在 Java 中使用 Rest API 上传文件
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/24890675/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
提示：将鼠标放在中文语句上可以显示对应的英文。显示中英文
时间：2020-08-14 15:23:58 来源:igfitidea点击:
File Upload using Rest API in Java
提问by JLyon
I am new to REST API. I want to upload user selected file to the user provided path(remote or local path) using REST API. My html file is having 1 text box and 1 file chooser. User will enter FilePath (local or remote machine folder location) in the text box. Please suggest how to solve this issue.
我是 REST API 的新手。我想使用 REST API 将用户选择的文件上传到用户提供的路径（远程或本地路径）。我的 html 文件有 1 个文本框和 1 个文件选择器。用户将在文本框中输入 FilePath（本地或远程机器文件夹位置）。请建议如何解决这个问题。
Here is my code:
这是我的代码：
FileUpload.html::
文件上传.html::
<body>
<form action="rest/file/upload" method="post" enctype="multipart/form-data">
<p>
Select a file : <input type="file" name="file" size="45" />
</p>
<p>Target Upload Path : <input type="text" name="path" /></p>
<input type="submit" value="Upload It" />
</form>
</body>
UploadFileService.java
上传文件服务.java
@Path("/file")
public class UploadFileService {
@POST
@Path("/upload")
@Consumes(MediaType.MULTIPART_FORM_DATA)
public Response uploadFile(
@FormDataParam("file") InputStream uploadedInputStream,
@FormDataParam("file") FormDataContentDisposition fileDetail,
@FormParam("path") String path) {
/*String uploadedFileLocation = "d://uploaded/" + fileDetail.getFileName();*/
/*String uploadedFileLocation = //10.217.14.88/Installables/uploaded/" + fileDetail.getFileName();*/
String uploadedFileLocation = path
+ fileDetail.getFileName();
// save it
writeToFile(uploadedInputStream, uploadedFileLocation);
String output = "File uploaded to : " + uploadedFileLocation;
return Response.status(200).entity(output).build();
}
// save uploaded file to new location
private void writeToFile(InputStream uploadedInputStream,
String uploadedFileLocation) {
try {
OutputStream out = new FileOutputStream(new File(
uploadedFileLocation));
int read = 0;
byte[] bytes = new byte[1024];
out = new FileOutputStream(new File(uploadedFileLocation));
while ((read = uploadedInputStream.read(bytes)) != -1) {
out.write(bytes, 0, read);
}
out.flush();
out.close();
} catch (IOException e) {
e.printStackTrace();
}
}
}
Web.xml
网页.xml
<display-name>JAXRSFileUploadJerseyExample</display-name>
<servlet>
<servlet-name>jersey-serlvet</servlet-name>
<servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>
<init-param>
<param-name>com.sun.jersey.config.property.packages</param-name>
<param-value>com.mkyong.rest</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>jersey-serlvet</servlet-name>
<url-pattern>/rest/*</url-pattern>
</servlet-mapping>
Exception:
例外：
HTTP Status 500 - com.sun.jersey.api.container.ContainerException: Exception obtaining parameters
type Exception report
message com.sun.jersey.api.container.ContainerException: Exception obtaining parameters
description The server encountered an internal error that prevented it from fulfilling this request.
exception
javax.servlet.ServletException: com.sun.jersey.api.container.ContainerException: Exception obtaining parameters
com.sun.jersey.spi.container.servlet.WebComponent.service(WebComponent.java:425)
com.sun.jersey.spi.container.servlet.ServletContainer.service(ServletContainer.java:537)
com.sun.jersey.spi.container.servlet.ServletContainer.service(ServletContainer.java:699)
javax.servlet.http.HttpServlet.service(HttpServlet.java:728)
root cause
com.sun.jersey.api.container.ContainerException: Exception obtaining parameters
com.sun.jersey.server.impl.inject.InjectableValuesProvider.getInjectableValues(InjectableValuesProvider.java:54)
com.sun.jersey.multipart.impl.FormDataMultiPartDispatchProvider$FormDataInjectableValuesProvider.getInjectableValues(FormDataMultiPartDispatchProvider.java:125)
com.sun.jersey.server.impl.model.method.dispatch.AbstractResourceMethodDispatchProvider$EntityParamInInvoker.getParams(AbstractResourceMethodDispatchProvider.java:153)
com.sun.jersey.server.impl.model.method.dispatch.AbstractResourceMethodDispatchProvider$ResponseOutInvoker._dispatch(AbstractResourceMethodDispatchProvider.java:203)
com.sun.jersey.server.impl.model.method.dispatch.ResourceJavaMethodDispatcher.dispatch(ResourceJavaMethodDispatcher.java:75)
com.sun.jersey.server.impl.uri.rules.HttpMethodRule.accept(HttpMethodRule.java:288)
com.sun.jersey.server.impl.uri.rules.RightHandPathRule.accept(RightHandPathRule.java:147)
com.sun.jersey.server.impl.uri.rules.ResourceClassRule.accept(ResourceClassRule.java:108)
com.sun.jersey.server.impl.uri.rules.RightHandPathRule.accept(RightHandPathRule.java:147)
com.sun.jersey.server.impl.uri.rules.RootResourceClassesRule.accept(RootResourceClassesRule.java:84)
com.sun.jersey.server.impl.application.WebApplicationImpl._handleRequest(WebApplicationImpl.java:1469)
com.sun.jersey.server.impl.application.WebApplicationImpl._handleRequest(WebApplicationImpl.java:1400)
com.sun.jersey.server.impl.application.WebApplicationImpl.handleRequest(WebApplicationImpl.java:1349)
com.sun.jersey.server.impl.application.WebApplicationImpl.handleRequest(WebApplicationImpl.java:1339)
com.sun.jersey.spi.container.servlet.WebComponent.service(WebComponent.java:416)
com.sun.jersey.spi.container.servlet.ServletContainer.service(ServletContainer.java:537)
com.sun.jersey.spi.container.servlet.ServletContainer.service(ServletContainer.java:699)
javax.servlet.http.HttpServlet.service(HttpServlet.java:728)
root cause
java.lang.IllegalStateException: The @FormParam is utilized when the content type of the request entity is not application/x-www-form-urlencoded
com.sun.jersey.server.impl.model.parameter.FormParamInjectableProvider$FormParamInjectable.getValue(FormParamInjectableProvider.java:81)
com.sun.jersey.server.impl.inject.InjectableValuesProvider.getInjectableValues(InjectableValuesProvider.java:46)
com.sun.jersey.multipart.impl.FormDataMultiPartDispatchProvider$FormDataInjectableValuesProvider.getInjectableValues(FormDataMultiPartDispatchProvider.java:125)
com.sun.jersey.server.impl.model.method.dispatch.AbstractResourceMethodDispatchProvider$EntityParamInInvoker.getParams(AbstractResourceMethodDispatchProvider.java:153)
com.sun.jersey.server.impl.model.method.dispatch.AbstractResourceMethodDispatchProvider$ResponseOutInvoker._dispatch(AbstractResourceMethodDispatchProvider.java:203)
com.sun.jersey.server.impl.model.method.dispatch.ResourceJavaMethodDispatcher.dispatch(ResourceJavaMethodDispatcher.java:75)
com.sun.jersey.server.impl.uri.rules.HttpMethodRule.accept(HttpMethodRule.java:288)
com.sun.jersey.server.impl.uri.rules.RightHandPathRule.accept(RightHandPathRule.java:147)
com.sun.jersey.server.impl.uri.rules.ResourceClassRule.accept(ResourceClassRule.java:108)
com.sun.jersey.server.impl.uri.rules.RightHandPathRule.accept(RightHandPathRule.java:147)
com.sun.jersey.server.impl.uri.rules.RootResourceClassesRule.accept(RootResourceClassesRule.java:84)
com.sun.jersey.server.impl.application.WebApplicationImpl._handleRequest(WebApplicationImpl.java:1469)
com.sun.jersey.server.impl.application.WebApplicationImpl._handleRequest(WebApplicationImpl.java:1400)
com.sun.jersey.server.impl.application.WebApplicationImpl.handleRequest(WebApplicationImpl.java:1349)
com.sun.jersey.server.impl.application.WebApplicationImpl.handleRequest(WebApplicationImpl.java:1339)
com.sun.jersey.spi.container.servlet.WebComponent.service(WebComponent.java:416)
com.sun.jersey.spi.container.servlet.ServletContainer.service(ServletContainer.java:537)
com.sun.jersey.spi.container.servlet.ServletContainer.service(ServletContainer.java:699)
javax.servlet.http.HttpServlet.service(HttpServlet.java:728)
采纳答案by JLyon
I have updated signature of method in below class and its working fine.
Instead of @FormParam, used @FormDataParam("path")String path and it solved my issue.
Below is the updated code:
我在下面的类中更新了方法的签名并且它工作正常。而不是@FormParam，使用@FormDataParam("path")字符串路径，它解决了我的问题。以下是更新后的代码：
import java.io.File;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import javax.ws.rs.Consumes;
import javax.ws.rs.FormParam;
import javax.ws.rs.POST;
import javax.ws.rs.Path;
import javax.ws.rs.core.MediaType;
import javax.ws.rs.core.Response;
import com.sun.jersey.core.header.FormDataContentDisposition;
import com.sun.jersey.multipart.FormDataParam;
@Path("/file")
public class UploadFileService {
@POST
@Path("/upload")
@Consumes(MediaType.MULTIPART_FORM_DATA)
public Response uploadFile(
@FormDataParam("file") InputStream uploadedInputStream,
@FormDataParam("file") FormDataContentDisposition fileDetail,
@FormDataParam("path") String path) {
// Path format //10.217.14.97/Installables/uploaded/
System.out.println("path::"+path);
String uploadedFileLocation = path
+ fileDetail.getFileName();
// save it
writeToFile(uploadedInputStream, uploadedFileLocation);
String output = "File uploaded to : " + uploadedFileLocation;
return Response.status(200).entity(output).build();
}
// save uploaded file to new location
private void writeToFile(InputStream uploadedInputStream,
String uploadedFileLocation) {
try {
OutputStream out = new FileOutputStream(new File(
uploadedFileLocation));
int read = 0;
byte[] bytes = new byte[1024];
out = new FileOutputStream(new File(uploadedFileLocation));
while ((read = uploadedInputStream.read(bytes)) != -1) {
out.write(bytes, 0, read);
}
out.flush();
out.close();
} catch (IOException e) {
e.printStackTrace();
}
}
}
回答by RuntimeException
Another possible solution
另一种可能的解决方案
@Consumes annotation and set it to application octet stream. Use PUT http method. Then in the client, read bytes from file and upload.
@Consumes 注释并将其设置为应用程序八位字节流。使用 PUT http 方法。然后在客户端，从文件中读取字节并上传。
My sample code for a REST service to zip a file is here - https://stackoverflow.com/a/32253028/15789
我用于压缩文件的 REST 服务的示例代码在这里 - https://stackoverflow.com/a/32253028/15789
Hope this helps.
希望这可以帮助。
