JAX-RS之上传文件
2011-08-08 20:29
155 查看
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:
Java代码
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:
Java代码
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.
Java代码
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:
Java代码
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?
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:
Java代码
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();
}
}
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:
Java代码
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";
}
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.
Java代码
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;
}
}
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:
Java代码
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();
}
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?
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