我已被學校分配創建一個應用程序,該應用程序包含一個包含20本不同書籍的書籍列表,並生成以下選項菜單:如何永久刪除一行數組並向上移動一行數組?
(a)列表 - 以列表格式顯示列表。每個展示應包含適當的標題和欄目標題; (b)搜索 - 使用ISBN在列表中搜索書籍記錄並打印該書籍的完整記錄;
(c)刪除 - 從列表中刪除現有書籍記錄;
(d)退出 - 停止程序。
這裏是我的方案的示例:
#include <iostream>
#include <iomanip>
#include <fstream>
#include <cstring>
#include <cctype>
using namespace std;
typedef struct
{
char code[50];
char author[50];
char name[50];
char edition[50];
char publish[50];
char price[50];
} BOOK_LIST;
void list (BOOK_LIST book[], int rows);
void showBook (BOOK_LIST book[], int rows);
void updateRecord (BOOK_LIST book[], int rows);
void advancedSearch (BOOK_LIST book[], int rows);
int deleteBook (BOOK_LIST book[], int rows);
int searchBook(BOOK_LIST book[], int rows);
int main()
{
ifstream inFile("list.txt");
if(!inFile)
cout << "Error opening input file\n";
else
{
BOOK_LIST books[50];
int index = -1, choice;
inFile.getline(books[++index].code, 50);
while(inFile)
{
if(inFile.peek() == '\n')
inFile.ignore(256, '\n');
inFile.getline(books[index].author, 50);
inFile.getline(books[index].name, 50);
inFile.getline(books[index].edition, 50);
inFile.getline(books[index].publish, 50);
inFile >> books[index].price;
// read next number
inFile >> books[++index].code;
}
inFile.close();
// menu starts
do
{
cout << "Do you want to:\n";
cout << "1. List all books\n";
cout << "2. Get details about a book\n";
cout << "3. Delete a book from the list\n";
cout << "4. Exit\n";
cout << "5. Advanced Search\n";
cout << "Enter choice: ";
cin >> choice;
switch (choice)
{
case 1 : list(books, index);
break;
case 2 : showBook(books, index);
break;
case 3 : updateRecord(books, index);
break;
case 5 : advancedSearch(books, index);
case 4 : break;
default: cout << "Invalid choice\n";
}
} while (choice != 4);
ofstream outFile("list.txt");
if(!outFile)
cout << "Error opening output file, records are not updated.\n";
else
{
for (int i = 0; i < index; i++)
{
outFile << books[i].code << endl;
outFile << books[i].author << endl;
outFile << books[i].name << endl;
outFile << books[i].edition << endl;
outFile << books[i].publish << endl;
outFile << books[i].price << endl;
}
outFile.close();
}
}
return 0;
}
void list(BOOK_LIST book[], int rows)
{
cout << fixed << setprecision(2);
cout << "ISBN\t Author BookName Edition\tPublisher\t Price\n";
for (int i = 0; i < rows; i++)
cout << book[i].code << "\t" << book[i].author << "\t"
<< book[i].name << "\t" << book[i].edition << "\t"
<< book[i].publish << "\t"
<< " " << book[i].price << endl;
return;
}
int searchBook(BOOK_LIST book[], int rows)
{
int i = 0;
bool found = false;
char code[50];
cout << "Enter an ISBN code of a book to search: ";
fflush(stdin);
cin.getline(code, 50);
while (i < rows && !found)
{
if (strcmp(code, book[i].code) == 0)
found = true;
else
i++;
}
if (found)
return i;
else
return -1;
}
void showBook(BOOK_LIST book[], int rows)
{
int pos = searchBook(book, rows);
if (pos != -1)
{
cout << "Author is " << book[pos].author << endl;
cout << "Book name is "<< book[pos].name << endl;
cout << book[pos].edition << " Edition" << endl;
cout << "The publisher of this book is " << book[pos].publish << endl;
