如何使用 HttpURL 连接仅发送一个请求

Question

我使用以下代码向服务器发出请求。我调用了该函数一次，但请求被发送了两次。我从服务器返回的结果来自第二个请求结果。

如何只向服务器发送一个请求？

我存储所有我的：Classrunnables

public class ServerRunnables implements Runnable {

    Account currentAccount = Account.getInstance();

    private String result = "";

    private Context context;

    private Boolean callSuccess = true;

    int serviceCalled;

    String email;

    String password;

    String username;

    int coinChanges;

    String urlString = "http://10.0.2.2/MobileCoin/" + services[serviceCalled];

    String requiredParams;

    public ServerRunnables(Context context,

                           int serviceCalled,

                           String email,

                           String password,

                           String username,

                           int coinChanges) {

        this.context = context;

        this.serviceCalled = serviceCalled;

        this.email = email;

        this.password = password;

        this.username = username;

        this.coinChanges = coinChanges;

    }

    public void run() {

        urlString = "http://10.0.2.2/MobileCoin/" + services[serviceCalled];

        switch (serviceCalled) {

            case SERVICE_RANDOM_COIN:

                generateCoinServiceCalled();

                break;

            default:

                break;

        }

    }

    private void generateCoinServiceCalled() {

        requiredParams = "email=" + email;

        urlString = urlString + "?" + requiredParams;

        getMethodCalled(urlString);

        callSuccess = true;

        String message1 = "";

        if (result.equals(ServerConstants.RANDOM_COIN_EMPTY)) {

            callSuccess = false;

            message1 = "No Data Received";

        } else if (result.equals(ServerConstants.RANDOM_COIN_FAIL)) {

            callSuccess = false;

            message1 = "Data Does not Exist.";

        } else if (result.equals(ServerConstants.CONNECTION_ERROR)) {

            callSuccess = false;

            message1 = "Connection Error, Check Server";

        } 

}

Answer 1

为什么不使用改造来轻松处理http请求？如果与 RxJava 集成，则可以异步管理请求。

Answer 2

我发现了错误：InputStream input = url.openStream();更改为InputStream input = hc.getInputStream();