Question

抱歉这个愚蠢的问题所以我在这里得到了这段代码：

function stat(x) {
    player.ATK = 0; player.DEF = 0; player.DEX = 0; player.crit = 0; player.HP = 0;
    player.weaponATK = weapon.x.atk;
    player.weaponDEF= weapon.x.def;
    player.weaponDEX = weapon.x.dex;
    player.weaponHP = weapon.x.hp;
    player.weaponCrit = weapon.x.crit;
    player.ATK = player.baseATK + player.weaponATK;
    player.DEF = player.baseDEF + player.weaponDEF;
    player.DEX = player.baseDEX + player.weaponDEX;
    player.HP = player.baseHP + player.weaponHP;
    player.crit = player.baseCrit + player.weaponCrit;
}

以下是我使用的全部内容：

var player = {
  HP: 0,
  baseHP: 100,
  weaponHP: 0,
  ATK: 0,
  baseATK: 0,
  weaponATK: 0,
  DEF: 0,
  baseDEF: 0,
  weaponDEF: 0,
  DEX: 0,
  baseDEX: 0,
  weaponDEX: 0,
  crit: 0,
  baseCrit: 5,
  weaponCrit: 0,
  level: 1,
  currentEXP: 0,
  expLeft: 10
};
var weapon = {
  hatchet: {
    atk: 2,
    def: -1,
    dex: 0,
    hp: 0,
    crit: 0
  },
  woodenSword: {
    atk: 5,
    dex: 0,
    def: 0,
    hp: 0,
    crit: 0,
  },
  ironSword: {
    atk: 10,
    crit: 5,
    dex: 0,
    def: 0,
    hp: 0
  },
  blade: {
    atk: 25,
    crit: 20,
    dex: 10,
    hp: 0,
    crit: 0,
  },
  mace: {
    atk: 30,
    def: 5,
    dex: -1,
    hp: 0,
    crit: 0,
  },
  battleAxe: {
    atk: 50,
    def: 5,
    dex: 0,
    hp: 0,
    crit: 0,
  },
  broadSword: {
    atk: 100,
    def: 20,
    dex: 0,
    crit: 0,
    hp: 0
  },
  woodenShield: {
    atk: 0,
    def: 10,
    dex: 0,
    hp: 0,
    crit: 0,
  },
  spikeShield: {
    def: 15,
    atk: 5,
    dex: 0,
    crit: 0,
    hp: 0,
  },
  bomb: {
    atk: 0,
    def: 0,
    crit: 0,
    hp: 0,
    dex: -5
  }
};

如果我运行stat(hatchet)，则假设使用x替换为hatchet来执行此功能。但相反，我得到一个错误：“x未定义”。有人能帮我吗？谢谢你的帮助。

Answer 1

我假设x是武器对象中的属性，因此您可以使用括号表示法从武器中访问它，如下所示。最后，当您调用stat函数时，请确保它是一个字符串。请参阅以下工作代码：

var player = {
  HP: 0,
  baseHP: 100,
  weaponHP: 0,
  ATK: 0,
  baseATK: 0,
  weaponATK: 0,
  DEF: 0,
  baseDEF: 0,
  weaponDEF: 0,
  DEX: 0,
  baseDEX: 0,
  weaponDEX: 0,
  crit: 0,
  baseCrit: 5,
  weaponCrit: 0,
  level: 1,
  currentEXP: 0,
  expLeft: 10
};
var weapon = {
  hatchet: {
    atk: 2,
    def: -1,
    dex: 0,
    hp: 0,
    crit: 0
  },
  woodenSword: {
    atk: 5,
    dex: 0,
    def: 0,
    hp: 0,
    crit: 0,
  },
  ironSword: {
    atk: 10,
    crit: 5,
    dex: 0,
    def: 0,
    hp: 0
  },
  blade: {
    atk: 25,
    crit: 20,
    dex: 10,
    hp: 0,
    crit: 0,
  },
  mace: {
    atk: 30,
    def: 5,
    dex: -1,
    hp: 0,
    crit: 0,
  },
  battleAxe: {
    atk: 50,
    def: 5,
    dex: 0,
    hp: 0,
    crit: 0,
  },
  broadSword: {
    atk: 100,
    def: 20,
    dex: 0,
    crit: 0,
    hp: 0
  },
  woodenShield: {
    atk: 0,
    def: 10,
    dex: 0,
    hp: 0,
    crit: 0,
  },
  spikeShield: {
    def: 15,
    atk: 5,
    dex: 0,
    crit: 0,
    hp: 0,
  },
  bomb: {
    atk: 0,
    def: 0,
    crit: 0,
    hp: 0,
    dex: -5
  }
};

function stat(x) {
    player.ATK = 0; player.DEF = 0; player.DEX = 0; player.crit = 0; player.HP = 0;
    player.weaponATK = weapon[x].atk;
    player.weaponDEF= weapon[x].def;
    player.weaponDEX = weapon[x].dex;
    player.weaponHP = weapon[x].hp;
    player.weaponCrit = weapon[x].crit;
    player.ATK = player.baseATK + player.weaponATK;
    player.DEF = player.baseDEF + player.weaponDEF;
    player.DEX = player.baseDEX + player.weaponDEX;
    player.HP = player.baseHP + player.weaponHP;
    player.crit = player.baseCrit + player.weaponCrit;
}

// make sure x is a string!
stat('hatchet');

console.log(player)

Answer 2

它是@brk或@ D-reaper建议的

player.weaponATK = weapon.x.atk;

和类似的需要是

player.weaponATK = weapon[x].atk;

或只是

player.weaponATK = x.atk;

Answer 3

由于x是传递给访问对象属性的密钥。它必须用作的 weapon[x].atk

我的函数参数未按预期工作

3 个答案: