[英]Resource not found Jersey 2.x
我正在嘗試創建一個Web服務,以將文件上傳到服務器。 我無法點擊該服務,並且總是收到404錯誤,即找不到資源。 我無法弄清楚代碼中的錯誤。 請幫忙。
以下是我的web.xml。 我還有其他資源(如登錄,注冊和跟蹤)也已注冊,這些資源通過ajax請求可以正常工作,但上載器Web服務無法正常工作。
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://java.sun.com/xml/ns/javaee" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd" id="WebApp_ID" version="3.0"> <display-name>mixtri</display-name>
<welcome-file-list>
<welcome-file>index.html</welcome-file>
<welcome-file>index.htm</welcome-file>
<welcome-file>index.jsp</welcome-file>
<welcome-file>default.html</welcome-file>
<welcome-file>default.htm</welcome-file>
<welcome-file>default.jsp</welcome-file>
</welcome-file-list>
<display-name>mixtri</display-name>
<servlet>
<servlet-name>Jersey REST Service</servlet-name>
<servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
<!-- Register resources and providers under com.vogella.jersey.first package. -->
<init-param>
<param-name>jersey.config.server.provider.packages</param-name>
<param-value>com.mixtri.tracks,com.mixtri.login,com.mixtri.signup,com.mixtri.uploader</param-value>
</init-param>
<init-param>
<param-name>jersey.config.server.provider.classnames</param-name>
<param-value>org.glassfish.jersey.media.multipart.MultiPartFeature</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>Jersey REST Service</servlet-name>
<url-pattern>/rest/*</url-pattern>
</servlet-mapping>
</web-app>
以下是我的html表單的標記:
<form method="post" action="/mixtri/rest/upload"
enctype="multipart/form-data">
<table align="center" border="1" bordercolor="black" cellpadding="0"
cellspacing="0">
<tr>
<td>Select Zip File :</td>
<td><input type="file" name="uploadFile" size="100" /></td>
</tr>
<tr>
<td><input type="submit" value="Upload File" /></td>
<td><input type="reset" value="Reset" /></td>
</tr>
</table>
</form>
這是我的資源類:
package com.mixtri.uploader;
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.GET;
import javax.ws.rs.POST;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;
import javax.ws.rs.core.Response;
import javax.ws.rs.core.Response.ResponseBuilder;
import org.glassfish.jersey.media.multipart.FormDataContentDisposition;
import org.glassfish.jersey.media.multipart.FormDataParam;
@Path("/rest")
public class FileServiceImpl implements IFileService {
public static final String UPLOAD_FILE_SERVER = "D:/Demo/upload/";
@POST
@Path("/upload")
@Consumes(MediaType.MULTIPART_FORM_DATA)
public Response uploadZippedFile(
@FormDataParam("uploadFile") InputStream fileInputStream,
@FormDataParam("uploadFile") FormDataContentDisposition fileFormDataContentDisposition) {
// local variables
String fileName = null;
String uploadFilePath = null;
try {
fileName = fileFormDataContentDisposition.getFileName();
uploadFilePath = writeToFileServer(fileInputStream, fileName);
}
catch(IOException ioe){
ioe.printStackTrace();
}
finally{
// release resources, if any
}
return Response.ok("File uploaded successfully at " + uploadFilePath).build();
}
/**
* write input stream to file server
* @param inputStream
* @param fileName
* @throws IOException
*/
private String writeToFileServer(InputStream inputStream, String fileName) throws IOException {
OutputStream outputStream = null;
String qualifiedUploadFilePath = UPLOAD_FILE_SERVER + fileName;
try {
outputStream = new FileOutputStream(new File(qualifiedUploadFilePath));
int read = 0;
byte[] bytes = new byte[1024];
while ((read = inputStream.read(bytes)) != -1) {
outputStream.write(bytes, 0, read);
}
outputStream.flush();
}
catch (IOException ioe) {
ioe.printStackTrace();
}
finally{
//release resource, if any
outputStream.close();
}
return qualifiedUploadFilePath;
}
}
任何指針都會有所幫助。
您可以將FileServiceImpl
中的@Path("/rest")
更改為@Path("/")
。 或者,您可以將其更改為@Path("/upload")
並將其從uploadZippedFile
方法中刪除。
您已經在此處將/rest
定義為Jersey應用程序的根映射。
<servlet-mapping>
<servlet-name>Jersey REST Service</servlet-name>
<url-pattern>/rest/*</url-pattern>
</servlet-mapping>
所以最終您的道路需要
action="/mixtri/rest/rest/upload"
如果您不更改注釋。
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.