Question

问题

我无法使用html / javascript在客户端之间进行通信，将html形式的数据发送到servlet，然后servlet重播json。我不知道我在哪里犯错。

想法是这样的：

JavaScript从表单获取数据，将其解析为json，然后将其发送到servlet。
在服务器端，servlet读取发送的json，并采取一些措施。产生另一个json，然后重播。
回到客户端，阅读json并在此基础上绘制一些html。（目前，我只是console.log()-正在使用它）

客户

从元素获取数据的javascript代码：

//first I add the listener
document.querySelector('#login_form').addEventListener('submit',(e)=>{
    e.preventDefault();
    login_send(e.target);
});
//then the function to run on submit
function login_send(form){
    console.log(form2json(form));
    //I get the content: {"email":"a@b.c","pass":"aoeu"}
    fetch('login',{
        method:'POST',
        headers:{
            'Accept':'application/json, text/plain, */*',
            'Content-type':'application/json'},
        body: form2json(form)
    })
    .then((res) =>res.json())
    .then((data) => {
        console.log("I got: "+data);//for now, just printing the data
    })
    .catch((err) => console.error(err));
}
//this is my handcrafted "form to json" string formatter,
//surely there is a better (correct?) way of doing it.
function form2json(form){
    let js="{";
    let last=form.length-1;
    for (i=0;i<last;i++){
        js+="\""+form.elements[i].name+"\":\""+form.elements[i].value+"\"";
        if(i+1<last) js+=",";
    }
    js+="}";
    return js;
}

服务器

web.xml 文件，用于将URL连接到Java类：

<?xml version="1.0" encoding="UTF-8"?>
<web-app version="3.1" xmlns="http://xmlns.jcp.org/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd">
    <servlet>
        <servlet-name>Login Servlet</servlet-name>
        <servlet-class>com.cactusstore-1.ui.Login</servlet-class>
    </servlet>
    <servlet-mapping>
        <servlet-name>Login Servlet</servlet-name>
        <url-pattern>/login</url-pattern>
    </servlet-mapping>
    <session-config>
        <session-timeout>30</session-timeout>
    </session-config>
</web-app>

Servlet 我将doGet（）和doPost（）重定向到processRequest（），这是在创建新的“ Web应用程序”时在Netbeans中默认提供的。

...
import javax.json.*;
...
public class Login extends HttpServlet {
    protected void processRequest(HttpServletRequest req,
            HttpServletResponse res) throws ServletException, IOException {
        res.setContentType("application/json;charset=UTF-8");
        try (PrintWriter out = res.getWriter()) {
            //an object with three fields.
            User u = new User(
                    req.getParameter("email"),
                    "10010099",
                    req.getParameter("pass")
            );
            //construct json answer.
            //based on https://www.youtube.com/watch?v=BPMVC999HTs
            JsonObject root;
            JsonObjectBuilder rootBuilder = Json.createObjectBuilder();
            JsonObjectBuilder userBuilder = Json.createObjectBuilder();
            userBuilder
                    .add("name",u.getName())
                    .add("email", u.getEmail())
                    .add("sid", u.getSId());

            root = rootBuilder.add("login", userBuilder).build();
            //write response to out
            out.println(root);
            out.flush();
            out.close();
        }
    }
}

我希望得到json，但出现此错误：

custom.js:23 SyntaxError: Unexpected end of JSON input

看来我的servlet没有返回任何信息。如果我将res.json()更改为res.text()，将一无所获。如果将out.println(root)更改为out.println("{\"name\":\"John\"}");，我仍然什么也没得到。

谢谢：）（请客气，这是我第一次使用所有这些语言。我已经很困惑了。）

编辑1

班级用户。添加班级以保持完整性。

public class User {
    private String email;
    private final String session_id;
    private String name;
    public User(String email, String id, String name) {
        this.email = email;
        this.session_id= id;
        this.name = name;
    }
    public String getName() {return name;}
    public String getEmail(){return email;}
    public String getSId()  {return session_id;}
    public void   setName (String name) {this.name=name;}
    public void   setEmail(String email){this.email=email;}
}

