I have a question please:
i would like to read multiple xml files (file1.xml, file2.xml....), for each file i read in memory then i edit and send request. for reading and editing its work fine without any problem, but when i send request:
//some declarations
.......
var payload=""
try{
myscenario= scenario ("scenario Name")
.exec{mysession : Session =>
//open files, list all files
for (file <-listofmyfile) {
//some regex that i reuse in payload
// payload contains the change that i would like to add for each xml file
payload=(// some function and text message.... "XXX")
}
mysession
}
.exec("protocol used".requestReply.message(StringBody( session =>
session(payload).as[String]))
.check(substring("XXX"))
)
}
setUp(myscenario.inject....)}
i get this 2 errors :
[ERROR] i.g.c.s.w.BufferedFileChannelWriter - Buffer overflow, you shouldn't be logging that much data. Truncating.
[ERROR] i.g.t.a.RequestReply - crashed on session Session
java.util.NoSuchElementException: No attribute named 'payload' is defined
Related
The user sends a post request, than based on that post body I create an Excel file (.xlsx) and want to send that file back, without storage of that file itself.
def writeAsync(out: OutputStream): Unit = {
Future {
val wb = new XSSFWorkbook
val sheet1: Sheet = wb.createSheet("sheet1");
val os = new ByteArrayOutputStream()
wb.write(os)
os.writeTo(out)
out.close()
wb.close()
}
}
...
pathPrefix("createAndDownloadExcel") {
post {
...
val generatedFileName = "customGeneratedFileName.xlsx" // <-- this name should the file name be like
val (out, source) = StreamConverters.asOutputStream().preMaterialize()
writeAsync(out)
complete(HttpEntity(ContentTypes.`application/octet-stream`, source))
}
}
The response has the excel content with the file name: "createAndDownloadExcel", but I would like it to have the file name based on the individual generated file name.
The name will be later manually generated based on the POST body, whereby a simple change in pathPrefix("fixedName.xlsx") does not work for my needs.
How can I solve this, being able to give a dynamic file name for that returned OutputStream?
"org.apache.poi" % "poi-ooxml" % "5.2.0"
Try adding response header Content-Disposition.
The first parameter in the HTTP context is either inline (default value, indicating it can be displayed inside the Web page, or as the Web page) or attachment (indicating it should be downloaded; most browsers presenting a 'Save as' dialog, prefilled with the value of the filename parameters if present).
import akka.http.scaladsl.model.headers.ContentDispositionTypes.attachment
import akka.http.scaladsl.model.headers.`Content-Disposition`
....
respondWithHeader(`Content-Disposition`(attachment, Map("filename" -> "customGeneratedFileName.xlsx"))) {
complete(HttpEntity(ContentTypes.`application/octet-stream`, source))
}
I have created a small Akka HTTP server to receive an uploaded file.
path("upload"){
uploadedFile("csv"){
case (metadata, file) =>{
println("file received " + file.length() );
complete("hahahah")
}
}
}
I can receive the file successfully but I cannot access other fields in this POST request. The field "csv" contains the file to be uploaded, while another field, "name", contains the user-defined name. I cannot access the data in "name". Can anybody give me some clues about how to get it?
You can use fromFields('user) to get user name. But unfortunately you will get this exception: java.lang.IllegalStateException: Substream Source cannot be materialized more than once It's known issue: https://github.com/akka/akka-http/issues/90
As workaround, you can use toStrictEntity directive:
toStrictEntity(3.seconds) {
formFields('user) { (user) =>
uploadedFile("csv") {
case (metadata, file) => {
println(s"file received by $user" + file.length())
complete("hahahah")
}
}
}
}
}
I don't think it's a good idea because you will read the entire request entity into memory and it works if you have the small entity.
As a better solution, you can implement your own uploadedFile directive that will extract needed parts and fields from your multipart form data, see uploadedFile source code as example: https://github.com/akka/akka-http/blob/v10.0.10/akka-http/src/main/scala/akka/http/scaladsl/server/directives/FileUploadDirectives.scala
I have run into a problem with a Play 2.1.0 form that contains a file upload and an additional input field. I use
def uploadTaxonomy() = Action(parse.multipartFormData) {
implicit request =>
request.body.file("xml").map { file =>
val xml = scala.io.Source.fromFile(file.ref.file).mkString
taxonomyForm.bindFromRequest().fold(
formWithErrors => BadRequest(views.html.index(formWithErrors)),
result => {
Taxonomies.create(result._1, xml)
Redirect(routes.Application.index())
}
)
}.getOrElse {
Redirect(routes.Application.index())
}
}
and my form is this:
val taxonomyForm = Form(
tuple(
"label" -> text,
"xml" -> text
)
)
The problem is that bindFromRequest() always fails (causing a bad request to be returned to the client).
