如何在Android中向HTTP GET请求添加参数?

I have a HTTP GET request that I am attempting to send. I tried adding the parameters to this request by first creating a BasicHttpParams object and adding the parameters to that object, then calling setParams( basicHttpParms ) on my HttpGet object. This method fails. But if I manually add my parameters to my URL (i.e. append ?param1=value1&param2=value2) it succeeds.

我正在尝试发送一个HTTP GET请求。我尝试向这个请求添加参数，首先创建一个BasicHttpParams对象，然后向该对象添加参数，然后在我的HttpGet对象上调用setParams(basicHttpParms)。这个方法失败。但是如果我手工将参数添加到我的URL(例如，追加?param1=value1&param2=value2)，它就成功了。

I know I'm missing something here and any help would be greatly appreciated.

我知道我在这里漏掉了一些东西，任何帮助都将非常感谢。

8 个解决方案

#1

224

I use a List of NameValuePair and URLEncodedUtils to create the url string I want.

我使用NameValuePair和urlencodeduty的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;
}

#2

To build uri with get parameters, Uri.Builder provides a more effective way.

要使用get参数uri构建uri，请使用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();

#3

As of HttpComponents 4.2+ there is a new class URIBuilder, which provides convenient way for generating URIs.

对于httpcomponent 4.2+，有一个新的类URIBuilder，它为生成uri提供了方便的方法。

You can use either create URI directly from String URL:

您可以直接使用字符串URL创建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();

Otherwise, you can specify all parameters explicitly:

否则，您可以显式地指定所有参数:

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();

Once you have created URI object, then you just simply need to create HttpGet object and perform it:

创建了URI对象之后，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

The method

该方法

setParams()

就像

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

only adds HttpProtocol parameters.

只会增加HttpProtocol参数。

To execute the httpGet you should append your parameters to the url manually

要执行httpGet，您应该手动将参数附加到url

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

or use the post request the difference between get and post requests are explained here, if you are interested

或者使用post请求get和post请求之间的区别在这里解释，如果你感兴趣的话

#5

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

Note: url = new URI(...) is buggy

注意:url =新URI(…)有错误

#6

    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");

#7

 class Searchsync extends AsyncTask<String, String, String> {

    @Override
    protected String doInBackground(String... params) {

        HttpClient httpClient = new DefaultHttpClient();/*write your url in Urls.SENDMOVIE_REQUEST_URL; */
        url = Urls.SENDMOVIE_REQUEST_URL; 

        url = url + "/id:" + m;
        HttpGet httpGet = new HttpGet(url);

        try {
            // httpGet.setEntity(new UrlEncodedFormEntity(nameValuePairs));

            // httpGet.setEntity(new StringEntity(json.toString()));
            HttpResponse response = httpClient.execute(httpGet);
            HttpEntity resEntity = response.getEntity();

            if (resEntity != null) {

                String responseStr = EntityUtils.toString(resEntity).trim();

                Log.d("Response from PHP server", "Response: "
                        + responseStr);
                Intent i = new Intent(getApplicationContext(),
                        MovieFoundActivity.class);
                i.putExtra("ifoundmovie", responseStr);
                startActivity(i);

            }
        } catch (UnsupportedEncodingException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        } catch (ClientProtocolException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        } catch (IOException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }

        return null;
    }//enter code here

}

#8

If you have constant URL I recommend use simplified http-request built on apache http.

如果您有常量URL，我建议使用构建在apache http上的简化http请求。

You can build your client as following:

你可以建立你的客户如下:

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());
}

Note: There are many useful methods to manipulate your response.

注意:有许多有用的方法可以操作您的响应。

#1

224