blocks|key|555266|text|也许问题就在这条线上：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|555267|else+if+(0.34+<+computerChoice+<+0.66)+{|code-block|syntax|javascript|555268|试试这个：|555269|else+if+(0.35+<+computerChoice+&&+computerChoice+<+0.67)+{|555270|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Maybe the problem is in this line:

<pre><code>else if (0.34 &lt; computerChoice &lt; 0.66) {
</code></pre>

Try this:

<pre><code>else if (0.35 &lt; computerChoice &amp;&amp; computerChoice &lt; 0.67) {
</code></pre>

blocks|key|1048750|text|最好是阅读一本编程书，或者阅读一些关于循环的在线教程，无论如何使用下面的代码来解决您的问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1048751|++++var+userChoice+=+prompt("Rock,+Paper,+or+Scissors?");

++++var+computerChoice+=+Math.random();

++++console.log(computerChoice);

++++if++(computerChoice+<=+0.33)+
++++{
++++++++console.log("Rock");
++++}

else+if(computerChoice>=0.34&&computerChoice<=0.66)
+++++{
+++++++console.log("Paper");+
+++++}
else+if(computerChoice>=0.67&&computerChoice<=1)
+++++{
+++++++console.log("Scissors");++++++
+++++}

+++else
++++{
++++++console.log("Invalid+input");++
++++}|code-block|syntax|javascript|1048752|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

its better to read a programming book , or read some online tutorial about the loops 
anyway use the below code it will solve your problem 

<pre><code> var userChoice = prompt("Rock, Paper, or Scissors?");

 var computerChoice = Math.random();

 console.log(computerChoice);

 if (computerChoice &lt;= 0.33) 
 {
 console.log("Rock");
 }

else if(computerChoice&gt;=0.34&amp;&amp;computerChoice&lt;=0.66)
 {
 console.log("Paper"); 
 }
else if(computerChoice&gt;=0.67&amp;&amp;computerChoice&lt;=1)
 {
 console.log("Scissors"); 
 }

 else
 {
 console.log("Invalid input"); 
 }
</code></pre>

blocks|key|555303|text|if(computerChoice>=0&&computerChoice<=0.33)
{
++console.log("Rock");
}
else+if(computerChoice>=0.34&&computerChoice<=0.66)
{
++console.log("Paper");+
}
else+if(computerChoice>=0.67&&computerChoice<=1)
{
++console.log("Scissors");++++++
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|555304|嗨，我想你验证的随机数范围是0到1。|unstyled|555305|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>if(computerChoice&gt;=0&amp;&amp;computerChoice&lt;=0.33)
{
 console.log("Rock");
}
else if(computerChoice&gt;=0.34&amp;&amp;computerChoice&lt;=0.66)
{
 console.log("Paper"); 
}
else if(computerChoice&gt;=0.67&amp;&amp;computerChoice&lt;=1)
{
 console.log("Scissors"); 
}
</code></pre>

HI, I assume that random number you verify range on from 0 to 1.

blocks|key|555325|text|每个人似乎都犯了一个错误，那就是你可以得到0.331，并且计算为“剪刀”。这样的逻辑更合理：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|555326|var+result+=+(computerChoice+>=+0.67+?+"Scissors"+:+(computerChoice+>=+0.34+?+"Paper"+:+"Rock"));
console.log(result);|code-block|syntax|javascript|555327|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

The error everyone seems to make is that you can have 0.331 and that evaluates to "Scissors". The logic is more reasonable to be the following:

<pre><code>var result = (computerChoice &gt;= 0.67 ? "Scissors" : (computerChoice &gt;= 0.34 ? "Paper" : "Rock"));
console.log(result);
</code></pre>

blocks|key|555366|text|我不想成为第十个答案，但所有的答案都是糟糕的。你明白吗|type|unstyled|depth|inlineStyleRanges|entityRanges|data|555367|else+if+(0.34+<+computerChoice+<+0.66)+{|code-block|syntax|javascript|555368|语法不好，对吧？好的，让我们也来解决偏压问题:+0.33小于三分之一，所有0.33到0.34之间的数字都默认为剪刀。如果您想要一个均匀分布，最简单的方法是使用早期退出创建一个函数：|555369|function+computerChoice()+{
++++var+v+=+Math.random();

++++if+(v+<+(1.0+/+3.0))+return+"rock";
++++if+(v+<+(2.0+/+3.0))+return+"paper";
++++return+"scissors";
}|555370|这是最好的双精度浮点将允许，这是足够好的大多数用途。|555371|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

I hate to be a tenth answerer, but all of the given answers are bad. You understand that

<pre><code>else if (0.34 &lt; computerChoice &lt; 0.66) {
</code></pre>

is bad syntax, right? OK, so let's also solve the bias problem: 0.33 is less than a third, and all the numbers between 0.33 and 0.34 are defaulting to scissors. If you want an even distribution, the simplest way is to make a function with early exits:

<pre><code>function computerChoice() {
 var v = Math.random();

 if (v &lt; (1.0 / 3.0)) return "rock";
 if (v &lt; (2.0 / 3.0)) return "paper";
 return "scissors";
}
</code></pre>

This is as good as double-precision floating point will allow, which is more than good enough for most purposes.

I am using codecademy courses and I'm learning how to make a rock paper scissors game out of randomizing numbers. It says I've done a "good job" but every time I try to properly get each one to properly show up to their respective conditions it messes up. 

Please leave a comment pinpointing the error and tips on avoiding it in the future.

Conditions 
- 0 to .33 = Rock
-.34 to .66 = Paper
-.67 to 1 = Scissors 

JS Code

<pre><code>var userChoice = prompt("Rock, Paper, or Scissors?");

var computerChoice = Math.random();

console.log(computerChoice);

if (computerChoice &lt;= 0.33) {
 console.log("Rock");
}

else if (0.34 &lt; computerChoice &lt; 0.66) {
 console.log("Paper"); 
}

else {
 console.log("Scissors"); 
}
</code></pre>

How to properly use if else and else if

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

我正在学习如何用随机数制作一个石头剪刀游戏。它说我做了一个“很好的工作”，但每次我试图让每个人正确地表现出他们各自的条件，它就会搞砸。请留下评论，指出错误和避免它在未来的提示。条件-0 to .33 = Rock -.34 to .66 =纸张-.67 to 1=剪刀JS码var userChoice = prompt...

问如何正确地使用“如果其他”和“如果”
EN

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何正确地使用“如果其他”和“如果”EN