在ES6中将字符串转换为模板文字 [英] Converting Strings to Template Literals in ES6

问题描述

假设我有一个字符串，如$ {a + b}，比如从JSON读取，我想将其解释为ES6模板文字。我认为这样的事情可能有用：

Suppose I have a string like "${a + b}", say read from JSON, that I'd like interpreted as an ES6 template literal. I thought something like this might work:

var x = {"add": "${a + b}"};
var a = 10, b = 20;
alert(`${x.add}`);

但此警报为 $ {a + b} ，所以它只做一个替换级别。

But this alerts as ${a + b}, so it just does one level of substitution.

通过再次解释它试图变得聪明：

Tried being clever by interpreting it again:

var a = 10, b = 20;
var x = {"add": "${a + b}"};
var i = `${x.add}`;
alert(`${i}`);

这仍然以 $ {a + b} 。

尝试更聪明：

var a = 10, b = 20;
var x = {"add": "${a} + ${b}"};
var i = `${x.add}`;
alert(`${i}`);

此提醒 $ {a} + $ {b} 。

以字符串开头，例如$ {a + b}，有没有办法让这个评估完成，好像它是模板文字一样？理想情况下没有 eval ！

Starting with a string, e.g. "${a + b}", is there any way to have this evaluated to completion as if it were a template literal? Ideally without eval!

推荐答案

是的，它们不是递归的。

Yes, they aren't recursive.

如果您的起点是包含这些占位符的字符串，据我所知，没有模板编译器功能。当然有 eval ; [插入关于使用 eval 的所有常见警告 —只有你信任的内容，而不是你可以避免它等等。 —这里]。

If your starting point is a string containing those placeholders, as far as I know there is no template compiler function. There's eval, of course; [insert all the usual caveats about using eval — only with content you trust, not if you can avoid it, etc., etc. — here].

例如：

"use strict";
var x = {"add": "${a + b}"};
var a = 10, b = 20;
console.log(eval("`" + x.add + "`"));

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