使用OkHttp

Question

我正在尝试通过get请求发送查询，问题是我不断获取java.lang.IllegalArgumentException: unexpected host

   HttpUrl url = new HttpUrl.Builder()
            .scheme("http")
            .host("10.0.2.2" + "/api/" + 7) // This is where the error is coming in
            .addQueryParameter("lat", deviceLat[0])
            .addQueryParameter("long", deviceLong[0])
            .build();


    Request request = new Request.Builder()
            .url(url)
            .build();

感谢您的帮助：）

Answer 1

我尝试使用此代码并且有效

HttpUrl url = new HttpUrl.Builder()
            .scheme("https")
            .host("www.google.com")
            .addPathSegment("search")
            .addQueryParameter("q", "polar bears")
            .build();


Request request = new Request.Builder()
            .url(url)
            .build();

所以，您的主机有问题。请在邮递员上测试您的主机或为其打开新端口。我也ping那个主机

Answer 2

您的问题是.host(string)方法仅期望网址的 host 部分。删除路径段即可。您的代码应如下所示：

HttpUrl url = new HttpUrl.Builder()
        .scheme("http")
        .host("10.0.2.2") //Just the host (like "google.com", not "google.com/api")
        .addPathSegment("api")
        .addPathSegment("7")
        .addQueryParameter("lat", deviceLat[0])
        .addQueryParameter("long", deviceLong[0])
        .build();


Request request = new Request.Builder()
        .url(url)
        .build();

Answer 3

HttpUrl有一个解析字符串的解析方法，我觉得它更短。

HttpUrl url = HttpUrl.parse("http://10.0.2.2/api/7")
                     .addQueryParameter("lat", deviceLat[0])
                     .addQueryParameter("long", deviceLong[0])
                     .build();

Answer 4

所以我不知道我不能构建URL的问题是什么，但是我手动在字符串中创建url的第二个问题，它工作得很好，一切都没问题。< / p>

基本上我改变了这个

HttpUrl url = new HttpUrl.Builder()
            .scheme("http")
            .host("10.0.2.2" + "/api/" + 7) // This is where the error is coming in
            .addQueryParameter("lat", deviceLat[0])
            .addQueryParameter("long", deviceLong[0])
            .build();


Request request = new Request.Builder()
        .url(url)
        .build();

到这个

Request request = new Request.Builder()
                                .url("10.0.2.2" + "/api/" + 7 + "?long=" + deviceLong[0] + "&lat=" + deviceLat[0])
                                .build();

它运作良好

Answer 5

试试这个：

HttpUrl.Builder httpUrl = new HttpUrl.Builder();
httpUrl.scheme("http");
httpUrl.host("10.0.2.2");
httpUrl.addPathSegment("api");
httpUrl.addPathSegment("7");
httpUrl.addQueryParameter("lat", deviceLat[0]);
httpUrl.addQueryParameter("long", deviceLong[0]);
HttpUrl build = httpUrl.build();

String result = build.toString();