使用HTTP的javacurl请求

我正在尝试使用Java执行一个CURL请求。CURL请求如下所示：

String stringUrl = "https://apis.sen.se/v2/feeds/N4hSBSpFlYzXT6ZN2IA1KadgSR9rTazv/events/?limit=1";
URL url = new URL(stringUrl);
URLConnection uc = url.openConnection();

uc.setRequestProperty("X-Requested-With", "Curl");

String userpass = "username" + ":" + "password";
String basicAuth = "Basic " + new String(new Base64().encode(userpass.getBytes()));
uc.setRequestProperty("Authorization", basicAuth);

InputStreamReader inputStreamReader = new InputStreamReader(uc.getInputStream());

int data = inputStreamReader.read();
char aChar = (char) data;
System.out.println(aChar);

curlhttps://apis.sen.se/v2/feeds/N4hSBSpFlYzXT6ZN2IA1KadgSR9rTazv/events/?limit=1 -u用户名：密码

我正在尝试执行以下请求：

String stringUrl = "https://apis.sen.se/v2/feeds/N4hSBSpFlYzXT6ZN2IA1KadgSR9rTazv/events/?limit=1";
URL url = new URL(stringUrl);
URLConnection uc = url.openConnection();

uc.setRequestProperty("X-Requested-With", "Curl");

String userpass = "username" + ":" + "password";
String basicAuth = "Basic " + new String(new Base64().encode(userpass.getBytes()));
uc.setRequestProperty("Authorization", basicAuth);

InputStreamReader inputStreamReader = new InputStreamReader(uc.getInputStream());

int data = inputStreamReader.read();
char aChar = (char) data;
System.out.println(aChar);

我正在尝试查看

inputStreamReader

的内容，如下所示：

String stringUrl = "https://apis.sen.se/v2/feeds/N4hSBSpFlYzXT6ZN2IA1KadgSR9rTazv/events/?limit=1";
URL url = new URL(stringUrl);
URLConnection uc = url.openConnection();

uc.setRequestProperty("X-Requested-With", "Curl");

String userpass = "username" + ":" + "password";
String basicAuth = "Basic " + new String(new Base64().encode(userpass.getBytes()));
uc.setRequestProperty("Authorization", basicAuth);

InputStreamReader inputStreamReader = new InputStreamReader(uc.getInputStream());

int data = inputStreamReader.read();
char aChar = (char) data;
System.out.println(aChar);

---

代码编译和运行良好，但没有返回任何内容。我哪里出错了？

我也在试着做那件事。我有一些解决办法，但它能看到所有东西

--这是密码---

---

--请注意，我使用此代码查看某个网站上是否存在某个帐户，因为它输出所有内容，所以我所做的是在代码上找到一个特定的规则，它可以告诉我该用户是否存在。嗯，我甚至不确定这是否能帮到你，但它可能是。祝你好运

我最终使用以下代码使其工作：

public static void main(String args[]) throws IOException {
        String stringUrl = "url";
        URL url = new URL(stringUrl);
        URLConnection uc = url.openConnection();
        uc.setRequestProperty("X-Requested-With", "Curl");
        String userpass = "username" + ":" + "password";
        String basicAuth = "Basic " + new String(new Base64().encode(userpass.getBytes()));
        uc.setRequestProperty("Authorization", basicAuth);
        StringBuilder html = new StringBuilder();
        BufferedReader input = null;
        try {
          input = new BufferedReader(new InputStreamReader(uc.getInputStream()));
          String htmlLine;
          while ((htmlLine = input.readLine()) != null) {
            html.append(htmlLine);
          }
        }
        catch (IOException e) {
          e.printStackTrace();
        }
        finally {
          try {
            input.close();
          }
          catch (IOException e) {
            e.printStackTrace();
          }
        }
        System.out.println(html.toString());
    }

---

请参阅此问题：是否可以提供临时用户名和密码？在任何时候都有抛出异常吗？没有抛出异常，我将尝试设置临时访问。我查看了他们的站点文档，它要求开发人员首先获取API密钥，如so-

curlhttps://apis.sen.se/v2/user/api_key/ \-d“username=demoone”\-d“password=\u您的Sen.se\u帐户\u密码”

。您应该使用

Authorization:token\u your\u token\u

在HTTPS请求上使用该令牌。你试过了吗？