從JSON數組到JSON對象轉換工作不正常

Question

我創建了這樣一個REST服務：

var rootURL = "http://localhost:8181/RestServiceProject/rest/WebService/getsummary"; 
function findBySummary() { 
    console.log("inside loading grid"); 
    var returnval1; 
    $.ajax({ 
     type: 'GET', 
     url: rootURL, 
     dataType: "json", 
     async: false, 
     success: function(json3){ 

      returnval1 = json3; 
      console.log(returnval1); 

     } 

    }); 
    return returnval1; 
} 

console.log(returnval1); 

回報：

@GET 
@Path("/getsummary") 
@Produces("application/json") 

public Response summary() 
{ 

     VoltDAOImpl voltDao = new VoltDAOImpl(); 
     Map<String ,List<HashMap<String,String>>> returnList=  voltDao.getOrderDetails("PEAKM" , "Hydra" , 
       "" , voltDao.client,"notional" ,1000); 
     List<HashMap<String,String>> totalSlpSummaryList = returnList.get("total_slp_summary"); 
     List<HashMap<String,String>> totalSlpSummaryBySideList = returnList.get("total_slp_summary_by_side"); 
     ObjectWriter ow = new ObjectMapper().writer().withDefaultPrettyPrinter(); 

      String json1 = null; 
      String json2 = null; 
      String json3 = null; 
      try { 
       json1 = ow.writeValueAsString(totalSlpSummaryList); 
       json2 = ow.writeValueAsString(totalSlpSummaryBySideList); 
       json3="["+json1+","+json2+"]"; 
       //json3 = json1 + "\n" + json2; 
       System.out.println(json1); 
       System.out.println(json2); 
      } catch (IOException e) { 
       // TODO Auto-generated catch block 
       e.printStackTrace(); 
      } 


    return Response.status(200).entity(json3).build(); 
} 

然後我在這樣一個javascript這個消費結果如下： returnval1

後來我打電話findBySummary();並執行以下操作：

var json3 = findBySummary(); 

     var json4 = json3[0]; 
console.log("json4"); 
     console.log(json4); 
     var json_temp = JSON.stringify(json4); 
     console.log("json_temp") 
     console.log(json_temp); 
     var json_temp1 = json_temp[0]; 
     console.log("json temp1"); 
     console.log(json_temp1); 

下面是此日誌輸出： log output

我想轉換json_temp，這就是：

[{"total_exec_qty":"286595","total_notional":"21820771.72","total_wt_arr_last_slp":"2.4364","total_num_ords":"1630","total_wt_ivwap_slp":"6.0969","total_wt_arr_slp":"1.7889","total_ord_qty":"576991"}] 

這個（json_temp1）：

{"total_exec_qty":"286595","total_notional":"21820771.72","total_wt_arr_last_slp":"2.4364","total_num_ords":"1630","total_wt_ivwap_slp":"6.0969","total_wt_arr_slp":"1.7889","total_ord_qty":"576991"} 

但是爲什麼我只爲json_temp1獲得[？

Answer 1

請注意，JSON.stringify(json4)返回一個字符串。

因此，代碼json_temp[0]將返回此字符串中的第一個元素。

你想要的是：

var json_temp1 = json4[0];

，而不是：

var json_temp1 = json_temp[0];

下面的代碼是不適合您的需要：

var json_temp = JSON.stringify(json4); 
console.log("json_temp") 
console.log(json_temp); 
var json_temp1 = json_temp[0];