A new expression requires (), [], or {} after type

Solution: EmployeProp empobj = new EmployeProp --Check here missed (); 
                    foreach (EmployeeInfo emps in empdetails)
                    {
                    }

Bad request 400 blobstorage Uploading Large Files

Add In Web.config:

<configuration>
   <system.webServer>
      <security>
         <requestFiltering>
            <requestLimits maxAllowedContentLength="<valueInBytes>"/>
         </requestFiltering>
      </security>
   </system.webServer>
</configuration>

Solution : Container name lowercase only

convert to verbating string

@"text" -- Text Before Add @

Saving a file in current working directory

Save : File.Copy(path.filename(uploaded Path), Appdomain.path);
Delete: File.Delete("Appdomain.path");

Find number of days between two given dates C++

// C++ program two find number of days between two given dates

#include<iostream>

using namespace std;

 

// A date has day 'd', month 'm' and year 'y'

struct Date

{

    int d, m, y;

};

 

// To store number of days in all months from January to Dec.

const int monthDays[12] = {31, 28, 31, 30, 31, 30,

                           31, 31, 30, 31, 30, 31};

 

// This function counts number of leap years before the

// given date

int countLeapYears(Date d)

{

    int years = d.y;

 

    // Check if the current year needs to be considered

    // for the count of leap years or not

    if (d.m <= 2)

        years--;

 

    // An year is a leap year if it is a multiple of 4,

    // multiple of 400 and not a multiple of 100.

    return years / 4 - years / 100 + years / 400;

}

 

// This function returns number of days between two given

// dates

int getDifference(Date dt1, Date dt2)

{

    // COUNT TOTAL NUMBER OF DAYS BEFORE FIRST DATE 'dt1'

 

    // initialize count using years and day

    long int n1 = dt1.y*365 + dt1.d;

 

    // Add days for months in given date

    for (int i=0; i<dt1.m - 1; i++)

        n1 += monthDays[i];

 

    // Since every leap year is of 366 days,

    // Add a day for every leap year

    n1 += countLeapYears(dt1);

 

    // SIMILARLY, COUNT TOTAL NUMBER OF DAYS BEFORE 'dt2'

 

    long int n2 = dt2.y*365 + dt2.d;

    for (int i=0; i<dt2.m - 1; i++)

        n2 += monthDays[i];

    n2 += countLeapYears(dt2);

 

    // return difference between two counts

    return (n2 - n1);

}

 

// Driver program

int main()

{

    Date dt1 = {1, 2, 2000};

    Date dt2 = {1, 2, 2004};

 

    cout << "Difference between two dates is " << getDifference(dt1, dt2);

 

    return 0;

}