cout << "Current price is " << book[pos].price << endl;
}
else
cout << "Product not found\n";
return;
}
void updateRecord(BOOK_LIST book[], int rows)
{
int pos = deleteBook(book, rows);
char code [50];
int i,j = 0;
for(i = 0; i < rows ; i++)
{
if(strcmp(code, book[i].code))
{
strcpy(book[j].code , book[i].code);
strcpy(book[j].author, book[i].author);
strcpy(book[j].name, book[i].name);
strcpy(book[j].edition, book[i].edition);
strcpy(book[j].publish, book[i].publish);
strcpy(book[j].price, book[i].price);
j++;
}//if
else
{
i++;
strcpy(book[j].code, book[i].code);
strcpy(book[j].author, book[i].author);
strcpy(book[j].name, book[i].name);
strcpy(book[j].edition, book[i].edition);
strcpy(book[j].publish, book[i].publish);
strcpy(book[j].price, book[i].price);
j++;
}//else
}//for
return;
}
int deleteBook (BOOK_LIST book[], int rows)
{
int i = 0;
bool found = false;
char code[50];
cout << "Enter an ISBN code of a book to delete: ";
fflush(stdin);
cin.getline(code, 50);
while (i < rows && !found)
{
if (strcmp(code, book[i].code) == 0)
found = true;
else
i++;
}
if (found)
return i;
else
return -1;
}
void advancedSearch (BOOK_LIST book[], int rows)
{
char advanced[50];
cout << "Please enter either the author's name or the book name to search: ";
fflush(stdin);
cin.getline(advanced, 50);
for(int i = 0; i < rows; i++)
{
if(strstr(book[i].author, advanced) || strstr(book[i].name, advanced))
{
cout << "ISBN is " << book[i].code << endl;
cout << "Author is " << book[i].author << endl;
cout << "Book name is " << book[i].name << endl;
cout << book[i].edition << " Edition" << endl;
cout << "Publisher is " << book[i].publish << endl;
cout << "Current price is " << book[i].price << endl;
}
}
return ;
}
這裏的問題開始: 當我想永久刪除本書記錄的整行。但刪除後該書記錄仍然存在。
首先,這是我的菜單,然後按1檢查IBSN的列表。然後,我按3進入刪除部分。那時候,我選擇TheHost刪除。在刪除後,以確保我已經刪除所選的書,所以我按1再次檢查之列,但不幸的是這本書仍然存在:
如果我能夠刪除一本書記錄,以及如何永久刪除記錄?刪除記錄後,如何向上移動剩餘的記錄,以便它不會留下空行?
爲刪除功能:
void updateRecord(BOOK_LIST book[], int rows)
{
int pos = deleteBook(book, rows);
char code [50];
int i,j = 0;
for(i = 0; i < rows ; i++)
{
if(strcmp(code, book[i].code))
{
strcpy(book[j].code , book[i].code);
strcpy(book[j].author, book[i].author);
strcpy(book[j].name, book[i].name);
strcpy(book[j].edition, book[i].edition);
strcpy(book[j].publish, book[i].publish);
strcpy(book[j].price, book[i].price);
j++;
}//if
else
{
i++;
strcpy(book[j].code, book[i].code);
strcpy(book[j].author, book[i].author);
strcpy(book[j].name, book[i].name);
strcpy(book[j].edition, book[i].edition);
strcpy(book[j].publish, book[i].publish);
strcpy(book[j].price, book[i].price);
j++;
}//else
}//for
return;
}
The Text file that I used in this program a.k.a the BOOK_LIST
請嘗試調試。 – 2015-04-05 13:57:12
扔掉那些固定和動態大小的數組,並且替換'string'和'vector'是我會做的第一件事。但除此之外,我沒有看到你改變行數的點。你需要這樣做:擦除,複製每個元素後刪除pos下來1,然後縮小行數1.另外,將io/ui和業務邏輯代碼拆分成不同的功能。 – Yakk 2015-04-05 14:01:40
你正試圖一次做幾件事。 **首先嚐試一些更簡單的方法**嘗試製作一個'int'數組,填充數字,然後刪除一個數字並移動其他數字以填補空白。一旦你有完美的工作,這將是容易的。一旦你解決了這個問題,你可以專注於學習1)[最小完整示例](http://stackoverflow.com/help/mcve),2)[basic_string](http://www.sgi.com/ tech/stl/basic_string.html)(比'char []'更容易處理),3)編寫一個Book類,並且集合* that *,而不是單獨收集書*所有東西。 – Beta 2015-04-05 14:02:12