如何在Servlet内的JSONObject中放置对象的JSONArray

Question

我正在尝试使用$ .ajax（）方法动态填充多个选择框。...这是我的html代码...

    <select id="imageSize" name="ImageSizeId"></select>
    <select id="circulation" name="circulationId"></select>

这是我的servlet代码.....

    protected void doGet(HttpServletRequest request, HttpServletResponse response)   throws ServletException, IOException {
    response.setContentType("text/json;charset=utf-8");
    response.setHeader("Cache-Control", "no-cache");
    PrintWriter out = response.getWriter();
    JSONObject usesfactors = new JSONObject();  
    JSONArray jArraysize = new JSONArray();
    JSONArray jArraycirculation = new JSONArray();
    Statement stmtsize,stmtcirculation;
    ResultSet rssize,rscirculation;
    int category_id = Integer.parseInt(request.getParameter("CategoryId"));
    int format_id = Integer.parseInt(request.getParameter("FormatId"));
    String size,display,des;
    int sizeid;



  try {
          if(category_id == 2 && format_id == 201){

    // Get image size factors

                     String sql1="SELECT SizeId,Size FROM Size WHERE Category_id = "+category_id+" AND Format_id = "+format_id+" ";                                          
                     PreparedStatement ps1 = conn.prepareStatement(sql1);
                     rssize =  ps1.executeQuery() ; 

                    if( rssize!=null){
                               System.out.println("Not Null");// its printing even if resultset rssize has no records.......why?
                            while(rssize.next())
                                {   
                                     System.out.println("inside resultset");            
                                     JSONObject jobj = new JSONObject();  
                                     sizeid = rssize.getInt("SizeId");
                                     size=rssize.getString("Size");  
                                     System.out.println(size);
                                     jobj.put("SizeId", sizeid);
                                     jobj.put("Size", size);                        
                                     jArraysize.add(jobj);
                                }
                              usesfactors.put("Size", jArraysize);
                              rssize.close();
                    }
                    else{
                              System.out.println("Does not have Size factors");
                    }

    // Get image circulation factors  

                    String sql2="SELECT circulationId,circulation FROM Circulation WHERE Category_id = "+category_id+" AND Format_id = "+format_id+" ";
                    PreparedStatement ps2 = conn.prepareStatement(sql2);                      
                    rscirculation =  ps2.executeQuery() ;    

                    if(rscirculation!=null){
                            while(rscirculation.next())
                                {      

                                     JSONObject jobj = new JSONObject();  
                                     display = rscirculation.getString("DisplayName");
                                     des=rscirculation.getString("Description");                                               
                                     jobj.put("DisplayName", display);
                                     jobj.put("Description", des);                        
                                     jArraycirculation.add(jobj);

                                }
                            usesfactors.put("Circulation", jArraycirculation);
                            rscirculation.close();
                  }
                    else{
                              System.out.println("Does not have Circulation factors");
                    }

                  out.println(usesfactors);
          }

我得到的是空的json结果...。错了吗？ {“ Size”：[]，“ Circulation”：[]}我不想执行此stmt“ usesfactors.put（” Size“，jArraysize）;” 如果resultset rssize为null，但是即使结果集rssize没有记录，此stmt也将执行。...总之，当且仅当结果集rssize有记录时，我想将json array（jArraysize）放在json对象（usesfactors）中。

Answer 1

The problem in your code might be your sql queries which are not returning any values.because i have done in my IDE in the same way you have done,only difference is that instead of getting values from database i have kept manually by iterating for loop.below is the code...
JSONObject usesfactors = new JSONObject();  
             JSONArray jArraysize = new JSONArray();
             JSONArray jArraycirculation = new JSONArray();

             for(int i=0;i<3;i++)
             {
             JSONObject jobj = new JSONObject();  
             jobj.put("SizeId", "1");
             jobj.put("Size", "2");                        
             jArraysize.put(jobj);                
             }
             usesfactors.put("Size", jArraysize);

             for(int i=0;i<3;i++)
             {
                  JSONObject jobj = new JSONObject();  
                  jobj.put("DisplayName", "3");
                  jobj.put("Description", "4");                        
                  jArraycirculation.put(jobj);
             }
             usesfactors.put("Circulation", jArraycirculation);
             System.out.println(usesfactors);

Answer 2

为了回答代码中有关为什么它进入rssize!=null块的问题，请参见executeQuery上的javadoc 。 它明确声明它永远不会返回null，因此null检查始终会通过。 这意味着一个空的结果集将跳过该部分中的循环，但仍会达到目标

usesfactors.put("Size", jArraysize);

应该重写此代码，以实际上避免在空返回集上插入。 应该进行类似的更改以更正流通部分，这也会导致无效的无效检查。

while(rssize.next())
{   
  System.out.println("inside resultset");            
  JSONObject jobj = new JSONObject();  
  sizeid = rssize.getInt("SizeId");
  size=rssize.getString("Size");  
  System.out.println(size);
  jobj.put("SizeId", sizeid);
  jobj.put("Size", size);                        
  jArraysize.add(jobj);
}
if (jArraysize.length > 0)
{
  usesfactors.put("Size", jArraysize);
}
else
{
  System.out.println("Does not have Size factors");
}
rssize.close();