编辑2

由于@vladwoguer，我找到了日志，现在我知道我在Json类上遇到了问题，奇怪的是Netbeans自动完成了函数名，并且在编辑器中未显示任何错误。

25-May-2019 01:11:57.354 SEVERE [http-nio-8080-exec-1] org.apache.catalina.core.StandardWrapperValve.invoke Servlet.service() for servlet [Login Servlet] in context with path [/cactusstore-1] threw exception [Servlet execution threw an exception] with root cause
    java.lang.ClassNotFoundException: javax.json.Json
        at org.apache.catalina.loader.WebappClassLoaderBase.loadClass(WebappClassLoaderBase.java:1363)
        at org.apache.catalina.loader.WebappClassLoaderBase.loadClass(WebappClassLoaderBase.java:1186)
        at com.cactusmania.ui.Login.processRequest(Login.java:66)
        at com.cactusmania.ui.Login.doPost(Login.java:93)
        at javax.servlet.http.HttpServlet.service(HttpServlet.java:660)
        at javax.servlet.http.HttpServlet.service(HttpServlet.java:741)
        at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:231)
        at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:166)
        at org.apache.tomcat.websocket.server.WsFilter.doFilter(WsFilter.java:53)
        at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:193)
        at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:166)
        at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:200)
        at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:96)
        at org.apache.catalina.authenticator.AuthenticatorBase.invoke(AuthenticatorBase.java:490)
        at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:139)
        at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:92)
        at org.apache.catalina.valves.AbstractAccessLogValve.invoke(AbstractAccessLogValve.java:678)
        at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:74)
        at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:343)
        at org.apache.coyote.http11.Http11Processor.service(Http11Processor.java:408)
        at org.apache.coyote.AbstractProcessorLight.process(AbstractProcessorLight.java:66)
        at org.apache.coyote.AbstractProtocol$ConnectionHandler.process(AbstractProtocol.java:836)
        at org.apache.tomcat.util.net.NioEndpoint$SocketProcessor.doRun(NioEndpoint.java:1839)
        at org.apache.tomcat.util.net.SocketProcessorBase.run(SocketProcessorBase.java:49)
        at java.base/java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1128)
        at java.base/java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:628)
        at org.apache.tomcat.util.threads.TaskThread$WrappingRunnable.run(TaskThread.java:61)
        at java.base/java.lang.Thread.run(Thread.java:834)

如果我注释掉所有Json包含的内容及其使用，请仅保留：

System.out.println("YOUR EMAIL: "+req.getParameter("email"));

我在catalina.out中获得了YOUR EMAIL: null。

再次感谢您的耐心和耐心。

Answer 1

第一个问题，我看到的是例外情况：java.lang.ClassNotFoundException: javax.json.Json 您需要将其包括在类路径中。对于我使用过javax.json的情况，请将YOUR_PROJECT/web/WEB-INF/lib jar包含在javax.json-1.0.jar文件夹中。

YOUR_PROJECT
|
|__web
   |
   |__WEB-INF
      |
      |__lib
         |  javax.json-1.0.jar

这样，当您导出war文件并将其部署在tomcat上时，jar将在类路径中可用。

第二个问题是您正在将JSON传递到服务器，并尝试使用req.getParameter获取参数，但真正需要的是解析json并获取类似的值：

 StringBuilder sb = new StringBuilder();
 BufferedReader br = req.getReader();
 String str = null;
 while ((str = br.readLine()) != null) {
     sb.append(str);
 }

 String json = sb.toString();


 JsonReader jsonReader = Json.createReader(new StringReader(json));
 JsonObject jsonObject = jsonReader.readObject();
 jsonReader.close();

 // an object with three fields.
 User u = new User(jsonObject.getString("email"), "10010099", jsonObject.getString("pass"));

完整代码：

public class Login extends HttpServlet {

    @Override
    protected void doPost(HttpServletRequest req, HttpServletResponse res) throws ServletException, IOException {
        res.setContentType("application/json;charset=UTF-8");
        try (PrintWriter out = res.getWriter()) {

            StringBuilder sb = new StringBuilder();
            BufferedReader br = req.getReader();
            String str = null;
            while ((str = br.readLine()) != null) {
                sb.append(str);
            }

            String json = sb.toString();


            JsonReader jsonReader = Json.createReader(new StringReader(json));
            JsonObject jsonObject = jsonReader.readObject();
            jsonReader.close();

            // an object with three fields.
            User u = new User(jsonObject.getString("email"), "10010099", jsonObject.getString("pass"));

            // construct json answer.
            // based on https://www.youtube.com/watch?v=BPMVC999HTs
            JsonObject root;
            JsonObjectBuilder rootBuilder = Json.createObjectBuilder();
            JsonObjectBuilder userBuilder = Json.createObjectBuilder();
            userBuilder.add("name", u.getName()).add("email", u.getEmail()).add("sid", u.getSId());

            root = rootBuilder.add("login", userBuilder).build();
            // write response to out
            out.println(root);
            out.flush();
            out.close();
        }
    }
}

Answer 2

不能为您建立json格式！
html中的每个表单都有一个称为onSubmit的函数。您可以使用它。

<form onsubmit="login_send()">
    <button type="submit"></button>
</form>

如何在javascript和servlet之间发送和接收json数据？

问题

客户

服务器

编辑1

编辑2

2 个答案: