Question

我正在尝试编写一个包含字符串的脚本，每当有字母时，下一个数字就是上标。为此，我将字符串分割成一个数组并循环遍历-每当我找到一个字母时，就在下一个数组元素上运行sup（）。

JS

var ec = "1s2 2s4".split("");
for (var i = 0; i < ec.length; i++) {
    if (ec[i].match(/[a-z]/i)) {
        ec[i + 1].sup();
    }
}

但是当我这样做时，我运行sup（）的数字没有任何反应。为什么会这样？

JSfiddle：http://jsfiddle.net/yxj143az/7/

Answer 1

请避免使用sup，这是如何完成此操作的快速示例：

var ec = "1s2 2s4".split("");
for (var i = 0; i < ec.length; i++) {
    if (ec[i].match(/[a-z]/i)) {
        // I removed the call to sup, even though it is only deprecated
        // and has been for awhile it is still accessible. Feel free to
        // use it if you would like, i just opted not to use it.
        // The main issue with your code was this line because you weren't
        // assigning the value of your call to sup back to the original variable,
        // strings are immutable so calling on a function on them doesn't change
        // the string, it just returns the new value
        ec[i + 1] = '<sup>' + ec[i + 1] + '</sup>';
        // if you want to continue to use sup just uncomment this
        // ec[i + 1] = ec[i + 1].sup();

        // This is a big one that I overlooked too.
        // This is very important because when your regex matches you reach
        // ahead and modify the next value, you should really add some checks
        // around here to make sure you aren't going to run outside the bounds
        // of your array
        // Incrementing i here causes the next item in the loop to be skipped. 
        i++
    }
}

console.log(ec.join(''));

根据评论中的一些有效反馈，

编辑/更新，我回去评论了答案，以确切说明我所做的更改以及原因。非常感谢@IMSoP向我指出这一点。

Answer 2

.sup() method不会在适当位置修改字符串，它需要一个字符串并返回一个新字符串。

因此，不仅仅是运行它...

 ec[i + 1].sup();

...您需要将其结果分配回您的字符串...

 ec[i + 1] = ec[i + 1].sup();

但是，正如其他用户指出的那样，该方法可能不应再使用，因为它被认为“已弃用”，并且可能被浏览器删除。幸运的是，替换非常简单，因为它所做的只是在字符串周围添加<sup>和</sup>，因此您可以不用它来重写行：

 ec[i + 1] = '<sup>' + ec[i + 1] + '</sup>';

JavaScript sup（）无法正常工作

2 个答案: