Library management using structure

Problem

Write a menu driven program that depicts the working of a library. The menu options should be:

1. Add book information
2. Display book information
3. List all books of given author
4. List the title of specified book
5. List the count of books in the library
6. List the books in the order of accession number
7. Exit

Create a structure called library to hold accession number, title of the book, author name, price of the book, and flag indicating whether book is issued or not.

Solution - Code

#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
#include<string.h>
struct library
{
 char bk_name[30];
 char arthur[30];
 int pages;
 float price;
};

int main()
{
 struct library l[3];
 char ar_nm,bk_nm;
 int i,j;

 printf ("1. Add book information\n2. Display book information\n3. List all books of given author\n4. List the title of specified book\n5. List the count of books in the library\n6. Exit");

 printf ("\n\nEnter one of the above : ");
 scanf("%d",&j);

 switch (j)
 {
  case 1:
   for (i=0;i<3;i++)
   {

    printf ("Enter book name = ");
    scanf ("%s",&l[i].bk_name);

    printf ("Enter arthur name = ");
    scanf ("%s",&l[i].arthur);

    printf ("Enter pages = ");
    scanf ("%d",&l[i].pages);

    printf ("Enter price = ");
    scanf ("%f",&l[i].price);
   }
   break;
  case 2:
   printf("you have entered the following information\n");
   for(i=0;i<3;i++)
   {
    printf ("book name = %s",l[i].bk_name);

    printf ("arthur name = %s",l[i].arthur);

    printf ("pages = %d",l[i].pages);

    printf ("price = %f",l[i].price);
   }
   break;

  case 3:
   printf ("Enter arthur name : ");
   scanf ("%s",&ar_nm);
   for (i=0;i<3;i++)
   {
    if (ar_nm == l[i].arthur)
     printf ("%s %s %d %f",l[i].bk_name,l[i].arthur,l[i].pages,l[i].price);
   }
   break;

  case 4:
   printf ("Enter book name : ");
   scanf ("%s",&bk_nm);
   for (i=0;i<3;i++)
   {
    if (bk_nm == l[i].bk_name)
     printf ("%s %s %d %f",l[i].bk_name,l[i].arthur,l[i].pages,l[i].price);
   }
   break;

  case 5:
   for (i=0;i<3;i++)
   {

   }
   break;
  case 6:
   exit (0);


 }
 return 0;
}

Sample Output