通过JSON对象RESTful Web服务的URL中的Andr​​oid？ [英] passing JSON object in URL of RESTFUL Web service in Android?

问题描述

我用JSON的REST Web serivces，我必须通过JSON对象中的服务URL工作。我已经成功创建了JSON对象，但得到异常时，我的网址创建与服务器的HTTP连接。

下面我不得不提到我的网址：

  http://72.5.167.50:8084/UpdateProfileInfo?{"ProfileEditId":"917","ContactsEmail":[{"Email":"dsfs","ContactId":""}],"ContactsPhone":[{"CountryId":"+1","Type":"2","Phone":"345345"}],"ProfileId":"290","LastName":"demo","GroupId":"1212","Title":"sdf","City":"dsf","TemplateId":"1212","State":"dsf","Auth$c$c":"9bcc6f63-2050-4c5b-ba44-b8103fbc377a","Address":"sdf","FirstName":"demo","ContactId":"","Zip":"23","Company":"tv"}
 

获得 java.lang.IllegalArgumentException：如果非法字符在code查询：

  INT TIMEOUT_MILLISEC = 100000; // 1000 milisec =1秒
INT SOCKET_TIMEOUT_MILISEC = 120000; //2分钟
的HttpParams的HttpParams =新BasicHttpParams（）;
HttpConnectionParams.setConnectionTimeout（的HttpParams，TIMEOUT_MILLISEC）;
HttpConnectionParams.setSoTimeout（的HttpParams，SOCKET_TIMEOUT_MILISEC）;
HttpClient的客户端=新DefaultHttpClient（的HttpParams）;
HttpPost请求=新HttpPost（URL）;
HTT presponse响应= client.execute（要求）;
responseString =请求（响应）;
 

请建议我，如果我做错了什么与我的网址。

*的编辑：的*试图用钥匙仍然得到Exeception：

<$p$p><$c$c>http://72.5.167.50:8084/UpdateProfileInfo?profileinof={"ProfileEditId":"917","ContactsEmail":[{"Email":"sdf","ContactId":""}],"ContactsPhone":[{"CountryId":"+1","Type":"2","Phone":"345345345"}],"ProfileId":"290","LastName":"demo","GroupId":"1212","Title":"dsf","City":"dsf","TemplateId":"1212","State":"dsf","Auth$c$c":"d968273a-0110-461b-8ecf-3f9c456d17ac","Address":"dsf","FirstName":"demo","ContactId":"","Zip":"23","Company":"tv"}

解决方案

目前，我们需要做的这种要求不同的HTTP请求的格式。

我有提到我下面的code这一点。

 公开的JSONObject getJSONObject（）{


    返回jsonObj;
    }
 

上面的方法我会返回一个JSON字符串被传递在下面的方法。

 公共静态的Htt presponse makeRequest（字符串URL）抛出异常
{
    // HttpClient的实例，使申请
    DefaultHttpClient的HttpClient =新DefaultHttpClient（）;

    // URL的POST数据
    HttpPost httpost =新HttpPost（URL）;

    //转换参数成JSON对象
    的JSONObject支架= getJSONObject（）;
    //将结果传递给一个字符串生成器/实体
    StringEntity本身=新StringEntity（holder.toString（））;
    //设置POST请求的结果字符串
    httpost.setEntity（SE）;
    httpost.setHeader（接受，应用/ JSON）;
    httpost.setHeader（内容型，应用/ JSON）;

    //手柄什么从页返回的
    ResponseHandler的ResponseHandler的=新BasicResponseHandler（）;
    返回httpclient.execute（httpost，ResponseHandler所）;
}
 

堆栈后帮我做这个任务...！

I am working with JSON Restful web serivces where I have to pass JSON object in the Service URL. I have created the JSON object successfully but getting exception when my URL created the HTTP connection with the SERVER.

Below I have mention my URL:

 http://72.5.167.50:8084/UpdateProfileInfo?{"ProfileEditId":"917","ContactsEmail":[{"Email":"dsfs","ContactId":""}],"ContactsPhone":[{"CountryId":"+1","Type":"2","Phone":"345345"}],"ProfileId":"290","LastName":"demo","GroupId":"1212","Title":"sdf","City":"dsf","TemplateId":"1212","State":"dsf","AuthCode":"9bcc6f63-2050-4c5b-ba44-b8103fbc377a","Address":"sdf","FirstName":"demo","ContactId":"","Zip":"23","Company":"tv"}

Getting java.lang.IllegalArgumentException: Illegal character in query in code :

int TIMEOUT_MILLISEC = 100000; // 1000 milisec = 1 seconds
int SOCKET_TIMEOUT_MILISEC = 120000; // 2 minutes
HttpParams httpParams = new BasicHttpParams();
HttpConnectionParams.setConnectionTimeout(httpParams, TIMEOUT_MILLISEC);
HttpConnectionParams.setSoTimeout(httpParams, SOCKET_TIMEOUT_MILISEC);
HttpClient client = new DefaultHttpClient(httpParams);
HttpPost request = new HttpPost(url);
HttpResponse response = client.execute(request);
responseString = request(response);

Please suggest me If I am doing something wrong with my URL.

*EDITED:*Tried with a key still getting Exeception:

http://72.5.167.50:8084/UpdateProfileInfo?profileinof={"ProfileEditId":"917","ContactsEmail":[{"Email":"sdf","ContactId":""}],"ContactsPhone":[{"CountryId":"+1","Type":"2","Phone":"345345345"}],"ProfileId":"290","LastName":"demo","GroupId":"1212","Title":"dsf","City":"dsf","TemplateId":"1212","State":"dsf","AuthCode":"d968273a-0110-461b-8ecf-3f9c456d17ac","Address":"dsf","FirstName":"demo","ContactId":"","Zip":"23","Company":"tv"}

解决方案

There is different format of HTTP request that we needed to make for this kind of REQUEST.

I have mention my code below for this.

public JSONObject getJSONObject(){


    return jsonObj;
    }

ABove method returns me a JSON String which is passed in the below method.

public static HttpResponse makeRequest(String url) throws Exception 
{
    //instantiates httpclient to make request
    DefaultHttpClient httpclient = new DefaultHttpClient();

    //url with the post data
    HttpPost httpost = new HttpPost(url);

    //convert parameters into JSON object
    JSONObject holder = getJSONObject();
    //passes the results to a string builder/entity
    StringEntity se = new StringEntity(holder.toString());
    //sets the post request as the resulting string
    httpost.setEntity(se);
    httpost.setHeader("Accept", "application/json");
    httpost.setHeader("Content-type", "application/json");

    //Handles what is returned from the page 
    ResponseHandler responseHandler = new BasicResponseHandler();
    return httpclient.execute(httpost, responseHandler);
}

Stack post helped me for doing this task...!!!

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