泛型集合转json语句方法、时间日期格式解析

public static string Obj2Json<T>(T data)
    {
        try
        {
            System.Runtime.Serialization.Json.DataContractJsonSerializer serializer = new System.Runtime.Serialization.Json.DataContractJsonSerializer(data.GetType());
            using (System.IO.MemoryStream ms = new System.IO.MemoryStream())
            {
                serializer.WriteObject(ms, data);
                return System.Text.Encoding.UTF8.GetString(ms.ToArray());
            }
        }
        catch
        {
            return null;
        }
    }  

用一个字符串接受

时间日期的解析：生成的json日期需要解析

var d = eval('new ' + str.substr(1, str.length - 2));

    var ar_date = [d.getFullYear(), d.getMonth() + 1, d.getDate()];
    function dFormat(i) {
        return i < 10 ? "0" + i.toString() : i;
    }
    for (var i = 0; i < ar_date.length; i++) ar_date[i] = dFormat(ar_date[i]);
    alert(ar_date.join('-'));

 或者

var str = '/Date(1333245600000+0800)/';
 
function data_string(str) {
    var d = eval('new ' + str.substr(1, str.length - 2));
    var ar_date = [d.getFullYear(), d.getMonth() + 1, d.getDate()];
    for (var i = 0; i < ar_date.length; i ++) ar_date[i] = dFormat(ar_date[i]);
    return ar_date.join('-');
     
    function dFormat(i) { return i < 10 ? "0" + i.toString() : i; }
}
 
alert(data_string(str));