Java 使用makehttp请求的HttpPost

嗨，朋友们，请帮帮我，我从15天开始就罢工了，我找不到我到处找过的，没有解决办法。 这是我用来发出http请求的代码，它在json中传递参数 在下面的字符串中，它将空字符串传递给

jurl

String jurl=calling.makeHttpRequest（url，“GET”，params）

public类loaditems扩展异步任务{
@凌驾
受保护的字符串背景（字符串…字符串）{
字符串url=”http://www.yell4food.com/json/data_standard_item_new.php?rname=standardtakeaway";
List params=new ArrayList（）；
参数添加（新的BasicNameValuePair（“cname”，项目））；
参数添加（新的BasicNameValuePair（“cids”，ID））；
调用=新调用（）；
String jurl=calling.makeHttpRequest（url，“GET”，params）；
日志d（“项目”，法律）；
试一试{
JSONArray数组=新的JSONArray（jurl）；
对于（int i=0；i

打电话，上课
公共类呼叫{

static InputStream is = null;
static JSONObject jObj = null;
static String json = "";

// constructor
public Calling() {

}

// function get json from url
// by making HTTP POST or GET mehtod
public String makeHttpRequest(String url, String method,
                              List<NameValuePair> params) {

    // Making HTTP request
    try {

        // check for request method
        if (method == "POST") {
            // request method is POST
            // defaultHttpClient
            DefaultHttpClient httpClient = new DefaultHttpClient();
            HttpPost httpPost = new HttpPost(url);
            httpPost.setEntity(new UrlEncodedFormEntity(params));

            HttpResponse httpResponse = httpClient.execute(httpPost);
            HttpEntity httpEntity = httpResponse.getEntity();
            is = httpEntity.getContent();

        } else if (method == "GET") {
            // request method is GET
            DefaultHttpClient httpClient = new DefaultHttpClient();
            String paramString = URLEncodedUtils.format(params, "utf-8");
            url += "?" + paramString;
            HttpGet httpGet = new HttpGet(url);

            HttpResponse httpResponse = httpClient.execute(httpGet);
            HttpEntity httpEntity = httpResponse.getEntity();
            is = httpEntity.getContent();
        }

    } catch (UnsupportedEncodingException e) {
        e.printStackTrace();
    } catch (ClientProtocolException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    }

    try {
        BufferedReader reader = new BufferedReader(new InputStreamReader(
                is, "iso-8859-1"), 8);
        StringBuilder sb = new StringBuilder();
        String line = null;
        while ((line = reader.readLine()) != null) {
            sb.append(line + "\n");
        }
        is.close();
        json = sb.toString();
    } catch (Exception e) {
        Log.e("Buffer Error", "Error converting result " + e.toString());
    }

    // return JSON String
    return json;

}

静态输入流为=null；
静态JSONObject jObj=null；
静态字符串json=“”；
//建造师
公共电话（）{
}
//函数从url获取json
//通过使用HTTP POST或GET方法
公共字符串makeHttpRequest（字符串url、字符串方法、，
列表参数）{
//发出HTTP请求
试一试{
//检查请求方法
如果（方法==“POST”）{
//请求方法为POST
//defaultHttpClient
DefaultHttpClient httpClient=新的DefaultHttpClient（）；
HttpPost HttpPost=新的HttpPost（url）；
setEntity（新的UrlEncodedFormEntity（参数））；
HttpResponse HttpResponse=httpClient.execute（httpPost）；
HttpEntity HttpEntity=httpResponse.getEntity（）；
is=httpEntity.getContent（）；
}else if（方法==“GET”）{
//请求方法是GET
DefaultHttpClient httpClient=新的DefaultHttpClient（）；
String paramString=URLEncodedUtils.format（params，“utf-8”）；
url+=“？”+参数字符串；
HttpGet HttpGet=新的HttpGet（url）；
HttpResponse HttpResponse=httpClient.execute（httpGet）；
HttpEntity HttpEntity=httpResponse.getEntity（）；
is=httpEntity.getContent（）；
}
}捕获（不支持的编码异常e）{
e、 printStackTrace（）；
}捕获（客户端协议例外e）{
e、 printStackTrace（）；
}捕获（IOE异常）{
e、 printStackTrace（）；
}
试一试{
BufferedReader reader=新的BufferedReader（新的InputStreamReader(
is，“iso-8859-1”），8）；
StringBuilder sb=新的StringBuilder（）；
字符串行=null；
而（（line=reader.readLine（））！=null）{
sb.追加（第+行“\n”）；
}
is.close（）；
json=sb.toString（）；
}捕获（例外e）{
Log.e（“缓冲区错误”，“错误转换结果”+e.toString（））；
}
//返回JSON字符串
返回json；
}

}
使用截击库…..我编辑了问题，在其中添加了调用类，检查一下是否太高兴了，兄弟，它解决了我的问题，但数据以不同的方式来了，兄弟，我一直尝试了将近15天，兄弟
static InputStream is = null;
static JSONObject jObj = null;
static String json = "";

// constructor
public Calling() {

}

// function get json from url
// by making HTTP POST or GET mehtod
public String makeHttpRequest(String url, String method,
                              List<NameValuePair> params) {

    // Making HTTP request
    try {

        // check for request method
        if (method == "POST") {
            // request method is POST
            // defaultHttpClient
            DefaultHttpClient httpClient = new DefaultHttpClient();
            HttpPost httpPost = new HttpPost(url);
            httpPost.setEntity(new UrlEncodedFormEntity(params));

            HttpResponse httpResponse = httpClient.execute(httpPost);
            HttpEntity httpEntity = httpResponse.getEntity();
            is = httpEntity.getContent();

        } else if (method == "GET") {
            // request method is GET
            DefaultHttpClient httpClient = new DefaultHttpClient();
            String paramString = URLEncodedUtils.format(params, "utf-8");
            url += "?" + paramString;
            HttpGet httpGet = new HttpGet(url);

            HttpResponse httpResponse = httpClient.execute(httpGet);
            HttpEntity httpEntity = httpResponse.getEntity();
            is = httpEntity.getContent();
        }

    } catch (UnsupportedEncodingException e) {
        e.printStackTrace();
    } catch (ClientProtocolException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    }

    try {
        BufferedReader reader = new BufferedReader(new InputStreamReader(
                is, "iso-8859-1"), 8);
        StringBuilder sb = new StringBuilder();
        String line = null;
        while ((line = reader.readLine()) != null) {
            sb.append(line + "\n");
        }
        is.close();
        json = sb.toString();
    } catch (Exception e) {
        Log.e("Buffer Error", "Error converting result " + e.toString());
    }

    // return JSON String
    return json;

}

See the code below

// request method is GET

DefaultHttpClient httpClient = new DefaultHttpClient();
String paramString = URLEncodedUtils.format(params, "utf-8");
url += "?" + paramString;
HttpGet httpGet = new HttpGet(url);
HttpResponse httpResponse = httpClient.execute(httpGet);
HttpEntity httpEntity = httpResponse.getEntity();
is = httpEntity.getContent();




line  url += "?" + paramString; 


adding a query string separator '?' in your url.
but in a valid url only query string separator '?' is allowed but you already include that one in your url

Try this solution
change your url from
 String url="http://www.yell4food.com/json/data_standard_item_new.php?rname=standardtakeaway";

to`enter code here`

String url="http://www.yell4food.com/json/data_standard_item_new.php";

and pass parameter rname in NameValuePairs list like this

    params.add(new BasicNameValuePair("rname", "standardtakeaway"));