在js中创建简单的Hangman游戏,如果语句正确读取则无法获取

时间:2018-02-24 06:41:34

标签: javascript if-statement

在下面的代码中,如果玩家猜测某个字母不在所选的密码中,我就无法执行代码。我把它设置为当提交按钮被击中时,我们首先检查猜测字母是否在拼图中,如果是字母被翻过来,那么我想说是否不在单词中远离玩家的生活。我已经尝试了几种方法,但无法使程序正确读取第二个if语句。

submit.addEventListener('click', function(){
    console.log("clicked");
    usedLetters.innerHTML += playersLetter.value + ' '

    for(i=0; i<checkWord.length; i++){
      console.log(wordLife)

      if(playersLetter.value == (checkWord[i])){
        console.log('in if#1')
        console.log(playersLetter.value)
        var answerLetter = document.getElementsByClassName('letter')[i]
        answerLetter.innerHTML = playersLetter.value
        wordLife--
      }
    }

  function scoreLetter(value){// To check if the letter is actually in the word
  for(let i = 0; i < checkWord.length; i++){
    if(value == (checkWord[i])){
      return true
    }else {
      return false
    }
  }
}
console.log('read function')

 if(scoreLetter(playersLetter.value) == false){
   console.log('in 2nd if')
   playerLife--

console.log(playerLife)
}    
})

0 个答案:

没有答案