system("CLS"); cout << "———————————————————————————————————————————————————查询图书信息—————————————————————————————————————————————————————" << endl; cout << "按类型查找图书信息" << endl; cout << "请输入要查询的书籍类型:"; cin >> type; // 修正变量名 sql = "select * from books where books_type = '" + type + "'"; if (mysql_query(&mysql, sql.c_str())) { cout << "Error: Query[查询成功] fail." << mysql_error(&mysql) << endl; break; } res = mysql_store_result(&mysql); while ((row = mysql_fetch_row(res))) { cout << setw(5) << "编号: " << row[0] << setw(10) << "书名: " << row[1] << setw(15) << "当前数量: " << row[2] << setw(15) << "价格:¥" << row[3] << setw(10) << "作者:" << row[4] << setw(10) << "类型:" << row[5] << endl; } cout << "——————————————————————————————————————————————————————————————————————————————————————————————————————————————————" << endl;详细描述解释上方代码
时间: 2023-08-25 08:05:35 浏览: 71
这段代码是用来查询图书信息的。首先使用系统命令`system("CLS")`清空控制台屏幕,然后输出一个标题。接着提示用户输入要查询的书籍类型,将用户输入的类型保存在变量`type`中。然后使用SQL语句查询数据库中`books`表中符合条件的所有书籍信息,将SQL语句保存在字符串变量`sql`中。如果查询失败,则输出错误信息并跳出循环。如果查询成功,则将查询结果保存在变量`res`中,并使用循环遍历每一行查询结果,输出每个书籍的编号、书名、当前数量、价格、作者和类型。最后再输出一个分隔线作为结尾。
相关问题
c++——文字rpg游戏
以下是一个简单的C++文字RPG游戏的示例代码,其中包括了基本的角色、战斗和物品系统:
```c++
#include <iostream>
#include <string>
#include <cstdlib>
#include <ctime>
using namespace std;
// 角色类
class Character {
public:
string name;
int hp;
int atk;
int def;
int gold;
Character(string n, int h, int a, int d, int g) {
name = n;
hp = h;
atk = a;
def = d;
gold = g;
}
// 攻击函数
void attack(Character& other) {
int damage = atk - other.def;
if (damage < 0) {
damage = 0;
}
other.hp -= damage;
cout << name << "攻击了" << other.name << ",造成了" << damage << "点伤害。" << endl;
}
// 是否死亡
bool isDead() {
return hp <= 0;
}
};
// 物品类
class Item {
public:
string name;
int price;
int hp;
int atk;
int def;
Item(string n, int p, int h, int a, int d) {
name = n;
price = p;
hp = h;
atk = a;
def = d;
}
};
// 商店类
class Shop {
public:
Item items[3];
Shop() {
items[0] = Item("草药", 10, 20, 0, 0);
items[1] = Item("铁剑", 50, 0, 10, 0);
items[2] = Item("铁甲", 100, 0, 0, 10);
}
// 显示商店物品
void showItems() {
cout << "欢迎光临!以下是本店的物品:" << endl;
for (int i = 0; i < 3; i++) {
cout << i + 1 << ". " << items[i].name << " - " << items[i].price << "金币" << endl;
}
}
// 购买物品
bool buy(Character& c, int choice) {
if (c.gold < items[choice - 1].price) {
cout << "金币不足,法购买!" << endl;
return false;
}
c.gold -= items[choice - 1].price;
c.hp += items[choice - 1].hp;
c.atk += items[choice - 1].atk;
c.def += items[choice - 1].def;
cout << "购买成功!" << endl;
return true;
}
};
// 战斗函数
void battle(Character& player, Character& enemy) {
cout << "你遇到了一只" << enemy.name << ",准备战斗!" << endl;
while (!player.isDead() && !enemy.isDead()) {
player.attack(enemy);
if (enemy.isDead()) {
cout << enemy.name << "被你打败了!" << endl;
player.gold += enemy.gold;
return;
}
enemy.attack(player);
if (player.isDead()) {
cout << "你被" << enemy.name << "打败了!" << endl;
return;
}
}
}
int main() {
srand(time(NULL)); // 初始化随机数种子
// 初始化角色和商店
Character player("勇者", 100, 10, 5, 50);
Character enemies[3] = {
Character("史莱姆", 30, 5, 2, 10),
Character("骷髅兵", 50, 10, 5, 20),
Character("巨龙", 100, 20, 10, 50)
};
Shop shop;
// 游戏循环
while (true) {
cout << "你的状态 - HP:" << player.hp << " ATK:" << player.atk << " DEF:" << player.def << " 金币:" << player.gold << endl;
cout << "请选择操作:" << endl;
cout << "1. 进入商店" << endl;
cout << "2. 进行战斗" << endl;
cout << "3. 离开游戏" << endl;
int choice;
cin >> choice;
switch (choice) {
case 1:
shop.showItems();
cout << "请选择要购买的物品(输入编号):" << endl;
cin >> choice;
shop.buy(player, choice);
break;
case 2:
battle(player, enemies[rand() % 3]);
break;
case 3:
cout << "游戏结束,欢迎再次光临!" << endl;
return 0;
default:
cout << "无效的操作!" << endl;
break;
}
}
return 0;
}
```
汉诺塔问题——循环算法
汉诺塔问题是一个经典的问题,源于印度一个古老传说。问题是将一根柱子上的64片黄金圆盘按照大小顺序重新摆放到另一根柱子上,且在任何时候,小圆盘上不能放大圆盘,且一次只能移动一个圆盘。
汉诺塔问题的循环算法可以使用迭代的方式来解决。下面是一个示例代码:
```cpp
#include <iostream>
#include <stack>
#include <string>
using namespace std;
void hanoiTowerIterative(int N, string from, string to, string help) {
stack<int> s1, s2, s3;
int moves = (1 << N) - 1;
for (int i = N; i > 0; i--) {
s1.push(i);
}
if (N % 2 == 0) {
swap(s2, s3);
}
for (int i = 1; i <= moves; i++) {
if (i % 3 == 1) {
if (s1.empty() || (!s2.empty() && s2.top() < s1.top())) {
cout << "Move " << s2.top() << " from " << to << " to " << from << endl;
s1.push(s2.top());
s2.pop();
} else {
cout << "Move " << s1.top() << " from " << from << " to " << to << endl;
s2.push(s1.top());
s1.pop();
}
} else if (i % 3 == 2) {
if (s1.empty() || (!s3.empty() && s3.top() < s1.top())) {
cout << "Move " << s3.top() << " from " << help << " to " << from << endl;
s1.push(s3.top());
s3.pop();
} else {
cout << "Move " << s1.top() << " from " << from << " to " << help << endl;
s3.push(s1.top());
s1.pop();
}
} else {
if (s2.empty() || (!s3.empty() && s3.top() < s2.top())) {
cout << "Move " << s3.top() << " from " << help << " to " << to << endl;
s2.push(s3.top());
s3.pop();
} else {
cout << "Move " << s2.top() << " from " << to << " to " << help << endl;
s3.push(s2.top());
s2.pop();
}
}
}
}
int main() {
hanoiTowerIterative(3, "A", "B", "C");
return 0;
}
```