Javascript 仅在另一个“if”失败时执行“if”

Question

I have started studying Javascript one month ago and I have a doubt. I'm programming a simple game that thinks of a random number and you have to guess it before you run out of attempts. To determine if the player guessed right or wrong the number, I have done the following (reduced code):

start(){
num=documentGetElementById("num").value;
attempts=10;
}
try(){
attempts=attempts-1;
if(num==random)documentGetElementById("message").innerHTML="You have won.";
if(attempts==0)documentGetElementById("message").innerHTML="You have lost.";
}

What happens here is that when you guess the number in the last attempt (1 attempt left) it says you have lost. So my question is: how to give priority to the if(num==random)...?

---EDIT---

Full code:

-HTML:

<div id="game">
  <div id="message"></div>
  <div id="status"></div>
  <div id="think"></div>
  <div id="start">
    <button id="start-button" class="boto-inici" onClick="start()"><span>START</span></button>
  </div>
  <div id="try">
    <label for="tryinput">Number:</label>
    <input type="text" id="tryinput"/>
    <button class="try-button" onClick="try()"><span>TRY</span></button>
  </div>
</div>

-JS:

function start(){
  document.getElementById("start").style.display="none";
  document.getElementById("try").style.display="block";
  document.getElementById("message").style.display="block";
  document.getElementById("status").style.display="block";
  document.getElementById("message").innerHTML="I've thought a number between 1 and 100";
  document.getElementById("status").innerHTML="You have 10 attempts.";
  random = Math.floor(Math.random()*100);
  attempts=10;
}

function try(){
  attempts=attempts-1;
  document.getElementById("status").innerHTML="You have " + attempts + " attempts left.";
  var num=document.getElementById("tryinput").value;
  if(num>random)document.getElementById("message").innerHTML="The number you chose is bigger.";
  if(num<random)document.getElementById("message").innerHTML="The number you chose is smaller.";
  if(num==random){
    document.getElementById("message").innerHTML="You won.";
    document.getElementById("status").innerHTML="The number I thought was the " + random + " .";
    document.getElementById("try").style.display="none";
    document.getElementById("start").style.display="block";
  }
  if(attempts==0){
    document.getElementById("try").style.display="none";
    document.getElementById("message").innerHTML="You lost.";
    document.getElementById("status").innerHTML="The number I thought was the " + random + " .";
    document.getElementById("start").style.display="block";
  }
  if(isNaN(num)){
    alert("Please write a number.");
  }
}

It may contain some errors with the id's because I translated it. Hope it helps.

Answer 1

This isn't a matter of priority . Both tests are running, in the order given.

Your problem is that you don't want to do the second test at all if the first one passes.

Use an else so the second test only runs if the first fails.