java - 这个错误是什么意思 : The method. ..is not applicable for the arguments

Question

我是新手，我不明白这个错误是什么意思。

我试过谷歌搜索，但无法理解我所发现的内容。

我需要与 API 交互以将票证发布到远程服务器，并且我从我正在学习的教程中获得了这段代码。

在这段代码中我遇到了这个错误:

The method postToRemoteServer(String) in the type HelpDeskTestService is not applicable for the arguments (String, new AsyncCallback(){})

sendButton.addClickHandler(new ClickHandler() {
        @Override
        public void onClick(ClickEvent event) {
            HelpDeskTestService.postToRemoteServer(
                    "http://xx.xx.xx.xx/sdpapi/request/", 
                    new AsyncCallback<String>() {
                        @Override
                        public void onFailure(Throwable caught)  {
                            Window.alert("Failure getting XML through proxy");
                        }

                        @Override
                        public void onSuccess(String result) {
                            processXML(result);
                        }
                    });
        }
    });

这是来自同步接口(interface)的代码:

public String postToRemoteServer(final String serviceUrl)
    throws HelpDeskTestException;

这是异步接口(interface)的代码:

void postToRemoteServer(
        final String serviceUrl,
        AsyncCallback<String> callback);

最后，这是实现类的代码:

@Override
public String postToRemoteServer(String serviceUrl)
        throws HelpDeskTestException {

    try {
        //dividing url into host: http://some.server
        //path: a/path/in/it
        //and parameters: this=that&those=others

        int hostStart= serviceUrl.indexOf("//");

        int pathStart= serviceUrl.substring(hostStart + 2).indexOf("/");

        int parameterStart= serviceUrl.substring(hostStart + 2 + pathStart).indexOf("?");

        final String serverHost= serviceUrl.substring(0, hostStart + pathStart + 2);

        final String serverPath= serviceUrl.substring(hostStart + 3, 
                hostStart + pathStart + 2 + parameterStart);

        final String serverParameters= serviceUrl.substring(hostStart + pathStart + 3 + parameterStart);

        final  URL url = new URL(serverHost);

        final URLConnection connection= url.openConnection();
        connection.setDoOutput(true);

        final OutputStreamWriter out= new OutputStreamWriter(connection.getOutputStream());

        final BufferedReader in= new BufferedReader(new InputStreamReader(
                connection.getInputStream()));

        out.write("POST " + serverPath + "\r\n");
        out.write("Host: " + serverHost + "\r\n");
        out.write("Accept-Encoding: identity\r\n");
        out.write("Connection: close\r\n");
        out.write("Content-Type: application/x-www-form-urlencoded\r\n");
        out.write("Content-Length: " + serverParameters.length() + "\r\n\r\n" +
            serverParameters + "\r\n");


        String result = "";
        String inputLine;

        while ((inputLine=in.readLine()) != null) {
            result+= inputLine;
        }

        in.close();
        out.close();

        return result;

    }  catch (final Exception e) {
        throw new HelpDeskTestException();

    }

我们将不胜感激。

Answer 1

调用服务时需要使用异步接口(interface)。您可以使用 GWT.create() 创建它的一个实例。假设您的异步接口(interface)称为“HelpDeskTestServiceAsync”，您将执行如下操作:

HelpDeskTestServiceAsync asyncService = GWT.create(HelpDeskTestService.class);

asyncService.postToRemoteServer(
                "http://xx.xx.xx.xx/sdpapi/request/", 
                new AsyncCallback<String>() {
                    @Override
                    public void onFailure(Throwable caught)  {
                        Window.alert("Failure getting XML through proxy");
                    }

                    @Override
                    public void onSuccess(String result) {
                        processXML(result);
                    }
                });