如何在for循环中迭代日期?
How to iterate date in for loop?
我有一些价值观。
DateTime date=04/03/2015(date)
Total Woring days=6 (total)
Rotation Days=2 (rotationday)
Shift Group=S1(this group contain two shift id 1 and 2)
我想运行轮班 6 天。但是在每 2 天 Shift id 1 旋转到 shift id 2 之后,又在两天后 shift id 2 旋转到 shift id 1 等等......
我的输出应该是
04/03/2015=1
05/03/2015=1
06/03/2015=2
07/03/2015=2
08/03/2015=1
09/03/2015=1
我正在通过 foreach 循环获取班次 ID。我尝试了下面提到的方式,但没有得到正确的结果。请帮我解决这个问题
SqlCommand cmd2 = new SqlCommand("select ShiftID from ShiftGroup where
ShiftName='" + ide + "'", conn2);
SqlDataAdapter sda2 = new SqlDataAdapter(cmd2);
DataSet ds4 = new DataSet();
var rows2 = ds4.Tables[0].Rows;
if (ds4.Tables[0].Rows.Count > 0)
{
foreach (DataRow row2 in rows2)
{
string shiftname = Convert.ToString(row2["ShiftID"]);
DateTime date=04/03/2015(date)
Total Woring days=6 (total)
Rotation Days=2 (rotationday)
DateTime rotation= date.AddDays(rotationday);//
DateTime totaldate = date.AddDays(workingdays);
for (DateTime rotation = date; rotation <= totaldate; rotation = rotation.AddDays(1))//total working days
{
//to check rotation date
if (rotation <= totaldate )
{
allocation.shiftallocation( rotation, shiftid);
}
}
}
我从昨天开始就被这个问题困住了,谁能帮帮我。无需考虑我的 for 循环,请提供生成上述输出的 for 循环
var date = DateTime.Parse("04/03/2015");
var totalWorkingDays = 6;
var rotationDays = 2;
var rotationId = 1;
var rotationCounter = 1;
for (DateTime rotation = date;
rotation <= date.AddDays(totalWorkingDays);
rotation = rotation.AddDays(1))
{
Console.WriteLine(rotation + " " + rotationId);
if (rotationCounter++ == 2)
{
rotationCounter = 1;
rotationId = 3 - rotationId;
}
}
或者,使用 linq 查询
var date = DateTime.Parse("04/03/2015");
var totalWorkingDays = 6;
var rotationDays = 2;
foreach (var d in Enumerable.Range(0, totalWorkingDays)
.Select((i, index) => date.AddDays(i) +
(((int)(index / rotationDays)) % 2 == 1 ? " 2" : " 1")))
{
Console.WriteLine(d);
}
你可以试试这个,
DateTime date=DateTime.ParseExact("04/03/2015","dd/MM/yyyy",CultureInfo.InvariantCulture);
DateTime totaldate = date.AddDays(6);
for (int i=0; date <= totaldate; date = date.AddDays(1),i++)//total working days
{
if((i/2)%2==0)
Console.WriteLine(date+" "+1);
else
Console.WriteLine(date+" "+2);
}
不要在循环周期中使用日期,使用抽象索引。你真的可以从具体数字中抽象出来,只使用变量来管理结果。因此,您可以轻松更改旋转天数甚至轮班标签数组,而无需更改周期。
var date = DateTime.Parse("04/03/2015");
var totalWorkingDays = 15;
var rotationDays = 2;
var workshifts = new[] { "S1", "S2" };
var currentWorkshiftIndex = 0;
for (int dayIndex = 0; dayIndex <= totalWorkingDays; dayIndex++) {
if (dayIndex != 0 && dayIndex % rotationDays == 0) currentWorkshiftIndex = (currentWorkshiftIndex + 1) % workshifts.Length;
Console.WriteLine(date.AddDays(dayIndex) + " " + workshifts[currentWorkshiftIndex]);
}
考虑到节假日的通用解决方案。
int totalDays = 10;
int rotationDays = 3;
int[] shifts = new[] { 1, 2 };
string startDate = "04/03/2015";
var holidays = new List<DayOfWeek>(new[] { DayOfWeek.Saturday, DayOfWeek.Sunday, DayOfWeek.Tuesday });
var shiftAllocations = new Dictionary<DateTime, int>();
int currentShiftIndex = 0;
DateTime current = DateTime.ParseExact(startDate, "dd/MM/yyyy", CultureInfo.InvariantCulture);
for (int i = 0; i < totalDays; i += rotationDays)
{
var currentShiftId = shifts[currentShiftIndex];
// For each day from the current day till the number of shift rotation days, allocate the shift.
for (int j = 0; j < rotationDays;)
{
// If the current day is a holiday, move to the next day by incrementing i.
if (holidays.Contains(current.AddDays(i + j).DayOfWeek))
{
i++;
continue;
}
shiftAllocations.Add(current.AddDays(i + j), currentShiftId);
j++;
}
// Increase the shift index if the value is within the bounds. If not reset the index to the beginning
if ((currentShiftIndex + 1) >= shifts.Length)
currentShiftIndex = 0;
else
currentShiftIndex++;
}
foreach (var kvp in shiftAllocations)
{
Console.WriteLine(kvp.Key.ToString("dd/MM/yyyy") + ":" + kvp.Value);
}
根据你对每周5个工作日的评论,我写了这个:
var start = new DateTime(2015, 04, 03);
var working_day_count = 6;
var rotation_interval = 2;
var shifts = new List<int> { 1, 2 };
var rotation_count = 1;
var shift_index = 0;
for (var i = 0; i < working_day_count; i++) {
while ((start.DayOfWeek == DayOfWeek.Saturday) ||
(start.DayOfWeek == DayOfWeek.Sunday)) {
start = start.AddDays(1);
}
Console.WriteLine(start.ToString() + " = " + shifts[shift_index].ToString());
start = start.AddDays(1);
rotation_count++;
if (rotation_count > rotation_interval) {
rotation_count = 1;
shift_index++;
if (shift_index >= shifts.Count) {
shift_index = 0;
}
}
}
只需根据需要更改前四个变量的值即可。
我试图让它易于理解,而不是高性能或紧凑。
我有一些价值观。
DateTime date=04/03/2015(date)
Total Woring days=6 (total)
Rotation Days=2 (rotationday)
Shift Group=S1(this group contain two shift id 1 and 2)
我想运行轮班 6 天。但是在每 2 天 Shift id 1 旋转到 shift id 2 之后,又在两天后 shift id 2 旋转到 shift id 1 等等...... 我的输出应该是
04/03/2015=1
05/03/2015=1
06/03/2015=2
07/03/2015=2
08/03/2015=1
09/03/2015=1
我正在通过 foreach 循环获取班次 ID。我尝试了下面提到的方式,但没有得到正确的结果。请帮我解决这个问题
SqlCommand cmd2 = new SqlCommand("select ShiftID from ShiftGroup where
ShiftName='" + ide + "'", conn2);
SqlDataAdapter sda2 = new SqlDataAdapter(cmd2);
DataSet ds4 = new DataSet();
var rows2 = ds4.Tables[0].Rows;
if (ds4.Tables[0].Rows.Count > 0)
{
foreach (DataRow row2 in rows2)
{
string shiftname = Convert.ToString(row2["ShiftID"]);
DateTime date=04/03/2015(date)
Total Woring days=6 (total)
Rotation Days=2 (rotationday)
DateTime rotation= date.AddDays(rotationday);//
DateTime totaldate = date.AddDays(workingdays);
for (DateTime rotation = date; rotation <= totaldate; rotation = rotation.AddDays(1))//total working days
{
//to check rotation date
if (rotation <= totaldate )
{
allocation.shiftallocation( rotation, shiftid);
}
}
}
我从昨天开始就被这个问题困住了,谁能帮帮我。无需考虑我的 for 循环,请提供生成上述输出的 for 循环
var date = DateTime.Parse("04/03/2015");
var totalWorkingDays = 6;
var rotationDays = 2;
var rotationId = 1;
var rotationCounter = 1;
for (DateTime rotation = date;
rotation <= date.AddDays(totalWorkingDays);
rotation = rotation.AddDays(1))
{
Console.WriteLine(rotation + " " + rotationId);
if (rotationCounter++ == 2)
{
rotationCounter = 1;
rotationId = 3 - rotationId;
}
}
或者,使用 linq 查询
var date = DateTime.Parse("04/03/2015");
var totalWorkingDays = 6;
var rotationDays = 2;
foreach (var d in Enumerable.Range(0, totalWorkingDays)
.Select((i, index) => date.AddDays(i) +
(((int)(index / rotationDays)) % 2 == 1 ? " 2" : " 1")))
{
Console.WriteLine(d);
}
你可以试试这个,
DateTime date=DateTime.ParseExact("04/03/2015","dd/MM/yyyy",CultureInfo.InvariantCulture);
DateTime totaldate = date.AddDays(6);
for (int i=0; date <= totaldate; date = date.AddDays(1),i++)//total working days
{
if((i/2)%2==0)
Console.WriteLine(date+" "+1);
else
Console.WriteLine(date+" "+2);
}
不要在循环周期中使用日期,使用抽象索引。你真的可以从具体数字中抽象出来,只使用变量来管理结果。因此,您可以轻松更改旋转天数甚至轮班标签数组,而无需更改周期。
var date = DateTime.Parse("04/03/2015");
var totalWorkingDays = 15;
var rotationDays = 2;
var workshifts = new[] { "S1", "S2" };
var currentWorkshiftIndex = 0;
for (int dayIndex = 0; dayIndex <= totalWorkingDays; dayIndex++) {
if (dayIndex != 0 && dayIndex % rotationDays == 0) currentWorkshiftIndex = (currentWorkshiftIndex + 1) % workshifts.Length;
Console.WriteLine(date.AddDays(dayIndex) + " " + workshifts[currentWorkshiftIndex]);
}
考虑到节假日的通用解决方案。
int totalDays = 10;
int rotationDays = 3;
int[] shifts = new[] { 1, 2 };
string startDate = "04/03/2015";
var holidays = new List<DayOfWeek>(new[] { DayOfWeek.Saturday, DayOfWeek.Sunday, DayOfWeek.Tuesday });
var shiftAllocations = new Dictionary<DateTime, int>();
int currentShiftIndex = 0;
DateTime current = DateTime.ParseExact(startDate, "dd/MM/yyyy", CultureInfo.InvariantCulture);
for (int i = 0; i < totalDays; i += rotationDays)
{
var currentShiftId = shifts[currentShiftIndex];
// For each day from the current day till the number of shift rotation days, allocate the shift.
for (int j = 0; j < rotationDays;)
{
// If the current day is a holiday, move to the next day by incrementing i.
if (holidays.Contains(current.AddDays(i + j).DayOfWeek))
{
i++;
continue;
}
shiftAllocations.Add(current.AddDays(i + j), currentShiftId);
j++;
}
// Increase the shift index if the value is within the bounds. If not reset the index to the beginning
if ((currentShiftIndex + 1) >= shifts.Length)
currentShiftIndex = 0;
else
currentShiftIndex++;
}
foreach (var kvp in shiftAllocations)
{
Console.WriteLine(kvp.Key.ToString("dd/MM/yyyy") + ":" + kvp.Value);
}
根据你对每周5个工作日的评论,我写了这个:
var start = new DateTime(2015, 04, 03);
var working_day_count = 6;
var rotation_interval = 2;
var shifts = new List<int> { 1, 2 };
var rotation_count = 1;
var shift_index = 0;
for (var i = 0; i < working_day_count; i++) {
while ((start.DayOfWeek == DayOfWeek.Saturday) ||
(start.DayOfWeek == DayOfWeek.Sunday)) {
start = start.AddDays(1);
}
Console.WriteLine(start.ToString() + " = " + shifts[shift_index].ToString());
start = start.AddDays(1);
rotation_count++;
if (rotation_count > rotation_interval) {
rotation_count = 1;
shift_index++;
if (shift_index >= shifts.Count) {
shift_index = 0;
}
}
}
只需根据需要更改前四个变量的值即可。 我试图让它易于理解,而不是高性能或紧凑。