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如何在Android中向HTTP GET请求添加参数?

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我有一个我试图发送的HTTP GET请求 . 我尝试通过首先创建一个 BasicHttpParams 对象并将参数添加到该对象,然后在我的 HttpGet 对象上调用 setParams( basicHttpParms ) 来尝试将参数添加到此请求中 . 此方法失败 . 但是如果我手动将我的参数添加到我的URL(即追加 ?param1=value1&param2=value2 ),它就会成功 .

我知道我在这里遗漏了一些东西,非常感谢任何帮助 .

7 回答

  • 27

    我使用NameValuePair和URLEncodedUtils列表来创建我想要的url字符串 .

    protected String addLocationToUrl(String url){
        if(!url.endsWith("?"))
            url += "?";
    
        List<NameValuePair> params = new LinkedList<NameValuePair>();
    
        if (lat != 0.0 && lon != 0.0){
            params.add(new BasicNameValuePair("lat", String.valueOf(lat)));
            params.add(new BasicNameValuePair("lon", String.valueOf(lon)));
        }
    
        if (address != null && address.getPostalCode() != null)
            params.add(new BasicNameValuePair("postalCode", address.getPostalCode()));
        if (address != null && address.getCountryCode() != null)
            params.add(new BasicNameValuePair("country",address.getCountryCode()));
    
        params.add(new BasicNameValuePair("user", agent.uniqueId));
    
        String paramString = URLEncodedUtils.format(params, "utf-8");
    
        url += paramString;
        return url;
    }
    
  • 94

    要使用get参数构建uri,Uri.Builder提供了一种更有效的方法 .

    Uri uri = new Uri.Builder()
        .scheme("http")
        .authority("foo.com")
        .path("someservlet")
        .appendQueryParameter("param1", foo)
        .appendQueryParameter("param2", bar)
        .build();
    
  • 8

    HttpComponents 4.2+ 开始,有一个新类URIBuilder,它提供了生成URI的便捷方法 .

    您可以直接从String URL使用create URI:

    List<NameValuePair> listOfParameters = ...;
    
    URI uri = new URIBuilder("http://example.com:8080/path/to/resource?mandatoryParam=someValue")
        .addParameter("firstParam", firstVal)
        .addParameter("secondParam", secondVal)
        .addParameters(listOfParameters)
        .build();
    

    否则,您可以显式指定所有参数:

    URI uri = new URIBuilder()
        .setScheme("http")
        .setHost("example.com")
        .setPort(8080)
        .setPath("/path/to/resource")
        .addParameter("mandatoryParam", "someValue")
        .addParameter("firstParam", firstVal)
        .addParameter("secondParam", secondVal)
        .addParameters(listOfParameters)
        .build();
    

    一旦创建了 URI 对象,那么您只需创建 HttpGet 对象并执行它:

    //create GET request
    HttpGet httpGet = new HttpGet(uri);
    //perform request
    httpClient.execute(httpGet ...//additional parameters, handle response etc.
    
  • 4

    方法

    setParams()
    

    喜欢

    httpget.getParams().setParameter("http.socket.timeout", new Integer(5000));
    

    只添加HttpProtocol参数 .

    要执行httpGet,您应该手动将参数附加到网址

    HttpGet myGet = new HttpGet("http://foo.com/someservlet?param1=foo&param2=bar");
    

    或者如果您有兴趣,可以使用帖子请求解释获取和发布请求之间的差异here

  • 224
    List<NameValuePair> params = new ArrayList<NameValuePair>();
    params.add(new BasicNameValuePair("param1","value1");
    
    String query = URLEncodedUtils.format(params, "utf-8");
    
    URI url = URIUtils.createURI(scheme, userInfo, authority, port, path, query, fragment); //can be null
    HttpGet httpGet = new HttpGet(url);
    

    URI javadoc

    注意: url = new URI(...) 是马车

  • 30
    HttpClient client = new DefaultHttpClient();
    
        Uri.Builder builder = Uri.parse(url).buildUpon();
    
        for (String name : params.keySet()) {
            builder.appendQueryParameter(name, params.get(name).toString());
        }
    
        url = builder.build().toString();
        HttpGet request = new HttpGet(url);
        HttpResponse response = client.execute(request);
        return EntityUtils.toString(response.getEntity(), "UTF-8");
    
  • 0

    如果你有 URL 我建议使用基于apache http的简化http-request .

    您可以按如下方式构建客户端:

    private filan static HttpRequest<YourResponseType> httpRequest = 
                       HttpRequestBuilder.createGet(yourUri,YourResponseType)
                       .build();
    
    public void send(){
        ResponseHendler<YourResponseType> rh = 
             httpRequest.execute(param1, value1, param2, value2);
    
        handler.ifSuccess(this::whenSuccess).otherwise(this::whenNotSuccess);
    }
    
    public void whenSuccess(ResponseHendler<YourResponseType> rh){
         rh.ifHasContent(content -> // your code);
    }
    
    public void whenSuccess(ResponseHendler<YourResponseType> rh){
       LOGGER.error("Status code: " + rh.getStatusCode() + ", Error msg: " + rh.getErrorText());
    }
    

    注意:有许多有用的方法可以处理您的响应 .

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