如何初始化具有开始日期和结束日期作为参数的t1对象?错误行是t1.store(" task1"," gui design",2014 01 01,2014 04 04,"已完成")。我如何将日期参数传递给存储方法...任何人都可以帮我找出来吗?
class Program
{
static void Main(string[] args)
{
Task t1 = new Task();
Task t2 = new Task();
t1.store("task1","gui design",2014 01 01,2014 04 04,"completed");
}
}
class Task
{
string _Tid;
string _tn;
DateTime _sdate;
DateTime _edate;
string _status;
public void store(string tid,string tname, Date start,Date end,string sts)
{
this._Tid = tid;
this._tn = tname;
this._sdate = start;
this._edate = end;
this._status = sts;
}
public void print()
{
Console.WriteLine("\n {0}\t{1}\t{2}\t{3}\t{4}",this._Tid,this._tn,this._sdate,this._edate,this._status);
}
}
}
答案 0 :(得分:1)
您可以通过这些方法创建DateTime
var dt1 = new DateTime(2014,10,25);
var dt2 = DateTime.Parse("2014/10/25");
并将方法签名更改为此
public void store(string tid,string tname, DateTime start,DateTime end,string sts)
如果您想确保只使用DateTime对象的Date部分,可以使用start.Date或end.Date
答案 1 :(得分:0)
t1.store("task1","gui design",Convert.ToDateTime(01/01/2014),Convert.ToDateTime(04/04/2014),"completed");
答案 2 :(得分:0)
您的参数不正确。 你必须使用这个数字: 新日期(2014 01 01)并通过拖曳法;
class Program
{
static void Main(string[] args)
{
Task t1 = new Task();
Task t2 = new Task();
t1.store("task1","gui design",new DateTime(2014, 01, 01),new DateTime(2014 ,04 ,04),"completed");
}
}
class Task
{
string _Tid;
string _tn;
DateTime _sdate;
DateTime _edate;
string _status;
public void store(string tid, string tname, DateTime start, DateTime end, string sts)
{
this._Tid = tid;
this._tn = tname;
this._sdate = start;
this._edate = end;
this._status = sts;
}
public void print()
{
Console.WriteLine("\n {0}\t{1}\t{2}\t{3}\t{4}", this._Tid, this._tn, this._sdate, this._edate, this._status);
}
}
答案 3 :(得分:0)
由于您使用的是custom Date
implementation,因此调用该方法的正确方法如下:
t1.store("task1","gui design", new Date(2014, 01, 01), new Date(2014, 04, 04) ,"completed");