Library Management System in python OOPs Concept

 class Library:

    def__init__(self,listofbooks,libraryname):

        self.lst = listofbooks

        self.name = libraryname

    def displaybook(self):

        return f"\t\t\t\tWelcome to {self.name} \

        \navaliable books are -\n {self.lst}"

    def lendbook(self,lendname,bookname):

        if bookname in lst and bookname not in d:

            print(f"{bookname} lend to {lendname}")

            d.__setitem__(bookname,lendname)

        elif bookname in d:

            lst.remove(bookname)

            print("Book is not available lend to-",d[bookname],"\nwe only have-",lst)

        else:

            print("Book is not in the library we only have-",lst)

    def addbook(self,bookname):

        print("Book added successfully in the library")

        lst.append(bookname)

    def returnbook(self,returenername,bookname):

        if returenername in d[bookname]:

            print("Book successfully returned")

            d.pop(bookname)

            

        else :

            print("Wrong data")


if __name__=='__main__':       

    lst = ['Python','Science','Mathematics',

    'Social Science']

    l = Library(lst,'World Library')

    print(l.displaybook())

    d = {"Python":"Ravi Dubey"}

    exit = 1

    while exit:

        print("For Lending book press 1\nFor add book press 2\nFor return book press 3\nFor exit press 4")

        n = int(input())

        if n==1:

            l.lendbook(str(input("Enter your name-")),str(input("Enter book name-")))

        elif n==2:

            l.addbook(str(input("Enter book name you want to add-")))

        elif n==3: 

            l.returnbook(str(input("Enter your name-")),str(input("Enter book name-")))

        elif n==4:

            exit = 0

        else:

            print("Wrong input")

Video of code is - 



Comments

Post a Comment

If you have any doubt you ask me in comment section

Popular posts from this blog

Create account in java |deposit money|withdraw money

Image
 class bank{ int ac_no; String name, type; float amount; void details (int n, String nm, String t, float a)  { ac_no=n; name=nm; type=t; amount=a; } void show()  { System. out. println(ac_no+" "+name+" "+type+" "+amount) ; } void deposit (float a)  {amount =amount+a; System.out.println(" Deposit amount:"+a) ; } void withdraw (float a)  { if(amount<a)  System.out.println("Balance is insufficient "); else{ amount=amount-a; System.out.println("Withdraw amount:"+a) ; } } void checkbalance()  { System.out.println("Balance is:"+amount); } } class test{ public static void main(String args[])  { bank b=new bank() ; b. details (123456, "Ankit", " Saving ", 1000) ; b. show() ; b. deposit (40000) ; b. checkbalance () ; b. withdraw (10000) ;  b. checkbalance () ; } } Output- 123456 Ankit Saving 1000.0 Deposit amount: 40000.0  Balance is :41000.0 Withdraw amount: 10000 Balance is: 31000
Read more

List comprehension hackerrank solution in python

Image
  Let's learn about list comprehensions! You are given three integers     and     representing the dimensions of a cuboid along with an integer   . Print a list of all possible coordinates given by     on a 3D grid where the sum of     is not equal to   . Here,   . Please use list comprehensions rather than multiple loops, as a learning exercise. Example All permutations of   are: . Print an array of the elements that do not sum to  . Input Format Four integers   and  , each on a separate line. Constraints Print the list in lexicographic increasing order. Sample Input 0 1 1 1 2 Sample Output 0 [[0, 0, 0], [0, 0, 1], [0, 1, 0], [1, 0, 0], [1, 1, 1]] Explanation 0 Each variable   and   will have values of   or  . All permutations of lists in the form  . Remove all arrays that sum to   to leave only the valid permutations. Sample Input 1 2 2 2 2 Sample Output 1 [[0, 0, 0], [0, 0, 1], [0, 1, 0], [0, 1, 2], [0, 2, 1], [0, 2, 2], [1, 0, 0], [1, 0, 2], [1, 1, 1], [1, 1, 2], [1, 2
Read more