为什么JavaScript不能生成函数的输出？

为什么javascript不能生成函数的输出？为什么JavaScript不能生成函数的输出？

var name = prompt("What is your name?"); 
var weight = prompt("what is your weight?"); 
function check(name ,weight){ 
     if (weight> 25) { 
      alert (name + "Your weight is " + weight + "which is not normal."+ "Sorry"); 
     } 
     else { 
      alert (name + "Your weight is " + weight + "which is normal." + "Welcome"); 
     } 

    return check; 
}check; 
</script>

来源

2014-11-24 Rumana

1.因为你有一个语法错误调用它。 2.因为你没有调用你定义的函数。 – icktoofay 2014-11-24 06:50:53

你似乎不打电话检查。你的功能还没关闭 – 2014-11-24 06:50:55

请再检查一次，编辑 – Rumana 2014-11-24 06:54:13

因为它格式不正确，您甚至不会结束该功能。

你必须返回一个值才能返回任何东西。

也许这就是你想要达到的目标：

var name = prompt("What is your name?"); 
var weight = prompt("what is your weight?"); 

function check(name ,weight){ 
    if (weight> 25) { 
    return name + "Your weight is " + weight + "which is not normal."+ "Sorry"; 
    } else { 
    return name + "Your weight is " + weight + "which is normal." + "Welcome"; 
    } 
} 

alert(check(name, weight));

注意我是如何传递参数name和weight到check -function

来源

2014-11-24 06:54:09 Ole

它应该是这样的：

 var name = prompt("What is your name?"); 
     var weight = prompt("what is your weight?"); 

     function check(name ,weight){ 
      if (weight> 25) { 
       alert (name + "Your weight is " + weight + "which is not normal."+ "Sorry"); 
      } 
      else { 
       alert (name + "Your weight is " + weight + "which is normal." + "Welcome"); 
      } 
     return true;    // check is invalid 
     }       // you forgot to close it 
     check(name,weight);   // calling the function

来源

2014-11-24 06:55:04 Pixeladed

当声明一个函数时，你声明的参数是：name and weight吨。这些参数只能在该函数的范围内使用。看起来你正试图将它们设置在范围之外。然后你返回变量检查。这是空的。你也不是从函数外部调用你的函数，所以函数check（）永远不会被触发。

这里一个例子（live example）：

var input1="peter"; input2=10; 
check(input1, input2); 
function check(name, weight) { 
    var check; 
    if (weight < 25) { //true 
     check = "Text"+name; 
    } else { 
     check = "No text"+name; 
    } 
    return alert(check); 
}

PS：没有测试此代码尚未。

来源

2014-11-24 07:08:10 Fleuv

'返回警报（...）;'意义不大，不是吗？ – algorhythm 2014-11-24 07:35:06

您的代码无法正常工作 – Rumana 2014-11-24 09:02:18

@Rumana，代码工作正常[请参阅此jsfiddle]（http://jsfiddle.net/aum07c6b/） – Fleuv 2014-11-24 09:08:50

该函数需要一个外部调用来执行该操作。所以，如果你想显示你可以在上面做的方式将结果添加

function check(name ,weight){ 
     if (weight> 25) { 
      alert (name + "Your weight is " + weight + "which is not normal."+ "Sorry"); 
     } 
     else { 
      alert (name + "Your weight is " + weight + "which is normal." + "Welcome"); 
     } 
     return true;    
    } 

    check(name,weight);

也可以通过其他方式

<p id="print"></p> 
<script> 
    function check(name ,weight){ 
     if (weight> 25) { 
      alert (name + "Your weight is " + weight + "which is not normal."+ "Sorry"); 
     } 
     else { 
      alert (name + "Your weight is " + weight + "which is normal." + "Welcome"); 
     } 
     return check;    
    }     

    var name = prompt("What is your name?"); 
    var weight = prompt("what is your weight?"); 

    document.getElementById("print").innerHTML = check(name, weight); 
</script>

来源

2014-11-24 07:34:52

为什么JavaScript不能生成函数的输出？

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