我有一个json数组,如下图所示:
我希望它是onyl json对象而不是数组。就像我做JSON.stringify一样,我得到了这个:[{"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"}]
但我只想这样:{"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"}
我如何得到它?
更新:现在就这样做:
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);
但是在console.log中获得以下内容: Getting this problem
答案 0 :(得分:2)
只需引用这样的对象:
var array = [{"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"}];
var object = array[0];
或者你可以复制对象并将其重新分配给同一个变量,如下所示,注意:第二种方法有点贵。
var data = [{"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"}];
data = JSON.parse(JSON.stringify(data[0]));
Here是一个有效的例子。