Javascript 在这个简单的代码中,为什么在函数中忽略提示值?
在以下代码中: 我也试着通过一个论点。我似乎看不出哪里出了问题Javascript 在这个简单的代码中,为什么在函数中忽略提示值?,javascript,Javascript,在以下代码中: 我也试着通过一个论点。我似乎看不出哪里出了问题 var ranThrow = ["empty", "rock", "paper", "scissors"]; var ranNum = Math.random(); var postRanNum = (ranNum * 3) + 1; var roundPostRanNum = Math.floor(postRanNum); var compThrow = ranThrow[roundPostRanNum]; var whatThr
var ranThrow = ["empty", "rock", "paper", "scissors"];
var ranNum = Math.random();
var postRanNum = (ranNum * 3) + 1;
var roundPostRanNum = Math.floor(postRanNum);
var compThrow = ranThrow[roundPostRanNum];
var whatThrow = prompt("Rock, Papper or Scissors?", "rock");
var rpsGame = function () {
if (whatThrow === "rock" && compThrow === "rock") {
return "You tie!";
}
else if (whatThrow === "rock" && compThrow === "paper") {
return "You lose!";
}
else if (whatThrow === "rock" && compThrow === "scissor") {
return "You win!";
}
else {
return "Error";
}
};
rpsGame();
console.log("The computer threw" + " " + compThrow);
函数中的值不会被忽略,调用函数的代码会忽略函数返回的内容 如果显示返回值,您将看到函数使用以下值:
console.log(rpsGame());
啊!当然这对我来说应该更加明显。非常感谢。我的印象是“return”仍然会记录字符串。