I have an assigment where im making a calendar, they have alot of options and theres 2 submenus.
when they are in a menu and they choose an option, it performs a task, after that the program closes, how do i get it to go back to the menu they was on.?
please help me?
do you have any code already or idea in the general direction which you want to go? if you do post what you have and i might be able to to help
The menu part is inside a loop. Or it could be in a function which is called in a loop.
basically, when the switch menu option is choose, like 1, it performs a function, i just want the menu to come back up after they have done the function so they can carry on performing tasks instead of it closing and haviong to reopen itCode:int main(void) { int value=0; printf("\nChoose from the list of options:\n"); printf("\n1.Calender Options"); printf("\n2.Appointment Options"); printf("\n3.Quit\n\n"); printf("Option: "); scanf("%d", &value); switch(value) { case 1: subm_cal(); break; case 2: subm_app(); break; case 3: // This Quits from the menu break; // default: // For when they type something not in range of 1-3 printf("\nInvalid Choice\n\n"); // } printf("\n\n"); system("pause"); return 0; } ////////////////////////////////////////////////////////////////////////// // Sub menus / ////////////////////////////////////////////////////////////////////////// void subm_cal() { int value=0; printf("\nCalender Menu:\n"); printf("\n1.Monthly Calender"); printf("\n2.Yearly Calender"); printf("\n3.Time Slots"); printf("\n4.Return To Main Menu\n\n"); printf("Option: "); scanf("%d", &value); switch(value) { case 1: int n, s; printf("\n\nEnter number of days in month:"); // scanf("%d", &n); // printf("\n\nEnter the day for the first of month:"); // This Area is calling the function to make the calender scanf("%d", &s); // makeCalendar(n,s); // break; // case 2: cal_year(); break; case 3: printf("Time Slots Here"); break; case 4: break; default: // For when they type something not in range of 1-3 printf("\nInvalid Choice\n\n"); // } } void subm_app() { int value=0; printf("\nAppointments Menu:\n"); printf("\n1.Add Appointment"); printf("\n2.Edit Appointment"); printf("\n3.View / Load Appointments"); printf("\n4.Return To Main Menu\n\n"); printf("Option: "); scanf("%d", &value); switch(value) { case 1: appointment(); break; case 2: printf("Edit Appointments Here"); break; case 3: ld_apps(); break; case 4: break; default: // For when they type something not in range of 1-3 printf("\nInvalid Choice\n\n"); // } }
It looks like you could place your switch statements into a couple of loops.
Jim
what a loop for each switchboard? or like 1 for main, and one for my submenus?
Sounds good, why don't you give it a try and see for yourself?
Jim
Code:while (1) { /* loop forever */ printf("\nChoose from the list of options:\n"); printf("\n1.Calender Options"); printf("\n2.Appointment Options"); printf("\n3.Quit\n\n"); printf("Option: "); scanf("%d", &value); switch(value) { case 1: subm_cal(); break; case 2: subm_app(); break; case 3: // This Quits from the menu break; // default: // For when they type something not in range of 1-3 printf("\nInvalid Choice\n\n"); // } printf("\n\n"); system("pause"); }
thanks nonoob
it works perfectly, i hate C to be 100% honest. but its soo pleasing when it all works =)
When you start succeeding more than failing, say, over 50% of the time, things start getting fun! Eventually you'll reach a point where you KNOW everything is solvable by yourself, given a bit of time. That's when you start wishing for really complicated challenges.
I'd say it takes at least a few months if this is your first programming language.
yeah thanks for your help.
the while loop you gave me worked, but i couldnt exit it, was very funny.
i latered it now, so that it all quits and stuff
made my while loop, not equal to a value which when entered quits.
=)
Oh sorry. Yes the nice way is to put a decision in the while() to exit. I saw the return 0 but I didn't pay close attention. If it's for the exit case 4 then the forever loop would have been OK.
Originally Posted by Conor1992
Code:while (1) { /* loop forever */ printf("\nChoose from the list of options:\n"); printf("\n1.Calender Options"); printf("\n2.Appointment Options"); printf("\n3.Quit\n\n"); printf("Option : "); scanf("%d", &value); switch(value) { case 1: subm_cal(); break; case 2: subm_app(); break; case 3: // This Quits from the menu Exit(1); default: // For when they type something not in range of 1-3 printf("\nInvalid Choice\n\n"); // } }
Not sure what Exit() is with a capital 'E' (is it some special function I don't know about?), but exit() will quit more than the menu. It will quit the entire program in great haste, skipping any "leave the terminal open" commands and any clean up code.Code:case 3: // This Quits from the menu Exit(1);
@OP: I find the best method for running a loop based on user input is to use a do while loop. Do while loops always execute once, which is good because you always want to display the menu and get input at least once. You also don't have to worry about initializing the user's input value to something non-quit. Example:
Code:do { print menu get input act on input } while (input != quit);
Hermmm... guilty with explaination...
That would be yer basic typo anduril. In my own defense... I am working with a messed up right hand (sprained fingers)...
Code:case 3: // This Quits from the menu // insert cleanup code here exit(1);
