Java中带有Json响应的Http请求

Question

我想对网站执行HTTP请求并接收JSON响应。

但是，我无法管理“连接到主机页面”。 我尝试了一些教程和技巧，但无法正常工作。

try{
    String thePath = "//Url";
    thePath = URLEncoder.encode(thePath, "UTF-8"); 

    HttpGet httpget = new HttpGet(thePath);

    DefaultHttpClient httpClient = new DefaultHttpClient();

    System.out.println("Executing HTTP Get...\n");
    HttpResponse response = httpClient.execute(httpget);

    if (response.getStatusLine().getStatusCode() != 200 && response.getStatusLine().getStatusCode() != 201) {
        throw new RuntimeException("Failed: HTTP error code: "
            + response.getStatusLine().getStatusCode());
    }

    BufferedReader br = new BufferedReader(
                    new InputStreamReader((response.getEntity().getContent())));

    String output = ""; Boolean keepGoing = true;
    while(keepGoing){
        String currentLine = br.readLine();

        if(currentLine == null)
            keepGoing = false;
        else
            output += currentLine;
    }

    System.out.println("Raw string result: \n" + output);

} catch (MalformedURLException e){
    System.out.println("Caught MalformedURLException: " + e.toString());

} catch (IOException e){
    System.out.println("Caught IOException: " + e.toString());

} catch (Exception e){
    System.out.println("Caught Exception: " + e.toString());
}

捕获的异常：java.lang.IllegalargumentException：主机名不能为null

Answer 1

删除最外面的try / catch块，让异常打印堆栈跟踪。 然后查看堆栈顶部的方法，在代码中找到导致错误方法调用的行号。 学习有关该方法的文档并提供正确的参数。 您很可能提供了错误的URL。