I'm trying to create an upload-image button and afterward showing the image on a different jsp page.
I want to do this by uploading into the app-root/data/images folder. This works with the below filepath: filePath = System.getenv("OPENSHIFT_DATA_DIR") + "images/";
But how can I show this image on my jsp? I tried using:
<BODY>
<h1>SNOOP PAGE</h1>
Ga weer terug
<% String filepath = System.getenv("OPENSHIFT_DATA_DIR") + "images/";
out.println("<img src='"+filepath+"logo21.jpg'/>");
%>
<img src="app-root/data/images/logo21.jpg"/>
</BODY>
Both these options don't work. I also read that I need to create a symbolic link. But when I'm in my app-root/data or app-root/data/images or in app-root the command ln -s returns missing file operand
The logo21.jpg does show up in my Git bash
#developercorey is right (gave you +1 👍), I just feel the need to explain why:
Your uploaded images ends up in a folder on your server
(String filepath = System.getenv("OPENSHIFT_DATA_DIR") + "images/" is the folder path in the server).
Your rendered HTML "<img src='"+filepath+"logo21.jpg'/> get sent to the client (the user's browser), with the server's filepath url.
Obviously, when the user's browser try to locate the image, using the path of the server, which doesn't exist on the local machine, it won't work.
The best solution, as #developercorey suggested, is to add a new servlet or a filter to serve photos from the OPENSHIFT_DATA_DIR folder:
You'll have a new url mapped to the servlet serving your photo, something like http://your-server/uploaded/
And you can use <img src="http://your-server/uploaded/logo21.jpg" /> in your jsp.
Here's the snippet from How-To: Upload and Serve files using Java Servlets on OpenShift
import java.io.File;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import javax.activation.MimetypesFileTypeMap;
import javax.servlet.ServletException;
import javax.servlet.annotation.MultipartConfig;
import javax.servlet.annotation.WebServlet;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import javax.servlet.http.Part;
#WebServlet(name = "uploads",urlPatterns = {"/uploads/*"})
#MultipartConfig
public class Uploads extends HttpServlet {
private static final long serialVersionUID = 2857847752169838915L;
int BUFFER_LENGTH = 4096;
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
PrintWriter out = response.getWriter();
for (Part part : request.getParts()) {
InputStream is = request.getPart(part.getName()).getInputStream();
String fileName = getFileName(part);
FileOutputStream os = new FileOutputStream(System.getenv("OPENSHIFT_DATA_DIR") + fileName);
byte[] bytes = new byte[BUFFER_LENGTH];
int read = 0;
while ((read = is.read(bytes, 0, BUFFER_LENGTH)) != -1) {
os.write(bytes, 0, read);
}
os.flush();
is.close();
os.close();
out.println(fileName + " was uploaded to " + System.getenv("OPENSHIFT_DATA_DIR"));
}
}
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
String filePath = request.getRequestURI();
File file = new File(System.getenv("OPENSHIFT_DATA_DIR") + filePath.replace("/uploads/",""));
InputStream input = new FileInputStream(file);
response.setContentLength((int) file.length());
response.setContentType(new MimetypesFileTypeMap().getContentType(file));
OutputStream output = response.getOutputStream();
byte[] bytes = new byte[BUFFER_LENGTH];
int read = 0;
while ((read = input.read(bytes, 0, BUFFER_LENGTH)) != -1) {
output.write(bytes, 0, read);
output.flush();
}
input.close();
output.close();
}
private String getFileName(Part part) {
for (String cd : part.getHeader("content-disposition").split(";")) {
if (cd.trim().startsWith("filename")) {
return cd.substring(cd.indexOf('=') + 1).trim()
.replace("\"", "");
}
}
return null;
}
}
The best way to serve user uploaded images that you are storing in your OPENSHIFT_DATA_DIR would be to use a servlet as described here: https://forums.openshift.com/how-to-upload-and-serve-files-using-java-servlets-on-openshift?noredirect
This servlet basically takes the path/name of the image that is being requested, reads it from the filesystem and then serves it to the requester.
The OPENSHIFT_DATA_DIR directory is not web-accessible. You can make images stored in the OPENSHIFT_DATA_DIR (aka app-root/data) directory web-accessible by creating a symlink to them from the publicly accessible OPENSHIFT_REPO_DIR.
For one-time use, as a proof of concept:
rhc ssh -a <your_app_name> -n <your_namespace>
ln -sf ${OPENSHIFT_DATA_DIR}images ${OPENSHIFT_REPO_DIR}images
You should now be able to access logo21.jpg at https://<your_app_name>-<your_namespace>.rhcloud.com/images/logo21.jpg, or <img src="/images/logo21.jpg"/>.
The contents of the OPENSHIFT_REPO_DIR are overwritten when you push changes, so you'll want to create the symlink with a deploy hook to re-create it each time you deploy. In .openshift/action_hooks/deploy:
#!/bin/bash
# This deploy hook gets executed after dependencies are resolved and the
# build hook has been run but before the application has been started back
# up again.
# create the images directory if it doesn't exist
if [ ! -d ${OPENSHIFT_DATA_DIR}images ]; then
mkdir ${OPENSHIFT_DATA_DIR}images
fi
# create symlink to uploads directory
ln -sf ${OPENSHIFT_DATA_DIR}images ${OPENSHIFT_REPO_DIR}images
You can upload the file to the DATA DIRECTORY, then copy the file from the DATA DIRECTORY to any folder in the HOME DIRECTORY.
Thereafter you should be able to reference the image as usual in your page but it appears Openshift only displays items from a previous deployment or git push, therefore perhaps it is best to save the file in a database then read it directly from that database.
Related
I have very little experience in JAVA (working on my first real program) been looking for a solution for hours. I have hacked together a small program to download PDF files from a link. It works fine for most links but some of them just don't work.
The connection type for all the links that works show up as application/pdf but some links show a connection of text/html for some reason.
I keep trying to rewrite the code using whatever I can find online but I keep getting the same result.
import java.io.BufferedInputStream;
import java.io.BufferedOutputStream;
import java.io.FileOutputStream;
import java.net.ConnectException;
import java.net.URL;
import java.net.URLConnection;
public class Main {
public static void main(String[] args) throws Exception {
String link = "https://www.menards.com/main/items/media/UNITE051/SDS/SpectracideVegetationKillerReadyToUse2-228-714-8845-SDS-Feb16.pdf";
String fileName = "File Name.pdf";
URL url1 = new URL(link);
try {
URLConnection urlConn = url1.openConnection();
byte[] buffer = new byte[1024];
double downloaded = 0.00;
int read = 0;
System.out.println(urlConn.getContentType()); // This shows as text/html but it should be PDF
FileOutputStream fos1 = new FileOutputStream(fileName);
BufferedInputStream is1 = new BufferedInputStream(urlConn.getInputStream());
BufferedOutputStream bout = new BufferedOutputStream(fos1, 1024);
try {
while ((read = is1.read(buffer, 0, 1024)) >= 0) {
bout.write(buffer, 0, read);
downloaded += read;
}
bout.close();
fos1.flush();
fos1.close();
is1.close();
} catch (Exception e) {}
} catch (Exception e) {}
}
}
I need to be able to download the PDF from the link in the code.
This is what is saved in a text document of the PDF:
<html>
<head>
<META NAME="robots" CONTENT="noindex,nofollow">
<script src="/_Incapsula_Resource?SWJIYLWA=5074a744e2e3d891814e9a2dace20bd4,719d34d31c8e3a6e6fffd425f7e032f3">
</script>
<body>
</body></html>
The website implemented a check to make sure I was using a browser. I copied the user agent from chrome and it allowed me to download the PDF.
The URL that you are fetching doesn't point to a PDF file. It is pointing to a HTML file which embeds the PDF file. You probably need to closely look at what is the URL to PDF file. You code seems alright.
Just do a cURL on the URL and see. It will most probably return a HTML file.
i want to get files inside a folder when my application is running, so i know that i need to get it as resouce, if i will get it as file, it wont work, so it what i did.
jaxbContext = JAXBContext.newInstance(Catalogo.class);
jaxbUnmarshaller = jaxbContext.createUnmarshaller();
InputStream resourceAsStream = getClass().getClassLoader().getResourceAsStream("catalogos/");
BufferedReader br = new BufferedReader(new InputStreamReader(resourceAsStream));
String line;
try {
while((line = br.readLine()) != null){
InputStream resourceAsStream1 = getClass().getClassLoader().getResourceAsStream("catalogos/"+line);
tempCat = (Catalogo) jaxbUnmarshaller.unmarshal(resourceAsStream1);
if(tempCat != null){
codigoCurso = String.valueOf(tempCat.getCourse().getId());
nomeDoCurso = dados.get(codigoCurso);
anoCatalogo = String.valueOf(tempCat.getAno());
if(nomeDoCurso == null){
dados.put(codigoCurso, tempCat.getCourse().getNome());
}
anos.add(anoCatalogo);
}
}
What i want to do is, get all files inside a folder (/catalogos/) and loop through and unmarshall each to an object so i will be able to access the property i need. So, when i run this with netbeans, works perfectly, but when i build and run the jar, i dont get the same result i've got using netbeans, i mean, the data is not where i expected.
The following example demonstrates how to get files from a directory in current runnable jar file and read these files contents.
Assume you have a NetBeans project with name "FolderTestApp". Do the following steps:
In your project root folder FolderTestApp\ create folder myFiles.
Copy your catalogos folder to the FolderTestApp\myFiles\
myFiles folder is necessary to preserve catalogos folder in your jar file structure when project jar is being generated. myFiles folder will disappear from jar file, but catalogos folder will remain.
If you don't do these steps, and place catalogos directly to the project folder (not as a child folder for myFiles), then your files from catalogos folder will be placed to the root of your jar file.
Add myFiles folder as a source folder in netbeans project properties.
Assume your property files contain the following contents:
file1.properties:
key11=value11
key12=value12
key13=value13
file2.properties:
key21=value21
key22=value22
key23=value23
Please note, that code below is not optimized. It is plain'n'dirty proof of concept to show, how to solve your task.
Add the following class to your project:
package folderapp;
import java.io.BufferedReader;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.net.URISyntaxException;
import java.util.Properties;
import java.util.Set;
import java.util.zip.ZipEntry;
import java.util.zip.ZipInputStream;
public class FolderTestApp {
public static void main(String[] args) throws URISyntaxException, IOException {
new FolderTestApp();
}
public FolderTestApp() throws URISyntaxException, IOException {
// determining the running jar file location
String jarFilePath = getClass().getProtectionDomain().
getCodeSource().getLocation().toURI().getPath();
// note, that the starting / is removed
// because zip entries won't start with this symbol
String zipEntryFolder = "catalogos/";
try (ZipInputStream zipInputStream
= new ZipInputStream(new FileInputStream(jarFilePath))) {
ZipEntry zipEntry = zipInputStream.getNextEntry();
while (zipEntry != null) {
System.out.println("processing: " + zipEntry.getName());
if (zipEntry.getName().startsWith(zipEntryFolder)) {
// directory "catalogos" will appear as a zip-entry
// and we're checking this condition
if (!zipEntry.isDirectory()) {
// adding symbol / because it is required for getResourceAsStream() call
printProperties("/" + zipEntry.getName());
}
}
zipEntry = zipInputStream.getNextEntry();
}
}
}
public void printProperties(String path) throws IOException {
try (InputStream is = getClass().getResourceAsStream(path)) {
InputStreamReader fr = new InputStreamReader(is);
BufferedReader br = new BufferedReader(fr);
Properties properties = new Properties();
properties.load(br);
System.out.println("contents from: " + path + "\n");
Set<Object> keySet = properties.keySet();
for (Object key : keySet) {
System.out.println(key + " = " + properties.get(key));
}
System.out.println("---------------------------------------");
}
}
}
Set this class as the main class in your project settings (Run section).
And build your project via menu: Run - Build.
As your project has been built, open FolderTestApp/dist folder, where your generated jar is located and run this jar file:
That's it :)
I've been trying for days to do this and got absolutely nowhere. I know it can be done, but I've been trawling SO for answers and got nothing working.
Upload a picture using my REST client
Insert that uploaded picture into the MySQL database.
What I have tried:
Following Load_File doesn't work, I'm using OS X so I don't know how to change ownership of folders etc... how do I do this? I never got an answer in my last post about this. How do I do this?
