是报这个错，一直解决不了。请问您遇到过么？
java.lang.RuntimeException: RESTEASY007500: Could find no Content-Disposition header within part

3 楼 Computer521System 2012-07-17

用的第一种方法、form表单里面还有其他的text。这样应该如何做呢、、？

2 楼 jackyrong 2012-05-18

搜索下吧，之前按老外例子搞的，搞完删除了

1 楼 570421779 2012-05-17

最近也在搞resteasy的上传下载，能把resteasy-multipart-provider.jar文件发给我吗？我的邮箱是570421779@qq.com,谢谢

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