Does anybody have an idea where the problem might lie?
Note: I am aware that there is a bug in 2.1.0 that manifests when no files are selected in an upload field; it does not seem to be related, however.
As far as I know the xml should not be part of the form definition as you get it directly from the request body.
I have a Mirth channel that set up as a web service listener, it receives an ID, build an HL7 query message and send this query and eventually get back an HL7 response.
Channel Name: QueryChanel
Source Connector Type: Web Service Listener
Destination Connector Name: QueryToVista
Destination connector Type:LLP Sender.
This is the typical HL7 response I receive back from my query is as follow:
MSH|~|\&|VAFC RECV|FACILITY|VAFC TRIGGER||20121011141136-0800||ADR~A19|58269|D|2.4|||NE|NE|USA
MSA|AA|1234|
QRD|20121011051137|R|I|500000001|||1^ICN|***500000001***|ICN|NI|
EVN|A1|20121004064809-0800||A1|0^^^^^^^^USVHA\\0363^L^^^NI^TEST FACILITY ID\050\L|20121004064809-0800|050
PID|1|500000001V075322|500000001V075322^^^USVHA\\0363^NI^VA FACILITY ID\050\L~123123123^^^USSSA\\0363^SS^TEST FACILITY ID\050\L~9^^^USVHA\\0363^PI^VA FACILITY ID\050\L||JOHN^DOE^^^^^L|""|19800502|M||""|""^""^""^""^""^^P^""^""~^^""^""^^^N|""|""|""||S|""||123123123|||""|""||||||""||
PD1|||SOFTWARE SERVICE^D^050
ZPD|1||||||||||||||||""
I can get all the above to return if I set my Source's Response From parameter to QueryToVista
However, I want to return only the value 500000001 from the above message. I've tried to play around with the transformer in the QueryChanel destination without success.
Update:
I tried to add a javascriptwriter connector after the QueryToVista connector in the same channel as follow:
var destination = responseMap.get('QueryToVista');
var responseMessage = destination.getMessage();
//Fails with following error: TypeError: Cannot read property "QRD.4" from undefined
var customack = ResponseFactory.getSuccessResponse(responseMessage['QRD']['QRD.4'] ['QRD.4.1'].toString())**
//work but send the whole HL7 message
var customack = ResponseFactory.getSuccessResponse(responseMessage.toString())**
responseMap.put('Barcode', customack);
I can't seem to use the normal transformation to retrieve the element at all.
Thank you.
You're on the right track, but your update illustrates a couple of issues. However, your basic approach of using two destinations is valid, so long as "Synchronize channel" is checked on the Summary tab.
Issue 1
In your example, the HL7 response you are wanting to parse is in pipe delimited HL7 form. In order to access the elements using E4X notation (eg. responseMessage['QRD']['QRD.4']['QRD.4.1']) you must first convert it into an E4X XML object. This can be done in two steps.
Convert the pipe delimited HL7 string into an XML string.
Convert the XML string into an E4X XML object
In a Javascript transformer of the JavaScript Writer (not the Javascript Writer script itself)
var response = responseMap.get("QueryToVista");
var responseStatus = response.getStatus();
// Get's the pipe delimited HL7 string
var responseMessageString = response.getMessage();
if (responseStatus == "SUCCESS")
{
// converts the pipe delimited HL7 string into an XML string
// note: the SerializeFactory object is available for use in transformer
// scripts, but not in the Javascript destination script itself
var responseMessageXMLString = SerializerFactory.getHL7Serializer(false,false,true).toXML(responseMessageString);
// convert the XML string into an E4X XML object
var responseMessageXMLE4X = new XML(responseMessageXMLString);
// grab the value you want
var ack_msg = responseMessageXMLE4X['QRD']['QRD.4']['QRD.4.1'].toString();
channelMap.put('ack_msg', ack_msg)
}
else
{
// responseStatus probably == "FAILURE" but I'm not sure of the full range of possibilities
// take whatever failure action you feel is appropriate
}
Edit**
I don't believe there is an Issue 2. After reviewing your own approach, I played a bit further, and believe I have confirmed that your approach was indeed correct for generating the SOAP reponse. I'm editing this section to reflect simpler code that still works.
