You’ll need to add a reference to the MSXML library: Dim sUrl As String Dim response As String Dim xmlhttp Set sUrl = “http://my.domain.com/service/operation/param” Set xmlhttp = Server.CreateObject(“MSXML2.ServerXMLHTTP”) xmlhttp.open “POST”, sURL, False xmlhttp.setRequestHeader “Content-Type”, “application/x-www-form-urlencoded” xmlhttp.send() Dim response As String = xmlhttp.responseText Set xmlhttp = Nothing
You can use ContentNegotiationConfigurer Firstly, you should override the configureContentNegotiation method in your configuration class: @Configuration @EnableWebMvc public class WebConfig extends WebMvcConfigurerAdapter { @Override public void configureContentNegotiation(ContentNegotiationConfigurer configurer) { configurer.favorPathExtension(false). favorParameter(true). defaultContentType(MediaType.APPLICATION_JSON). mediaType(“xml”, MediaType.APPLICATION_XML); } } favorParameter(true) – enabling favoring path expressions over parameter or accept headers. defaultContentType(MediaType.APPLICATION_JSON) – sets the default content type. this … Read more
You could use the AsyncClient() from the httpx library, as described in this answer, as well as this answer and this answer (have a look at those answers for more details on the approach demonstrated below). You can spawn a Client inside the startup event handler, store it on the app instance—as described here, as … Read more
Methods like getServerSideProps and getStaticProps are for fetching data on the server but they only work for page components inside the pages folder (the initial way of setting up routes in Next.js). Since Next.js 13, in the app directory we have Server Components, where you can fetch data directly in the component body. In your … Read more
Returning a File Response First, to return a file that is saved on disk from a FastAPI backend, you could use FileResponse (in case the file was already fully loaded into memory, see here). For example: from fastapi import FastAPI from fastapi.responses import FileResponse some_file_path = “large-video-file.mp4” app = FastAPI() @app.get(“/”) def main(): return FileResponse(some_file_path) … Read more
android.jar contains only stub implementation of the classes. It provides the means for you app to build, once you have your APK you must run it on an android device or emulator. If I’m not wrong you are trying to run on host’s JVM.
In FastAPI, as described in this answer, because endpoints are evaluated in order (see FastAPI’s about how order matters), it makes sure that the endpoint you defined first in your app—in this case, that is, /project/{project_id}/…—will be evaluated first. Hence, every time you call one of the other two endpoints, i.e., /project/details/… and /project/metadata/…, the … Read more
Introduce a new class as below @XmlRootElement(name = “responseList”) public class ResposeList { private List<Object> list; public List<Object> getList() { return list; } public void setList(List<Object> list) { this.list = list; } } and set the list as below @GET @Path(“/get”) @Produces(MediaType.APPLICATION_XML) public ResposeList addObjects() { Book book = new Book(); book.setName(“Here is the Game”); … Read more
Please find some code that might help you to pass a file along with its details in a single call to the REST service: First you would need something called a MultipartParser as shown below: using System; using System.Collections.Generic; using System.IO; using System.Text; using System.Text.RegularExpressions; namespace SampleService { public class MultipartParser { private byte[] requestData; … Read more
I think you should change your java JDK change jvm v8 for jdk7. This link can help you: Is it possible to use Java 8 for Android development? Other possible issue its dependency error, clean gradle before build. And change your jackson library for this: compile ‘com.fasterxml.jackson.core:jackson-databind:2.2.+’ compile ‘com.fasterxml.jackson.core:jackson-core:2.2.+’ compile ‘com.fasterxml.jackson.core:jackson-annotations:2.2.+’