I've also tried doing it another way: http://examples.javacodegeeks.com/enterprise-java/rest/jersey/jersey-file-upload-example/
This does not work at all. I keep getting the error described in this post: Jersey REST WS Error: "Missing dependency for method... at parameter at index X", but the answer doesn't help me as I still don't know what it should be...
Can anyone please guide me through it?
I'm using a Jersey REST client in Java. Many of the tutorials to do this mention a pom.xml file, I don't have one or know what it is.
Thank you,
Omar
EDIT:
This is the file upload:
package com.omar.rest.apimethods;
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.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("/files")
public class FileUpload {
private String uploadLocationFolder = "/Users/Omar/Pictures/";
#POST
#Path("/upload")
#Consumes(MediaType.MULTIPART_FORM_DATA)
public Response uploadFile(
#FormDataParam("file") InputStream fileInputStream,
#FormDataParam("file") FormDataContentDisposition contentDispositionHeader) {
String filePath = "/Users/Omar/Pictures/" + contentDispositionHeader.getFileName();
// save the file to the server
saveFile(fileInputStream, filePath);
String output = "File saved to server location : " + filePath;
return Response.status(200).entity(output).build();
}
// save uploaded file to a defined location on the server
private void saveFile(InputStream uploadedInputStream,
String serverLocation) {
try {
OutputStream outpuStream = new FileOutputStream(new File(serverLocation));
int read = 0;
byte[] bytes = new byte[1024];
outpuStream = new FileOutputStream(new File(serverLocation));
while ((read = uploadedInputStream.read(bytes)) != -1) {
outpuStream.write(bytes, 0, read);
}
outpuStream.flush();
outpuStream.close();
} catch (IOException e) {
e.printStackTrace();
}
}
}
Schema for the table (one I created for testing):
image_id: int auto-incrementing PK, picture: BLOB.
I could make it a file link and just load the image on my website but I can't even get that far yet.
I would recommend storing your image in some kind of cheap, well permissioned flat storage like network storage, and then storing a path to that storage location in the database. If you're storing your image as a blob, the database is going to do something similar to this already anyways, but I believe there will be some overhead involved with making the database manage storing and retrieving these images. These images will eat through a lot of your database's disk space, and if you want to add more space for images, adding space to flat storage should be easier than adding space to a database.
I am trying to read a file on HDFS and copy the content of the file into a newly created local file using the following java program. FYI, I have installed hadoop single node cluster on my machine.
HdfsCli.java
package com;
import java.io.BufferedOutputStream;
import java.io.File;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.OutputStream;
import org.apache.hadoop.conf.Configuration;
import org.apache.hadoop.fs.FSDataInputStream;
import org.apache.hadoop.fs.FileSystem;
import org.apache.hadoop.fs.Path;
public class HdfsCli {
public void readFile(String file) throws IOException {
Configuration conf = new Configuration();
String hadoopConfPath = "\\opt\\hadoop\\etc\\hadoop\\";
conf.addResource(new Path(hadoopConfPath + "core-site.xml"));
conf.addResource(new Path(hadoopConfPath + "hdfs-site.xml"));
conf.addResource(new Path(hadoopConfPath + "mapred-site.xml"));
FileSystem fileSystem = FileSystem.get(conf);
// For the join type of queries, output file in the HDFS has 'r' in it.
