I am a Scala/PlayFramework noob here, so please be easy on me :).
I am trying to create an action (serving a GET request) so that when I enter the url in the browser, the browser should download the file. So far I have this:
def sepaCreditXml() = Action {
val data: SepaCreditTransfer = invoiceService.sepaCredit()
val content: HtmlFormat.Appendable = views.html.sepacredittransfer(data)
Ok(content)
}
What it does is basically show the XML in the browser (whereas I actually want it to download the file). Also, I have two problems with it:
I am not sure if using Play's templating "views.html..." is the best idea to create an XML template. Is it good/simple enough or should I use a different solution for this?
I have found Ok.sendFile in the Play's documentation. But it needs a java.io.File. I don't know how to create a File from HtmlFormat.Appendable. I would prefer to create a file in-memory, i.e. no new File("/tmp/temporary.xml").
EDIT: Here SepaCreditTransfer is a case class holding some data. Nothing special.
I think it's quite normal for browsers to visualize XML instead of downloading it. Have you tried to use the application/force-download content type header, like this?
def sepaCreditXml() = Action {
val data: SepaCreditTransfer = invoiceService.sepaCredit()
val content: HtmlFormat.Appendable = views.html.sepacredittransfer(data)
Ok(content).withHeaders("Content-Type" -> "application/force-download")
}
Scalding has a great utility to run an integration test for the job flow.
In this way the inputs and outputs are the in-memory buffer
val input = List("0" -> "This a a day")
val expectedOutput = List(("This", 1),("a", 2),("day", 1))
JobTest(classOf[WordCountJob].getName)
.arg("input", "input-data")
.arg("output", "output-data")
.source(TextLine("input-data"), input)
.sink(Tsv("output-data")) {
buffer: mutable.Buffer[(String, Int)] => {
buffer should equal(expectedOutput)
}
}.run
How can I transfare/write another code that will read input and write output to the real local file? Like FileTap/LFS in cascading - and not an in-memory approach
You might check out HadoopPlatformJobTest and the TypedParquetTupleTest.scala example which uses a local mini-cluster.
This unit test writes to a "MiniLocalCLuster" - While it's not directly a file, but accessible via reading the local minicluster with Hadoop filesystem.
Given you local file scenario, maybe you can copies the files with local reads to the mini-HDFS.
How can I change the current working directory from within a Java program? Everything I've been able to find about the issue claims that you simply can't do it, but I can't believe that that's really the case.
I have a piece of code that opens a file using a hard-coded relative file path from the directory it's normally started in, and I just want to be able to use that code from within a different Java program without having to start it from within a particular directory. It seems like you should just be able to call System.setProperty( "user.dir", "/path/to/dir" ), but as far as I can figure out, calling that line just silently fails and does nothing.
I would understand if Java didn't allow you to do this, if it weren't for the fact that it allows you to get the current working directory, and even allows you to open files using relative file paths....
There is no reliable way to do this in pure Java. Setting the user.dir property via System.setProperty() or java -Duser.dir=... does seem to affect subsequent creations of Files, but not e.g. FileOutputStreams.
The File(String parent, String child) constructor can help if you build up your directory path separately from your file path, allowing easier swapping.
An alternative is to set up a script to run Java from a different directory, or use JNI native code as suggested below.
The relevant OpenJDK bug was closed in 2008 as "will not fix".
If you run your legacy program with ProcessBuilder, you will be able to specify its working directory.
There is a way to do this using the system property "user.dir". The key part to understand is that getAbsoluteFile() must be called (as shown below) or else relative paths will be resolved against the default "user.dir" value.
import java.io.*;
public class FileUtils
{
public static boolean setCurrentDirectory(String directory_name)
{
boolean result = false; // Boolean indicating whether directory was set
File directory; // Desired current working directory
directory = new File(directory_name).getAbsoluteFile();
if (directory.exists() || directory.mkdirs())
{
result = (System.setProperty("user.dir", directory.getAbsolutePath()) != null);
}
return result;
}
public static PrintWriter openOutputFile(String file_name)
{
PrintWriter output = null; // File to open for writing
try
{
output = new PrintWriter(new File(file_name).getAbsoluteFile());
}
catch (Exception exception) {}
return output;
}
public static void main(String[] args) throws Exception
{
FileUtils.openOutputFile("DefaultDirectoryFile.txt");
FileUtils.setCurrentDirectory("NewCurrentDirectory");
FileUtils.openOutputFile("CurrentDirectoryFile.txt");
}
}
It is possible to change the PWD, using JNA/JNI to make calls to libc. The JRuby guys have a handy java library for making POSIX calls called jnr-posix. Here's the maven info
As mentioned you can't change the CWD of the JVM but if you were to launch another process using Runtime.exec() you can use the overloaded method that lets you specify the working directory. This is not really for running your Java program in another directory but for many cases when one needs to launch another program like a Perl script for example, you can specify the working directory of that script while leaving the working dir of the JVM unchanged.
