java.lang.IllegalArgumentException：主机名可能不为空，同时触发获取请求

Question

感谢您对此提供的帮助，以下是正在尝试执行的代码，但得到的只是这个异常，我做了很多更改，但无法解决这个问题。

如果您有任何指示，请告诉我，我在 android 4.4.4 上运行

HttpGet request = new HttpGet("https://s3­-eu-­west-­1.amazonaws.com/developer-application-­test/cart/list");
resp = client.execute(request);

01-22 22:25:03.885: W/System.err(14697): java.lang.IllegalArgumentException: Host name may not be null
01-22 22:25:03.886: W/System.err(14697):    at org.apache.http.HttpHost.<init>(HttpHost.java:83)
01-22 22:25:03.886: W/System.err(14697):    at org.apache.http.impl.client.AbstractHttpClient.determineTarget(AbstractHttpClient.java:508)
01-22 22:25:03.886: W/System.err(14697):    at org.apache.http.impl.client.AbstractHttpClient.execute(AbstractHttpClient.java:498)
01-22 22:25:03.886: W/System.err(14697):    at org.apache.http.impl.client.AbstractHttpClient.execute(AbstractHttpClient.java:476)

Answer 1

正如@atish shimpi 所提到的，这很可能是由于 URL 格式不正确。我输入了以下代码，并在开发手机上对其进行了调试：

HttpClient httpClient = new DefaultHttpClient();
URI url = new URI("https://s3-eu­west­1.amazonaws.com/developer-application­test/cart/list");
URI url1 = new URI("https://www.google.com/");
HttpGet httpGet = new HttpGet(url);
HttpResponse httpResponse = httpClient.execute(httpGet);
HttpEntity httpEntity = httpResponse.getEntity();

如您所见，我添加了另一个指向https://www.google.com/ 的URI 对象用作比较。当我调试时，我在 URI 对象的创建上设置了断点。为您提供的地址创建相应的URI 对象后，host 字段为null...

但是，当我为 Google 地址创建类似的 URI 对象时，host 字段不是 null，这意味着您的地址有问题...

我仍然不太清楚为什么URI(String spec) 方法无法解析正确的字段。这可能是一个错误，也可能只是与您的特定 URL 有关。无论如何，我最终能够通过获取您提供的链接并手动创建URI 对象来处理请求，如下所示：

URI uri = new URI("https", "s3-eu-west-1.amazonaws.com", "/developer-application-test/cart/list", null, null);

使用这个手动创建的URI，我能够下载你创建的列表：

"products" : [
    {
    "product_id" : "1",
    "name" : "Apples",
    "price" : 120,
    "image" : "https://s3-eu-west-1.amazonaws.com/developer-application-test/images/1.jpg"
    },
    {
    "product_id" : "2",
    "name" : "Oranges",
    "price" : 167,
    "image" : "https://s3-eu-west-1.amazonaws.com/developer-application-test/images/2.jpg"
    },
    {
    "product_id" : "3",
    "name" : "Bananas",
    "price" : 88,
    "image" : "https://s3-eu-west-1.amazonaws.com/developer-application-test/images/3.jpg"
    },
etc....

作为参考，这是我的最终工作代码：

try
{
    HttpClient httpClient = new DefaultHttpClient();
    URI uri = new URI("https", "s3-eu-west-1.amazonaws.com", "/developer-application-test/cart/list", null, null);
    HttpGet httpGet = new HttpGet(uri);
    HttpResponse httpResponse = httpClient.execute(httpGet);
    HttpEntity httpEntity = httpResponse.getEntity();
    if (httpEntity != null)
    {
        InputStream inputStream = httpEntity.getContent();
        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(inputStream));
        StringBuilder stringBuilder = new StringBuilder();
        String currentLine = null;
        while ((currentLine = bufferedReader.readLine()) != null)
        {
            stringBuilder.append(currentLine + "\n");
        }
        String result = stringBuilder.toString();
        Log.v("Http Request Results:",result);
        inputStream.close();
    }
}
catch (Exception e)
{
    e.printStackTrace();
}

Answer 2

java.lang.IllegalArgumentException: Host name may not be null

问题是因为你的 URL 检查你的 URL "https://s3­eu­west­1.amazonaws.com/developer­application­test/cart/list"