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.
Related
My scala-play api provides endpoints to return a file as a stream via the Ok.chunked-function.
I now want to be able to allow the download of multiple files as a zip archive.
I want to create a zip file as a stream which play should directly return as a filestream.
Meaning without the need to temporarly save the zip-file on the disc and serving it while it is being created.
What would be a good way to implemente a function that creates this stream?
I solved the issue by using akkas alpakka.
import akka.stream.alpakka.file.ArchiveMetadata
import akka.stream.alpakka.file.scaladsl.Archive
import akka.stream.scaladsl.Source
import akka.util.ByteString
val fileSource: Source[ByteString, _] = FileIO.fromPath(path)
val tupelWithMetaData = (ArchiveMetadata(s"file.txt"), fileSource)
val stream: Source[ByteString, _] = Source(List(tupelWithMetaData)).via(Archive.zip())
First I create a akka ByteString source. The source is used inside the Tupel2 with some ArchiveMetadata. This tuple can than be used to create a new source which can be connected to alpakkas Archive.zip().
The resulting stream can than be used with plays Ok.chunked.
I hope this solution might help you if you have the same question.
I have the following piece of scala code that will write text to a txt file sitting locally.
// PrintWriter
import java.io._
val pw = new PrintWriter(new File("resources/myfile.txt"))
pw.write("Test text")
pw.close
How do i get this to work on azure blob storage?
I have tried:
val pw = new PrintWriter(new File("wasb://[my container name]#[My Storage account].blob.core.windows.net/resources/myfile.txt"))
But it doesn't work.
What am i doing wrong? By the way, for the sake of this example, I'm keeping it simple. In reality, I am outputting more meaningful data.
Thanks
Con
You cannot create a file on Azure blog storage using the java.io or nio package. You need to use their REST API or their SDK.
https://learn.microsoft.com/en-us/rest/api/storageservices/create-file
https://learn.microsoft.com/en-us/azure/storage/blobs/storage-java-how-to-use-blob-storage
Iam currently having a json object say student.json. The Structure looks something like this
{"serialNo":"1","name":"Rahul"}
{"serialNo":"2","name":"Rakshith"}
case class Student(serialNo:Int,name:String)
student.json is a huge file which Iam planning to parse through a spark job. And the snippet :
import play.api.libs.json.{ Json, JsObject, JsString }
.....
.....
for(jsonLine <-sc.textFile("student.json")
student<- Json.parse(jsonLine).asOpt[Student])
yield(student.serialNumber -> student.name)
Is there a better way to do this??
If student.json is a huge file, and each line is just a valid json object, you should do:
val myRdd = sc.textFile("student.json").map(l=> Json.parse(l).asOpt[Student])
If you want to get the RDD to your local master, you can:
val students = myRdd.collect()..// then you can do operate it in the old fashion way.
I saw you are importing play.api.libs.json which is from the Play Framework. I don't think running a Spark program in a web application is a good idea...
http://spark.apache.org/docs/latest/sql-programming-guide.html#interoperating-with-rdds
The link shows how to change txt file into RDD, and then change to Dataframe.
So how to deal with binary file ?
Ask for an example ,Thank you very much .
There is a similar question without answer here : reading binary data into (py) spark DataFrame
To be more detail, I don't know how to parse the binary file .for example , I can parse txt file into lines or words like this:
JavaRDD<Person> people = sc.textFile("examples/src/main/resources/people.txt").map(
new Function<String, Person>() {
public Person call(String line) throws Exception {
String[] parts = line.split(",");
Person person = new Person();
person.setName(parts[0]);
person.setAge(Integer.parseInt(parts[1].trim()));
return person;
}
});
It seems that I just need the API that could parse the binary file or binary stream like this way:
JavaRDD<Person> people = sc.textFile("examples/src/main/resources/people.bin").map(
new Function<String, Person>() {
public Person call(/*stream or binary file*/) throws Exception {
/*code to construct every row*/
return person;
}
});
EDIT:
The binary file contains structure data (relational database 's table,the database is a self-made database) and I know the meta info of the structure data.I plan to change the structure data into RDD[Row].
And I could change every thing about the binary file when I use FileSystem's API (http://hadoop.apache.org/docs/current/api/org/apache/hadoop/fs/FileSystem.html) to write the binary stream into HDFS .And The binary file is splittable. I don't have any idea to parse the binary file like the example code above . So I cann't try anything so far.
There is a binary record reader that is already available for spark (I believe available in 1.3.1, atleast in the scala api).
sc.binaryRecord(path: string, recordLength: int, conf)
Its on you though to convert those binaries to an acceptable format for processing.
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")
}