I am new to scalajs, I have a finch endpoint in my backend project serving a scalatag html generated from frontend project
val apiEndpoints: Endpoint[Response] = get("index") {
val res = Response()
res.setContentString(scalaTagUI.toString())
Future(res)
}
in my Html I have the script tag
script(attr("type"):="text/javascript", attr("src"):="./target/scala-2.12/classes/assets/frontend-jsdeps.js")
The /index is working but it can't access the javascript files giving 404 - I can see the fastopt.js in my target/scala-2.12.....
The solution at the end was simple - I have added a specific endpoint in finch exclusively for js/css files. Differently if you use something like jetty you would add a folder like "webapp" containing js/css to your server context.
Anyway It works now but I wonder if is there a different maybe cleaner approach.
Related
I am trying to serve a PDF file that my SvelteKit app generates and allow a user to download it from an endpoint.
My project structure looks like this:
---------------------
/src/routes/downloads
---------------------
[file].ts
ABC.pdf
XYZ.pdf
My [file].ts endpoint looks like this:
import fs from 'fs'
// ----- GET -----
export async function get({ params }){
//console.log(params.file) -> ABC
var pdf = fs.readFileSync('./src/routes/downloads/'+params.file+'.pdf')
return{
status:200,
headers: {
"Content-type" : "application/pdf",
"Content-Disposition": "attachment; filename="+params.file+".pdf"
},
body: pdf
}
}
So then when I hit http://localhost:3000/downloads/ABC, the PDF file named ABC.pdf downloads.
But my readFileSync path isn't something that's going to work on production. As far as I know, there is no /src/routes folder.
How do I serve my file from a http://localhost:3000 url? Everything I've tried yields a 404 and it can't find the file. I'm also open to a different way of handling this scenario. This is just my best guess of how to do this in SvelteKit.
The recommended way to do this, for adapter-node, is to place your application data in a new folder under your project's root directory (ie. alongside /src and /static). You can then read files with a relative path: fs.readFile('./my-app-data/foo.txt').
For deployment, you just have to make sure to execute node build from the project root, as this guarantees that you have the same working directory during both development and production.
The static folder works, but it is not meant to carry application data—files in this folder represent new routes that are served directly to users, so this is not desirable if your generated files must be protected in any way. Even if they're meant to be public files, it still blurs what is supposed to be production and source data: should a new deploy overwrite all the files in static? If you're not careful, a naming clash could mean overwriting production data.
You can use import.meta.glob for this.
export async function get({ params }){
const file = `./${params.file}.pdf`;
const pdfs = import.meta.glob(('./*.pdf', { as: 'raw' });
const pdf = pdfs[file];
return {
status:200,
headers: {
"Content-type" : "application/pdf",
"Content-Disposition": "attachment; filename="+params.file+".pdf"
},
body: pdf
}
}
The import.meta.glob in combination with the as: 'raw' option will effectively embed the contents of each file in your resulting code. (this is purely server side so no worries about shipping to much to the client)
Note that this of course means that only files present during build can be served this way.
As #Stephane suggest, put your files under statics folder. This way you can serve directly through a reverse proxy, like Nginx
I have Swagger UI for API documentation, I use the same approach like in official specification for accessing it I use next URL:
http://localhost:9000/docs/swagger-ui/index.html?url=/assets/swagger.json
But I want to use http://localhost:9000/docs/ instead. I won't want to use WS for delegating, I would like to use single line in routes, like this:
GET /docs controllers.Assets.at(path:String="/public/lib/swagger-ui", file:String="index.html?url=/assets/swagger.json")
Or
GET /docs controllers.Assets.at(path:String="/public/lib/swagger-ui", file:String="index.html")
and http://localhost:9000/docs?url=/assets/swagger.json
What shold I change so it work?
You can't make shortness in route file for the URL /docs/swagger-ui/index.html?url=/assets/swagger.json because index.html is generated by swagger-ui plugin to public directory and requires access to files nearby (like js and css files). Play swagger-ui uses javascript for fetching json based description of your routes via URL parameter for further parsing this document to swagger-ui, in your case it's /assets/swagger.json endpoint.
I tried to make the mapping swagger's index.html file, so pass json location like URL parameter directly:
GET /swagger-ui controllers.Assets.at(path = "/public/lib/swagger-ui", file = "index.html")
Play couldn't render this page, and CSS wasn't found. I appended dummy mapping to every file in swagger's default directory /public/lib/swagger-ui:
GET /*file controllers.Assets.at(path = "/public/lib/swagger-ui", file)
Even after that Play couldn't properly render index.html.
How it can be solved without Play-Swagger:
Create directory public\swagger in your project;
Download all files from https://github.com/swagger-api/swagger-ui/tree/master/dist and copy them to public\swagger;
Copy your swagger.json (it's specification, right?) to public\specification;
Add to your routes file next line:
GET /docs
controllers.Assets.versioned(path="/public/specification", file: Asset
= "swagger.json")
I'm trying to make a Spring Boot Soap WebService application, and was following the Get Started (https://spring.io/guides/gs/producing-web-service/) example to learn how to do this.
I've created what I want, but I have two URL problems with this setup and I could not find what configuration should I change to fix this :
WSDL URL basic is localhost:8080/ws/countries.wsdl but anything like localhost:8080/ws/whatever/countries.wsdl is correct
service URL for SoapUI request is localhost:8080/ws but anything like localhost:8080/ws/whatever is correct
I know that this is a feature for Spring WS, but I want a fixed URL (without 'whatever' in both cases) and could not find what to change for this
There is no straight forward way to restrict the way you want.
SOAP service is not URL based.
SOAP message body describe the endpoint.
The thing you wanted is possible following way.
Changing URL mapping in ServletRegistrationBean to restrict URL access
Existing /ws/* mapping is the reason why all the /ws/whatever url successfully responded.
Change as new ServletRegistrationBean(servlet, "/ws");
Effect will be you can not request other than /ws URL
Now the problem is, you can not get WSDL by this mapping.
Solution to get WSDL
The DefaultWsdl11Definition is actually generating WSDL from XSD on every request.
Save countries.wsdl to resource folder as static WSDL file.
Remove DefaultWsdl11Definition bean.
Create a new SimpleWsdl11Definition bean as like
#Bean(name = "countries")
public SimpleWsdl11Definition orders() {
SimpleWsdl11Definition wsdl11Definition = new SimpleWsdl11Definition();
wsdl11Definition.setWsdl(new ClassPathResource("countries.wsdl"));
return wsdl11Definition;
}
Now add another static URL mapping in ServletRegistrationBean. As it will be finally look like new ServletRegistrationBean(servlet, "/ws", "/ws/countries.wsdl");
This practice is good for development phase as you can publish easily the changed definition. But it is recommended to use static-wsdl for production environment. Details ** here
Just change
return new ServletRegistrationBean(servlet, "/ws/*");
for example to
return new ServletRegistrationBean(servlet, new String[]{
"/ws/v1/countries.wsdl",
"/ws/v2/countries.wsdl"
});
I have the following Scala code to setup a Jetty server with Scalatra.
val server = new Server(8080)
val context = new WebAppContext()
context.setResourceBase("visualization")
context.addServlet(new ServletHolder(new CallTreeServlet(dataProvider)), "/*")
context.addServlet(new ServletHolder(new DataLoadingServlet(dataProvider)), "/*")
server.setHandler(context)
My problem is that it seems to work only if I register a single servlet.
If I register more than one, like I do in the code I posted, it loads only one of them.
Is it possible to load multiple servlets? I guess it is, but I can't figure out how.
If I try to load a page from the first servlet I got this error message that references only pages belonging to the second servlet:
Requesting "GET /callTrees" on servlet "" but only have:
GET /components
POST /load
POST /searchCallTrees
POST /selectPlugIn
To troubleshoot this, you should verify the servlet lifecycle. One convenient way to do this is to peruse the servlet container's logs to see what it reports while starting up the web application. It should tell you about each web app ( servlet context ) and each servlet . . .
However, I think I see what your problem is. Your servlet path mappings are kind of funky. It looks to me that you are mapping both servlets to receive ALL requests. This can't work, from a practical point of view, and might not work in terms of the servlet rules. From the servlet specification:
SRV.11.2
Specification of Mappings
In the Web application deployment descriptor, the following syntax is used to define
mappings:
• A string beginning with a ‘/’ character and ending with a ‘/*’ suffix is used
for path mapping.
• A string beginning with a ‘*.’ prefix is used as an extension mapping.
• A string containing only the ’/’ character indicates the "default" servlet of
the application. In this case the servlet path is the request URI minus the con-
text path and the path info is null.
• All other strings are used for exact matches only.
I suggest you make them both unique. As it looks now, you have them both at "/*" which is kind of like the "default servlet", but not . . .
Why not try "/first/" and "/second/" as a sanity check. Then move from there toward getting the configuration how you like.
I am trying to use HTTP to POST a file to an outside API from within a grails service. I've installed the rest plugin and I'm using code like the following:
def theFile = new File("/tmp/blah.txt")
def postBody = [myFile: theFile, foo:'bar']
withHttp(uri: "http://picard:8080/breeze/project/acceptFile") {
def html = post(body: postBody, requestContentType: URLENC)
}
The post works, however, the 'myFile' param appears to be a string rather than an actual file. I have not had any success trying to google for things like "how to post a file in grails" since most of the results end up dealing with handling an uploaded file from a form.
I think I'm using the right requestContentType, but I might have missed something in the documentation.
POSTing a file is not as simple as what you have included in your question (sadly). Also, it depends on what the API you are calling is expecting, e.g. some API expect files as base64 encoded text, while others accept them as mime-multipart.
Since you are using the rest plugin, as far as I can recall it uses the Apache HttpClient, I think this link should provide enough info to get you started (assuming you are dealing with mime-multipart). It shouldn't be too hard to change it around to work with your API and perhaps make it a bit 'groovy-ier'