In the Javascript Writer script
var barcode = channelMap.get('ack_msg');
var mirthResponse = ResponseFactory.getSuccessResponse(barcode);
responseMap.put('Barcode', mirthResponse);
Thank you very much csj,
I played around and got mine to work and looking at your solution, you pointed out my bottle neck to the issue as well which is the XML part, I did not realize you have to cast it into XML as per the new XML when you already call toXML function :)
Here is my script, though basic I thought I post it up for anyone find it useful down the road.
var destination = responseMap.get('QueryToVista');
var responseMessage = destination.getMessage();
var Xmsg = new XML(SerializerFactory.getHL7Serializer().toXML(responseMessage));
var xml_msg = '<?xml version="1.0" encoding="utf-8" ?>'+
'<XML><Patient Name="'+Xmsg['PID']['PID.5']['PID.5.1']+
'" Barcode="'+Xmsg['QRD']['QRD.8']['QRD.8.1']+'" /></XML>';
var sResp = ResponseFactory.getSuccessResponse(xml_msg)
responseMap.put('Response', sResp);
What is the preferred method to serve a static file from a Play Framework 2 Scala controller?
The file is bundled with my application, so it's not possible to hardcode a filesystem absolute /path/to/the/file, because its location depends on where the Play app happens to be installeld.
The file is placed in the public/ dir, but not in app/assets/, because I don't want Play to compile it.
(The reason I don't simply add a route to that file, is that one needs to login before accessing that file, otherwise it's of no use.)
Here is what I've done so far, but this breaks on my production server.
object Application ...
def viewAdminPage = Action ... {
... authorization ...
val adminPageFile = Play.getFile("/public/admin/index.html")
Ok.sendFile(adminPageFile, inline = true)
}
And in my routes file, I have this line:
GET /-/admin/ controllers.Application.viewAdminPage
The problem is that on my production server, this error happens:
FileNotFoundException: app1/public/admin/index.html
Is there some other method, rather than Play.getFile and OK.sendFile, to specify which file to serve? That never breaks in production?
(My app is installed in /some-dir/app1/ and I start it from /some-dir/ (without app1/) — perhaps everything would work if I instead started the app from /some-dir/app1/. But I'd like to know how one "should" do, to serve a static file from inside a controller? So that everything always works also on the production servers, regardless of from where I happen to start the application)
Check Streaming HTTP responses doc
def index = Action {
Ok.sendFile(
content = new java.io.File("/tmp/fileToServe.pdf"),
fileName = _ => "termsOfService.pdf"
)
}
You can add some random string to the fileName (individual for each logged user) to avoid sharing download link between authenticated and non-authinticated users and also make advanced download stats.
I did this: (but see the Update below!)
val fileUrl: java.net.URL = this.getClass().getResource("/public/admin/file.html")
val file = new java.io.File(adminPageUrl.toURI())
Ok.sendFile(file, inline = true)
(this is the controller, which is (and must be) located in the same package as the file that's being served.)
Here is a related question: open resource with relative path in java
Update
Accessing the file via an URI causes an error: IllegalArgumentException: URI is not hierarchical, if the file is then located inside a JAR, which is the case if you run Play like so: play stage and then target/start.
So instead I read the file as a stream, converted it to a String, and sent that string as HTML:
val adminPageFileString: String = {
// In prod builds, the file is embedded in a JAR, and accessing it via
// an URI causes an IllegalArgumentException: "URI is not hierarchical".
// So use a stream instead.
val adminPageStream: java.io.InputStream =
this.getClass().getResourceAsStream("/public/admin/index.html")
io.Source.fromInputStream(adminPageStream).mkString("")
}
...
return Ok(adminPageFileString) as HTML
Play has a built-in method for this:
Ok.sendResource("public/admin/file.html", classLoader)
You can obtain a classloader from an injected Environment with environment.classLoader or from this.getClass.getClassLoader.
The manual approach for this is the following:
val url = Play.resource(file)
url.map { url =>
val stream = url.openStream()
val length = stream.available
val resourceData = Enumerator.fromStream(stream)
val headers = Map(
CONTENT_LENGTH -> length.toString,
CONTENT_TYPE -> MimeTypes.forFileName(file).getOrElse(BINARY),
CONTENT_DISPOSITION -> s"""attachment; filename="$name"""")
SimpleResult(
header = ResponseHeader(OK, headers),
body = resourceData)
The equivalent using the assets controller is this:
val name = "someName.ext"
val response = Assets.at("/public", name)(request)
response
.withHeaders(CONTENT_DISPOSITION -> s"""attachment; filename="$name"""")
Another variant, without using a String, but by streaming the file content:
def myStaticRessource() = Action { implicit request =>
val contentStream = this.getClass.getResourceAsStream("/public/content.html")
Ok.chunked(Enumerator.fromStream(contentStream)).as(HTML)
}