设置 onclick 以在循环中使用变量的当前值 [英] Setting onclick to use current value of variable in loop

问题描述

标题可能没有意义，单独的描述也没有意义，所以这里是代码示例:

The title may not make sense and nor would a description alone, so here's a sample of the code:

for(var a=0;a<10;a++) {
    var b=document.createElement('b')
    b.onclick=function() {
        alert(a)
    }
    b.innerHTML=a
    document.body.appendChild(b)
}

奇怪的是，当我将innerHTML 设置为a 时，它使用a 的当前值.因此，此代码创建了十个 元素，其值为 0、1、2、3、4、5、6、7、8、&9. 这完美地工作.然而，onclick 函数没有.当单击这些元素中的任何一个时，它返回 a (10) 的最终值.我尝试将另一个变量设置为 a，然后在 onclick 事件中使用该其他变量，但它仍然返回 a 的最终值.设置 onclick 时，如何让它使用 a 的值?

What's odd is that when I set innerHTML to a, it uses the current value of a. So, this code creates ten <b> elements with the values 0, 1, 2, 3, 4, 5, 6, 7, 8, & 9. This works perfectly. What doesn't, however, is the onclick function. It returns the final value of a (10) when any of these elements is clicked. I've tried setting another variable to a and then using that other variable in the onclick event, but still it returns the final value of a. How would I make it use the value of a when onclick is set?

推荐答案

试试下面的

b.onclick= function(arg) {
    return function() {
        alert(arg);
    }
}(a);

您遇到的问题是 javascript 不使用块作用域.相反，所有变量都有其包含函数的生命周期.所以在所有的 onclick 函数中只捕获了一个 a 并且他们都看到了它的最终值.

The problem you were hitting is that javascript doesn't use block scope. Instead all variables have a lifetime of their containing function. So there was only ever one a captured in all of the onclick functions and they all see it's final value.

解决方案通过为每个作用域创建一个新函数和值，然后立即将该值设置为 a 的当前值来解决此问题.

The solution works around this by creating a new function and value per scope and then immediately setting that value to the current value of a.

这是一个稍微扩展的版本，可能更具可读性

Here's a version that's a bit expanded and perhaps a bit more readable

var createClickHandler = function(arg) {
  return function() { alert(arg); };
}

for(var a=0;a<10;a++) {
    var b=document.createElement('b')
    b.onclick = createClickHandler(a);
    b.innerHTML=a
    document.body.appendChild(b)
}

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