// String type="r";
Path path = new Path(file);
if (!fileSystem.exists(path)) {
System.out.println("File " + file + " does not exists");
return;
}
FSDataInputStream in = fileSystem.open(path);
String filename = file.substring(file.lastIndexOf('/') + 1,
file.length());
OutputStream out = new BufferedOutputStream(new FileOutputStream(
new File("/home/DAS_Pig/" + filename)));
byte[] b = new byte[1024];
int numBytes = 0;
while ((numBytes = in.read(b)) > 0) {
out.write(b, 0, numBytes);
}
conf.clear();
in.close();
out.close();
fileSystem.close();
}
public static void main(String[] args) throws IOException {
// TODO Auto-generated method stub
HDFSClient hc = new HDFSClient();
hc.readFile("hdfs://localhost:9000//DasData//salaries.txt");
System.out.println("Successfully Done!");
}
}
However, when I am running this code, the following error is coming:
Exception in thread "main" org.apache.hadoop.ipc.RemoteException: Server IPC version 9 cannot communicate with client version 4
at org.apache.hadoop.ipc.Client.call(Client.java:1066)
at org.apache.hadoop.ipc.RPC$Invoker.invoke(RPC.java:225)
at com.sun.proxy.$Proxy1.getProtocolVersion(Unknown Source)
at org.apache.hadoop.ipc.RPC.getProxy(RPC.java:396)
at org.apache.hadoop.ipc.RPC.getProxy(RPC.java:379)
at org.apache.hadoop.hdfs.DFSClient.createRPCNamenode(DFSClient.java:119)
at org.apache.hadoop.hdfs.DFSClient.<init>(DFSClient.java:238)
at org.apache.hadoop.hdfs.DFSClient.<init>(DFSClient.java:203)
at org.apache.hadoop.hdfs.DistributedFileSystem.initialize(DistributedFileSystem.java:89)
at org.apache.hadoop.fs.FileSystem.createFileSystem(FileSystem.java:1386)
at org.apache.hadoop.fs.FileSystem.access$200(FileSystem.java:66)
at org.apache.hadoop.fs.FileSystem$Cache.get(FileSystem.java:1404)
at org.apache.hadoop.fs.FileSystem.get(FileSystem.java:254)
at org.apache.hadoop.fs.FileSystem.get(FileSystem.java:123)
at com.HDFSClient.readFile(HDFSClient.java:22)
at com.HdfsCli.main(HdfsCli.java:57)
I am newb in hadoop development. Can anyone guide me in resolving this?
Thank you!
Server and client versions are different. Looks like server version is 4.i, and client is 3.i. You have to upgrade client classpath libraries up to server version.
Delete all current hadoop jars depencies included in your project. Download a newer version of Hadoop. Configure a build path of your project with new jars.
In my auto updater application i am downloading a zipped file that contains the new MyApp.app application file. So i am downloading MyApp.zip.. Then i use this following class to try and unzip it:
package update;
import java.io.BufferedOutputStream;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.util.Enumeration;
import java.util.List;
import java.util.zip.ZipEntry;
import java.util.zip.ZipFile;
import java.util.zip.ZipInputStream;
public class UnZip {
public static final void copyInputStream(InputStream in, OutputStream out)
throws IOException
{
byte[] buffer = new byte[1024];
int len;
while((len = in.read(buffer)) >= 0)
out.write(buffer, 0, len);
in.close();
out.close();
}
public static final void unZipIt(String F1, String F2) {
Enumeration entries;
ZipFile zipFile;
try {
zipFile = new ZipFile(F1);
entries = zipFile.entries();
while(entries.hasMoreElements()) {
ZipEntry entry = (ZipEntry)entries.nextElement();
if(entry.isDirectory()) {
// Assume directories are stored parents first then children.
System.err.println("Extracting directory: " + entry.getName());
// This is not robust, just for demonstration purposes.
(new File(entry.getName())).mkdirs();
continue;
}
System.err.println("Extracting file: " + entry.getName());
copyInputStream(zipFile.getInputStream(entry),
new BufferedOutputStream(new FileOutputStream(entry.getName())));
}
zipFile.close();
} catch (IOException ioe) {
System.err.println("Unhandled exception:");
ioe.printStackTrace();
return;
}
}
}
However after the unzip the application wont launch.. any ideas?
Your executable file is most likely not flagged as executable. The trick is that .app "files" are in fact directories, so making them executable serves no practical purpose, you need to find the actual binary.
To do that, you need to open ./myApp.app/Contents/Info.plit and look for the CFBundleExecutable key: the associated string is the path of the executable file, relative to ./myApp.app/Contents/MacOS, I believe.
Once you've found that file, chmod +x it, and check whether your application still fails to start.
If it doesn't, problem solved.
If it does, try and open your application from the terminal through the open ./myApp.app command. If anything odd is printed, update your question with it and let us know what that was.
If all else fails, look into the Console application for interesting log entries - you can search for your application's name, see if anything comes up.