See Runtime.exec javadocs
Specifically,
public Process exec(String[] cmdarray,String[] envp, File dir) throws IOException
where dir is the working directory to run the subprocess in
If I understand correctly, a Java program starts with a copy of the current environment variables. Any changes via System.setProperty(String, String) are modifying the copy, not the original environment variables. Not that this provides a thorough reason as to why Sun chose this behavior, but perhaps it sheds a little light...
The working directory is a operating system feature (set when the process starts).
Why don't you just pass your own System property (-Dsomeprop=/my/path) and use that in your code as the parent of your File:
File f = new File ( System.getProperty("someprop"), myFilename)
The smarter/easier thing to do here is to just change your code so that instead of opening the file assuming that it exists in the current working directory (I assume you are doing something like new File("blah.txt"), just build the path to the file yourself.
Let the user pass in the base directory, read it from a config file, fall back to user.dir if the other properties can't be found, etc. But it's a whole lot easier to improve the logic in your program than it is to change how environment variables work.
I have tried to invoke
String oldDir = System.setProperty("user.dir", currdir.getAbsolutePath());
It seems to work. But
File myFile = new File("localpath.ext");
InputStream openit = new FileInputStream(myFile);
throws a FileNotFoundException though
myFile.getAbsolutePath()
shows the correct path.
I have read this. I think the problem is:
Java knows the current directory with the new setting.
But the file handling is done by the operation system. It does not know the new set current directory, unfortunately.
The solution may be:
File myFile = new File(System.getPropety("user.dir"), "localpath.ext");
It creates a file Object as absolute one with the current directory which is known by the JVM. But that code should be existing in a used class, it needs changing of reused codes.
~~~~JcHartmut
You can use
new File("relative/path").getAbsoluteFile()
after
System.setProperty("user.dir", "/some/directory")
System.setProperty("user.dir", "C:/OtherProject");
File file = new File("data/data.csv").getAbsoluteFile();
System.out.println(file.getPath());
Will print
C:\OtherProject\data\data.csv
You can change the process's actual working directory using JNI or JNA.
With JNI, you can use native functions to set the directory. The POSIX method is chdir(). On Windows, you can use SetCurrentDirectory().
With JNA, you can wrap the native functions in Java binders.
For Windows:
private static interface MyKernel32 extends Library {
public MyKernel32 INSTANCE = (MyKernel32) Native.loadLibrary("Kernel32", MyKernel32.class);
/** BOOL SetCurrentDirectory( LPCTSTR lpPathName ); */
int SetCurrentDirectoryW(char[] pathName);
}
For POSIX systems:
private interface MyCLibrary extends Library {
MyCLibrary INSTANCE = (MyCLibrary) Native.loadLibrary("c", MyCLibrary.class);
/** int chdir(const char *path); */
int chdir( String path );
}
The other possible answer to this question may depend on the reason you are opening the file. Is this a property file or a file that has some configuration related to your application?
If this is the case you may consider trying to load the file through the classpath loader, this way you can load any file Java has access to.
If you run your commands in a shell you can write something like "java -cp" and add any directories you want separated by ":" if java doesnt find something in one directory it will go try and find them in the other directories, that is what I do.
Use FileSystemView
private FileSystemView fileSystemView;
fileSystemView = FileSystemView.getFileSystemView();
currentDirectory = new File(".");
//listing currentDirectory
File[] filesAndDirs = fileSystemView.getFiles(currentDirectory, false);
fileList = new ArrayList<File>();
dirList = new ArrayList<File>();
for (File file : filesAndDirs) {
if (file.isDirectory())
dirList.add(file);
else
fileList.add(file);
}
Collections.sort(dirList);
if (!fileSystemView.isFileSystemRoot(currentDirectory))
dirList.add(0, new File(".."));
Collections.sort(fileList);
//change
currentDirectory = fileSystemView.getParentDirectory(currentDirectory);
This is the first time i am integrating Email service with liftweb
I want to send Email with attachments(Like:- Documents,Images,Pdfs)
my code looking like below
case class CSVFile(bytes: Array[Byte],filename: String = "file.csv",
mime: String = "text/csv; charset=utf8; header=present" )
val attach = CSVFile(fileupload.mkString.getBytes("utf8"))
val body = <p>Please research the enclosed.</p>
val msg = XHTMLPlusImages(body,
PlusImageHolder(attach.filename, attach.mime, attach.bytes))
Mailer.sendMail(
From("vyz#gmail.com"),
Subject(subject(0)),
To(to(0)),
)
this code is taken from LiftCookbook its not working like my requirement
its working but only the Attached file name is coming(file.csv) no data in it(i uploaded this file (gsy.docx))
Best Regards
GSY
You don't specify what type fileupload is, but assuming it is of type net.liftweb.http. FileParamHolder then the issue is that you can't just call mkString and expect it to have any data since there is no data in the object, just a fileStream method for retrieving it (either from disk or memory).
