文件输入没有遍历文件

File input not looping through file

本文关键字：文件 遍历 输入      更新时间：2023-10-16

我从一个名为Enemies.txt的文本文件中输入。但是有一个问题，它只会打印出文件中的第一个敌人。

下面是我的代码: 
void Enemies :: loadEnemies()
{
    string fileName = "Enemies\Enemies.txt";
    ifstream infile(fileName);
    string garbage;
    int loadEnemyID;
    string loadEnemyName;
    int loadEnemyStrength;
    int loadExperienceGain;
    string loadWeapon;
    string loadRoom;
    while(infile >> garbage >> garbage >> loadEnemyID
     >> garbage >> garbage >> garbage >> loadEnemyName
     >> garbage >> garbage >> loadEnemyStrength
     >> garbage >> garbage >> garbage >> loadExperienceGain
     >> garbage >> garbage >> garbage >> loadWeapon
     >> garbage >> garbage >> loadRoom >> garbage)
    {
    cout << "Enemy ID: tt" << loadEnemyID <<"n";
    cout << "Enemy Name: tt"<< loadEnemyName << "n";
    cout << "Enemy Strength: t"<< loadEnemyStrength << "n";
    cout << "Experience Gain: t" << loadExperienceGain << "n";
    cout << "Weapon: tt" << loadWeapon << "n";
    cout << "Room: ttt" << loadRoom << "n";
    int id = weapon.getWeaponID();
    int weight = weapon.getWeight();
    int damage = weapon.getDamage();
    Weapons w1 (id,loadWeapon, weight,damage);
    int roomID = room.getRoomID();
    string roomExits = room.getRoomExits();
    Rooms r1 (roomID,loadRoom, roomExits);
    Enemies e1 (loadEnemyID,loadEnemyName,loadEnemyStrength,loadExperienceGain,w1,r1);
    enemiesList.Append(e1);
    }
}

Enemies.txt 

Enemy ID: 1.
Enemy Name: Wolves.
Enemy Strength: 5.
Experience Gain: 186.
Weapon Name: Claws.
Room Name: 1.
Enemy ID: 2.
Enemy Name: Cave Bear.
Enemy Strength: 10.
Experience Gain: 302.
Weapon Name: Claws.
Room Name: 4.
Enemy ID: 3.
Enemy Name: Viking.
Enemy Strength: 6.
Experience Gain: 254.
Weapon Name: Longsword.
Room Name: 0.
Enemy ID: 4.
Enemy Name: Criminal.
Enemy Strength: 6.
Experience Gain: 198.
Weapon Name: War Axe.
Room Name: 0.
Enemy ID: 5.
Enemy Name: Ninja.
Enemy Strength: 6.
Experience Gain: 211.
Weapon Name: Katana.
Room Name: 0.
Enemy ID: 6.
Enemy Name: Wild Boar.
Enemy Strength: 3.
Experience Gain: 111.
Weapon Name: Claws.
Room Name: 0.

但有趣的是，我有类似的武器代码，它工作完美。

void Weapons :: loadWeapons()
    {
    string fileName = "Weapons\Weapons.txt";
    ifstream infile(fileName);
    string garbage;
    int loadWeaponID;
    string loadWeaponName;
    int loadDamage;
    int loadWeight;
    while(infile >> garbage >> loadWeaponID >> garbage >> garbage
        >> garbage >> loadWeaponName >> garbage >> loadDamage >> garbage
        >> garbage >> loadWeight >> garbage)
    {
        //cout << "Weapon ID: tt"<< loadWeaponID<< "n";
        //cout << "Weapon Name: tt"<< loadWeaponName << "n";
        //cout << "Damage: tt" << loadDamage <<"n";
        //cout << "Weight: tt" << loadWeight << "n";
        Weapons w1 (loadWeaponID,loadWeaponName,loadDamage,loadWeight);
        weaponsList.Append(w1);
    }
    }

Weapons.txtID: 1。武器名称:武士刀。伤害:20。重量:6 . 

ID: 2.
Weapon Name: Longsword.
Damage: 17.
Weight: 9.
ID: 3.
Weapon Name: WarAxe.
Damage: 22.
Weight: 20.
ID: 4.
Weapon Name: Staff.
Damage: 9.
Weight: 6.
ID: 5.
Weapon Name: Staff.
Damage: 3.
Weight: 0.

在你的第二个也是最后一个敌人中，你有一个由两个词组成的名字:

Enemy Name: Cave Bear.

当它试图读入下一行的强度时，这会使解析混乱。

我建议使用不同的解析方法，如getline和split，而不是试图滥用operator>>。这样，你就可以把任何你想要的放到右边，而不会弄乱解析。

编辑:

另外，由于某种原因，您需要从解析中删除最后一个>> garbage，否则它也无法工作。然而，我仍然强烈建议你换一种别的东西，因为这种奇怪的事情发生的事实表明它是多么容易被打破。

要执行类似getline和split的解析，您可以这样做:

std::string line;
while(std::getline(file, line)) {
    std::stringstream ss(line);
    std::string left, right;
    std::getline(ss, left, ':'); // Read until the colon
    std::getline(ss, right, '.'); // Read until the period
    std::cout << left << ": " << right << std::endl;
}
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