Javascript if / else語句有幫助，這段代碼我在做什么錯？

Question

以下代碼來自我的課程，但無法正常工作。

我盡力想出一切可能

 var eatSteak = confirm("Do you eat steak?"); var confirmSteak = confirm("You like Steak!"); var confirmNoSteak = confirm("Here's a tofu stir-fry"); var howOld = prompt("How old are you?"); var howNotOld = confirm("No Sake for you!"); var sakeForYou = confirm("Sake for you!"); var tryAgain = confirm("TRY AGAIN!"); if (eatSteak == true) { alert(confirmSteak); } else { alert(confirmNoSteak); } if (howOld < 21) { alert(howNotOld); } else if (howOld >= 21) { alert(sakeForYou); } else { alert(tryAgain); } 

控制台未顯示任何錯誤消息

Answer 1

confirm()僅應用於提問。 如果只想為消息設置變量，則不應調用confirm() 。 所以像

var confirmSteak = confirm("You like Steak!");

應該只是

var confirmSteak = "You like Steak!";

另外，在測試howOld ，請檢查它們是否小於21，然后檢查它們是否至少為21。這是僅有的兩種可能性，它永遠無法到達else塊。 因此，您可以只使用if和else －它們是否獲得緣故。

 var eatSteak = confirm("Do you eat steak?"); var confirmSteak = "You like Steak!"; var confirmNoSteak = "Here's a tofu stir-fry"; var howOld = prompt("How old are you?"); var howNotOld = "No Sake for you!"; var sakeForYou = "Sake for you!"; if (eatSteak == true) { alert(confirmSteak); } else { alert(confirmNoSteak); } if (howOld < 21) { alert(howNotOld); } else { alert(sakeForYou); } 

Answer 2

一個簡單的規則。 僅在您希望獲得用戶輸入時使用prompt並confirm 。

 if (confirm("Do you eat steak?") == true) alert('You like Steak!') else alert('Heres a tofu stir-fry'); //Here is a variable which stores the user input var age = prompt("How old are you?"); if (age < 21) alert('No Sake for you!') else if (age > 21) alert('Sake for you!') else alert('TRY AGAIN!');