The easiest to accomplish what you want would be to use a ByteArrayInputStream and copy the data to it. I haven't tested it, but the code below should solve your issue. For brevity, it uses Apache IO Commons to copy the streams, but you could just as easily do it natively.
val data = {
val os = new ByteArrayOutputStream()
IOUtils.copy(fileupload.fileStream, os)
os.toByteArray
}
val attach = CSVFile(data)
BTW, you say you are uploading a Word (DOCX) file and expecting it to automatically be CSV when the extension is changed? You will just get a DOCX file with a csv extension unless you actually do some conversion.
I am trying to read a property file from classpath using scala. But it looks like it won't work, it is different from java. The following 2 code snippet, one is java (working), another is scala (not working). I don't understand what is the difference.
// working
BufferedReader reader = new BufferedReader(new InputStreamReader(
Test.class.getResourceAsStream("conf/fp.properties")));
// not working
val reader = new BufferedReader(new InputStreamReader(
getClass.getResourceAsStream("conf/fp.properties")));
Exception in thread "main" java.lang.NullPointerException
at java.io.Reader.<init>(Reader.java:78)
at java.io.InputStreamReader.<init>(InputStreamReader.java:72)
at com.ebay.searchscience.searchmetrics.fp.conf.FPConf$.main(FPConf.scala:31)
at com.ebay.searchscience.searchmetrics.fp.conf.FPConf.main(FPConf.scala)
This code finally worked for me:
import java.util.Properties
import scala.io.Source
// ... somewhere inside module.
var properties : Properties = null
val url = getClass.getResource("/my.properties")
if (url != null) {
val source = Source.fromURL(url)
properties = new Properties()
properties.load(source.bufferedReader())
}
And now you have plain old java.util.Properties to handle what my legacy code actually needed to receive.
I am guessing that your BufferedReader is a java.io.BufferedReader
In that case you could simply do the following:
import scala.io.Source.fromUrl
val reader = fromURL(getClass.getResource("conf/fp.properties")).bufferedReader()
However, this leaves the question open as to what you are planning to do with the reader afterwards. scala.io.Source already has some useful methods that might make lots of your code superfluous .. see ScalaDoc
My prefered solution is with com.typesafe.scala-logging. I did put an application.conf file in main\resources folder, with content like:
services {
mongo-db {
retrieve = """http://xxxxxxxxxxxx""",
base = """http://xxxxxx"""
}
}
and the to use it in a class, first load the config factory from typesafe and then just use it.
val conf = com.typesafe.config.ConfigFactory.load()
conf.getString("services.mongo-db.base"))
Hope it helps!
Ps. I bet that every file on resources with .conf as extension will be read.
For reading a Properties file i'd recommend to use java.util.ResourceBundle.getBundle("conf/fp"), it makes life a little easier.
The NullPointerException you are seeing is caused by a bug in the underlying Java code. It could be caused by a mistyped file name.
Sometimes you get this error also if you're trying to load the resource with the wrong classloader.
Check the resource url carefully against your classpath.
Try Source.fromInputStream(getClass.getResourceAsStream(...))
Try Source.fromInputStream(getClass.getClassLoader.getResourceAsStream())
Maybe you are using other classloaders you can try?
The same story goes for Source.fromUrl(...)
If you're trying to load configuration files and you control their format, you should have a look at Typesafe's Config utility.
The Null Pointer Exception you are getting is from getResourceAsStream returning null. The following junit.scala snippet shows how there is a difference in class vs classloader. see What is the difference between Class.getResource() and ClassLoader.getResource()?. Here I assume fileName is the name of a file residing in the class path, but not a file next to the class running the test.
assertTrue(getClass.getClassLoader().getResourceAsStream(fileName) != null)
assertTrue(getClass.getClassLoader().getResourceAsStream("/" + fileName) == null)
assertTrue(getClass.getResourceAsStream(fileName) == null)
assertTrue(getClass.getResourceAsStream("/" + fileName) != null)