如果你有一个java.io.InputStream对象,你应该如何处理该对象并产生一个String?
假设我有一个InputStream包含文本数据,并且我想将其转换为a String,所以例如我可以将其写入日志文件.
采取InputStream并将其转换为最简单的方法是String什么?
public String convertStreamToString(InputStream is) {
// ???
}
Run Code Online (Sandbox Code Playgroud) 我创建了一个Undertow服务器和一个处理程序来记录请求.我在检索请求体时遇到问题HttpServerExchange.
在LoggingHandler课堂上,我的身体没有问题.但在TestEndpoint身体空洞.
如果我删除了检索请求体的行,LoggingHandler则会填充正文TestEndpoint.
有谁知道这样做的方法?
我的服务器类:
package com.undertow.server;
import com.undertow.server.endpoints.TestEndpoint;
import org.jboss.resteasy.plugins.server.undertow.UndertowJaxrsServer;
import org.jboss.resteasy.spi.ResteasyDeployment;
import io.undertow.Undertow;
import io.undertow.Undertow.Builder;
import io.undertow.server.HandlerWrapper;
import io.undertow.server.HttpHandler;
import io.undertow.server.handlers.BlockingHandler;
import io.undertow.servlet.api.DeploymentInfo;
public class UndertowServer {
private UndertowJaxrsServer server;
public UndertowServer() {
this.server = new UndertowJaxrsServer();
}
public void start() {
Builder builder = Undertow.builder().addHttpListener(8000, "0.0.0.0");
this.server.start(builder);
this.configureEndpoints();
}
private void configureEndpoints() {
ResteasyDeployment deployment = new ResteasyDeployment();
deployment.getActualResourceClasses().add(TestEndpoint.class);
DeploymentInfo deploymentInfo = this.server.undertowDeployment(deployment) //
.setClassLoader(ClassLoader.getSystemClassLoader()).setContextPath("/gateway/") //
.setDeploymentName("gateway.war"); …Run Code Online (Sandbox Code Playgroud) java ×2
httphandler ×1
inputstream ×1
io ×1
rest ×1
resteasy ×1
stream ×1
string ×1
undertow ×1