url是要访问的地址，method是请求方式可以是POST或者GET,params是后面的参数比如?aa=11&bb=22 public String connctionURL_Params(String url,String method,String params){ StringBuffer bufferRes = new StringBuffer(); url = url + params;...
http://www.cnblogs.com/lori/p/4045633.html 看里面的dopost这一块
在Android/java平台上实现POST一个json数据： JSONObject jsonObj = new JSONObject(); jsonObj.put("username", username); jsonObj.put("apikey", apikey); // Create the POST object and add the parameters HttpPost httpPost = new HttpPo...
header('Content-type:application/json;charset=UTF-8');echo json_encode($json);
大哥，那是前端概念， 都不知道你想说什么
RequestParams requestParams =new RequestParams("http://"+ SpeechApp.url+"/interfaceAndroid/Project/AddServiceItemQuote"); ArrayList arrayList =new ArrayList(); arrayList.add(new Dingdan_Model("1","2","3","4","5")); arrayList.ad...
1、编程语言 //----1、HttpPost request = new HttpPost(url);// 先封装一个 JSON 对象JSONObject param = new JSONObject();param.put("name", "rarnu");param.put("password", "123456");// 绑定到请求 EntryStringEntity se = new StringEnti...
在Android/java平台上实现POST一个json数据： JSONObject jsonObj = new JSONObject(); jsonObj.put("username", username); jsonObj.put("apikey", apikey); // Create the POST object and add the parameters HttpPost httpPost = new HttpPo...
最近有遇到这样的问题，希望对你有帮助，不知道你是什么情况。 1.使用json格式化工具格式化你的json报文，确认是json无误 2.查找json报文中是否存在以下两个字符"=","&"，这两个字符是用get/post请求中用来拼接查询参数用的特殊符号，如果有就去...
Public Function Ajax_Post(ByVal StrUrl As String, Optional ByVal StrData As String, Optional ByVal Index As Long) As Variant On Error GoTo MyError: Dim Object As Object, S As String, B() As Byte Set Object = CreateObject("Micro...