Question

如何使用ArrayList创建此JSON以在Android中发布json

{
"user" : 
    {
        "nickname" : "nickname6",
        "password" : "1234",
    }
}

我只得到一个扁平的JSON

    ArrayList<BasicNameValuePair> nameValuePairs = new ArrayList<BasicNameValuePair>();

    nameValuePairs.add(new BasicNameValuePair("nickname","nickname6"));
    nameValuePairs.add(new BasicNameValuePair("password", "1234"));

Answer 1

您需要创建一个JSON，为此您必须在POST方法中添加该JSON作为参数。要做到这一点，你可以试试这个：

RewriteCond %{HTTP_HOST} !^www\.
RewriteRule ^(.*)$ https://www.%{HTTP_HOST}/$1 [R=301,L]

RewriteCond $1 !\.(gif|jpe?g|png)$ [NC]
RewriteCond %{REQUEST_FILENAME} !-f 
RewriteCond %{REQUEST_FILENAME} !-d 
RewriteRule ^(.*)$ /index.php?/$1 [L]

RewriteCond %{HTTPS} off
RewriteRule .* https://%{HTTP_HOST}%{REQUEST_URI} [L,R=301]
RewriteCond %{HTTP_HOST} !^www\.
RewriteRule .* https://www.%{HTTP_HOST}%{REQUEST_URI} [L,R=301]

RewriteBase / 
RewriteCond %{THE_REQUEST} ^GET.*index\.php\?/ [NC] 
RewriteRule (.*?)index\.php\?/ / [R=301,NE,L]

Answer 2

如您所知，HttpClasses，NameValuePair和BasicNameValuePair已在最新的android中弃用。我们现在应该避免它。

如果你想创建

{
   "user":{
        "nickname":"nickname6"
        "password":"1234",             
    }
}

您可以使用下面的代码示例使用JSONObject类创建相同的json。

JSONObject jObj = new JSONObject();

try {       
    JSONObject userCredentials = new JSONObject();
    userCredentials.put("nickname","nickname6");
    userCredentials.put("password","1234");

    jObj.put("user", userCredentials);
} catch(Exception e) {
   e.printStackTrace();
}

Answer 3

尝试像这样使用ContentValues

ContentValues values=new ContentValues();
values.put("username",name);
values.put("password",password);

OR 使用MultipartEntity

MultipartEntity multi = new MultipartEntity();
multi.addPart("name", new StringBody("your data"));
multi.addPart("Id", new StringBody("123"));

Answer 4

这里是使用AsyncTask的http帖子的一个例子：

public class httpSendrequest extends AsyncTask<Void,Void,String> {

        ArrayList<NameValuePair> nvPairs=new ArrayList<>();

    Boolean error=false;

        @Override
        protected void onPreExecute() {
            super.onPreExecute();

            //here you initialize your json object and add it to value pairs
            JSONObject jObj = new JSONObject();

    try {       
            JSONObject userCredentials = new JSONObject();
            userCredentials.put("nickname","nickname6");
            userCredentials.put("password","1234");

            jObj.put("user", userCredentials);
    nvPairs.add(new BasicNameValuePair("jsonstring",jObj.toString()));
    } catch(Exception e) {
      error=true;
    }

        }

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

    if(!error)
            try{

                String link ="http://"+ip+"/QSystem.asmx/insert_answer" ;
                HttpClient httpclient = new DefaultHttpClient();
                HttpPost httppost = new HttpPost(link); //adding URL to the http post
                httppost.setEntity(new UrlEncodedFormEntity(nvPairs, HTTP.UTF_8)); //adding the value pairs and encoding to the http post request
                HttpResponse response = httpclient.execute(httppost);
                HttpEntity resEntity = response.getEntity();


            } catch (Exception e){
                error=true;
            }

            return "str";
        }

        @Override
        protected void onPostExecute(String s) {
            super.onPostExecute(s);

            if(!error) {
                Toast.makeText(getApplicationContext(), "Sent", Toast.LENGTH_SHORT).show();

            }
        }

    }

然后你可以从onCreate或OnClickListener中调用它来获取一个按钮，如下所示：

new httpSendrequest().execute();

希望这会有所帮助:)

使用BasicNameValuePair创建复杂的JSON

4 个答案: