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：Classrunnablespublic 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();