I am trying to create a temp directory and file underneath it. Here is my code snippet:
var tempPath = System.getProperty("java.io.tmpdir")
val myDir = new File(tempPath.concat(scala.util.Random.nextString(10).toString))
myDir.mkdir()
val tempFile = new File(myDir.toString+"/temp.log")
This code is working fine. However I am wondering if there is any better way of doing this, please provide your comments.
Java has existing methods that can do this for you, like Files.createTempFile, Files.createTempDirectory and their overloads.
You can find some examples in this blog post.
Related
I am currently writing tests for a function that takes file paths and loads a dataset from them. I am not able to change the function. To test it currently I am creating files for each run of the test function. I am worried that simply making files and then deleting them is a bad practice. Is there a better way to create temporary test files in Scala?
import java.io.{File, PrintWriter}
val testFile = new File("src/main/resources/temp.txt" )
val pw = new PrintWriter(testFile)
val testLines = List("this is a text line", "this is the next text line")
testLines.foreach(pw.write)
pw.close
// test logic here
testFile.delete()
I would generally prefer java.nio over java.io. You can create a temporary file like so:
import java.nio.Files
Files.createTempFile()
You can delete it using Files.delete. To ensure that the file is deleted even in the case of an error, you should put the delete call into a finally block.
I have a zip file in hdfs and i need to add a file to the zip and save in the same HDFS location. Any examples would be appreciated.
I have the following code.
val filePattern = s"${hdfsFolderPath}/${filePath}.txt"
val zipFilePath = hdfsWrapper.getFileNameFromPattern(s"${targetFilePath}/*.zip")
if (hdfsWrapper.filter(filePattern).size() > 0)
{
Try
{
val zipEntry = new ZipEntry(filePattern)
val zos: ZipOutputStream = new ZipOutputStream(new FileOutputStream(zipFilePath))
zos.putNextEntry(zipEntry)
zos.closeEntry()
zos.close()
}
}
Would like to know if above code is right?
I believe that your code would result in the zip file being replaced with a new one containing just the new file. See Appending files to a zip file with Java for an example of adding a file to a zip archive.
I'm not extremely familiar with HDFS, but I suspect that you can't write to that directly either, you probably have to create the new zip file and then replace it in HDFS
I want to read .doc file in scala. I tried using apache.poi library for this but the method HWPFDocument(java.io.InputStream istream) accepts java io stream.
If anyone can shed some light on this, that would great!
So, here is a teaser to get you started:
val fis = new FileInputStream("/path/to/file/doc.doc")
val doc = new HWPFDocument(fis)
val we = new WordExtractor(doc)
val paras = we.getParagraphText()
You can use InputStream in Scala, just as any other Java class/interface.
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")
}
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.