javascript - 为什么我的简单 JS 脚本返回错误？

Question

如果 maryGames 设置为此处显示的，她将获胜。但是如果将 97 更改为 7，则代码将 no if 语句返回为 true，这是 draw 的一部分，在这种情况下不存在。帮助将不胜感激:)

var jGames, mGames, maryGames;

jGames = (89 + 120 + 103) / 3;
mGames = (116 + 94 + 123) / 3;
maryGames = (97 + 134 + 105) / 3;

if (jGames > mGames && jGames > maryGames) {
  console.log("John wins with " + jGames);
} else if (mGames > jGames && mGames > mGames) {
  console.log("Marc wins with " + mgames);
} else if (maryGames > jGames && maryGames > mGames) {
  console.log("Mary Wins the game with " + maryGames);
} else {
  console.log("Draw");
}

标签： javascriptdebugging

Answer 1

if (jGames > mGames && jGames > maryGames) {
  console.log("John wins with " + jGames);
} else if (mGames > jGames && mGames > mGames) {
  console.log("Marc wins with " + mgames);
} else if (maryGames > jGames && maryGames > mGames) {
  console.log("Mary Wins the game with " + maryGames);
} else {
  console.log("Draw");
}

In this case else if (mGames > jGames && mGames > mGames) { should have been else if (mGames > jGames && mGames > maryGames) {