Android httpget不适用于查询字符串

我正在通过查询字符串获取数据。我已经完成了httppost，但还没有完成，我正在尝试使用httpget。但我不能

 List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
            nameValuePairs.add(new BasicNameValuePair("key", "android developer"));

            String k = "android developer";
             URI uri = new URI("http://10.0.2.2:8080/r/r.php");


             HttpClient httpclient = new DefaultHttpClient();
             HttpGet httpget = new HttpGet(uri);
             //httpget.setEntity(new UrlEncodedFormEntity(nameValuePairs));
             HttpResponse response = httpclient.execute(httpget);
             HttpEntity entity = response.getEntity();
             is = entity.getContent();

List nameValuePairs=new ArrayList（）；
添加（新的BasicNameValuePair（“key”，“android开发者”）；
String k=“android开发者”；
URI=新的URI（“http://10.0.2.2:8080/r/r.php");
HttpClient HttpClient=新的DefaultHttpClient（）；
HttpGet HttpGet=新的HttpGet（uri）；
//setEntity（新的UrlEncodedFormEntity（nameValuePairs））；
HttpResponse response=httpclient.execute（httpget）；
HttpEntity=response.getEntity（）；
is=entity.getContent（）；

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请专家帮助我

您可以尝试以下HTTP获取

public String[] doHttpGetWithCode(String url, HashMap<String, String> headerParam) throws Exception {
    String[] result = new String[2];
    HttpClient httpclient = new DefaultHttpClient();
    HttpGet httpget = new HttpGet(url);     
    httpget.addHeader("language", Constants.DEVICE_LANGUAGE);       
    Iterator myIterator = headerParam.keySet().iterator();
    while(myIterator.hasNext()) {
        String key=(String)myIterator.next();
        String value=(String)headerParam.get(key);        
        httpget.addHeader(key, value);
    }       
    HttpResponse response;
    response = httpclient.execute(httpget);
    HttpEntity entity = response.getEntity();
    InputStream is = entity.getContent();
    BufferedReader reader = new BufferedReader(new InputStreamReader(is));
    StringBuilder sb = new StringBuilder();
    String line = null;
    try {
        while ((line = reader.readLine()) != null) {
            sb.append(line + "\n");
        }
    } catch (IOException e) {
        e.printStackTrace();
    } finally {
        try {
            is.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
    result[0] = response.getStatusLine().getStatusCode()+"";
    result[1] = sb.toString();
    return result;
}

---

响应为字符串类型

您可以尝试以下HTTP get

public String[] doHttpGetWithCode(String url, HashMap<String, String> headerParam) throws Exception {
    String[] result = new String[2];
    HttpClient httpclient = new DefaultHttpClient();
    HttpGet httpget = new HttpGet(url);     
    httpget.addHeader("language", Constants.DEVICE_LANGUAGE);       
    Iterator myIterator = headerParam.keySet().iterator();
    while(myIterator.hasNext()) {
        String key=(String)myIterator.next();
        String value=(String)headerParam.get(key);        
        httpget.addHeader(key, value);
    }       
    HttpResponse response;
    response = httpclient.execute(httpget);
    HttpEntity entity = response.getEntity();
    InputStream is = entity.getContent();
    BufferedReader reader = new BufferedReader(new InputStreamReader(is));
    StringBuilder sb = new StringBuilder();
    String line = null;
    try {
        while ((line = reader.readLine()) != null) {
            sb.append(line + "\n");
        }
    } catch (IOException e) {
        e.printStackTrace();
    } finally {
        try {
            is.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
    result[0] = response.getStatusLine().getStatusCode()+"";
    result[1] = sb.toString();
    return result;
}

---

响应为字符串类型

您可以通过两种方式执行此操作：

使用URI生成器：

自己创建url字符串：

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您可以通过两种方式完成此操作：

使用URI生成器：

自己创建url字符串：

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我正在使用SalesForce API，需要将参数“q”设置为“从帐户中选择名称、Id、帐户\人员”。使用HttpClient版本4.2.5设置参数会导致错误页面，而使用带有3.1 API的“setQueryString”会导致JSON。。。有什么线索吗？请不要把SQL查询放在URL中。这将是一个灾难时，去生活。SQL注入！如果您有问题，发布一个问题，评论其他人的线程不方便。我正在使用SalesForce API，需要将参数“q”设置为“从帐户中选择名称、Id、帐户\人员”。使用HttpClient版本4.2.5设置参数会导致错误页面，而使用带有3.1 API的“setQueryString”会导致JSON。。。有什么线索吗？请不要把SQL查询放在URL中。这将是一个灾难时，去生活。SQL注入！如果你有问题，发表一个问题，评论别人的帖子是不方便的。

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

String paramString = URLEncodedUtils.format(nameValuePairs, "utf-8");
uri = new Uri("http://10.0.2.2:8080/r/r.php?" + paramString);