由于模板字符串必须动态地(在运行时)获得对b变量的引用，所以答案是:不，没有动态代码生成是不可能做到的。

但是，使用eval非常简单:

let tpl = eval('`'+a+'`');

你应该试试这个小JS模块，由Andrea Giammarchi，来自github: https://github.com/WebReflection/backtick-template

/*! (C) 2017 Andrea Giammarchi - MIT Style License */
function template(fn, $str, $object) {'use strict';
  var
    stringify = JSON.stringify,
    hasTransformer = typeof fn === 'function',
    str = hasTransformer ? $str : fn,
    object = hasTransformer ? $object : $str,
    i = 0, length = str.length,
    strings = i < length ? [] : ['""'],
    values = hasTransformer ? [] : strings,
    open, close, counter
  ;
  while (i < length) {
    open = str.indexOf('${', i);
    if (-1 < open) {
      strings.push(stringify(str.slice(i, open)));
      open += 2;
      close = open;
      counter = 1;
      while (close < length) {
        switch (str.charAt(close++)) {
          case '}': counter -= 1; break;
          case '{': counter += 1; break;
        }
        if (counter < 1) {
          values.push('(' + str.slice(open, close - 1) + ')');
          break;
        }
      }
      i = close;
    } else {
      strings.push(stringify(str.slice(i)));
      i = length;
    }
  }
  if (hasTransformer) {
    str = 'function' + (Math.random() * 1e5 | 0);
    if (strings.length === values.length) strings.push('""');
    strings = [
      str,
      'with(this)return ' + str + '([' + strings + ']' + (
        values.length ? (',' + values.join(',')) : ''
      ) + ')'
    ];
  } else {
    strings = ['with(this)return ' + strings.join('+')];
  }
  return Function.apply(null, strings).apply(
    object,
    hasTransformer ? [fn] : []
  );
}

template.asMethod = function (fn, object) {'use strict';
  return typeof fn === 'function' ?
    template(fn, this, object) :
    template(this, fn);
};

演示(以下所有测试返回true):

const info = 'template';
// just string
`some ${info}` === template('some ${info}', {info});

// passing through a transformer
transform `some ${info}` === template(transform, 'some ${info}', {info});

// using it as String method
String.prototype.template = template.asMethod;

`some ${info}` === 'some ${info}'.template({info});

transform `some ${info}` === 'some ${info}'.template(transform, {info});

TLDR: https://jsfiddle.net/bj89zntu/1/

每个人似乎都在担心访问变量的问题。为什么不直接通过呢?我相信在调用者中获取变量context并将其传递下去不会太难。使用 Ninjagecko从obj那里得到道具的答案。

function renderString(str,obj){
    return str.replace(/\$\{(.+?)\}/g,(match,p1)=>{return index(obj,p1)})
}

以下是完整的代码:

function index(obj,is,value) {
    if (typeof is == 'string')
        is=is.split('.');
    if (is.length==1 && value!==undefined)
        return obj[is[0]] = value;
    else if (is.length==0)
        return obj;
    else
        return index(obj[is[0]],is.slice(1), value);
}

function renderString(str,obj){
    return str.replace(/\$\{.+?\}/g,(match)=>{return index(obj,match)})
}

renderString('abc${a}asdas',{a:23,b:44}) //abc23asdas
renderString('abc${a.c}asdas',{a:{c:22,d:55},b:44}) //abc22asdas

仍然是动态的，但似乎比使用裸eval更可控:

const vm = require('vm')
const moment = require('moment')


let template = '### ${context.hours_worked[0].value} \n Hours worked \n #### ${Math.abs(context.hours_worked_avg_diff[0].value)}% ${fns.gt0(context.hours_worked_avg_diff[0].value, "more", "less")} than usual on ${fns.getDOW(new Date())}'
let context = {
  hours_worked:[{value:10}],
  hours_worked_avg_diff:[{value:10}],

}


function getDOW(now) {
  return moment(now).locale('es').format('dddd')
}

function gt0(_in, tVal, fVal) {
  return _in >0 ? tVal: fVal
}



function templateIt(context, template) {
  const script = new vm.Script('`'+template+'`')
  return script.runInNewContext({context, fns:{getDOW, gt0 }})
}

console.log(templateIt(context, template))

https://repl.it/IdVt/3

这个解决方案不需要ES6:

function render(template, opts) {
  return new Function(
    'return new Function (' + Object.keys(opts).reduce((args, arg) => args += '\'' + arg + '\',', '') + '\'return `' + template.replace(/(^|[^\\])'/g, '$1\\\'') + '`;\'' +
    ').apply(null, ' + JSON.stringify(Object.keys(opts).reduce((vals, key) => vals.push(opts[key]) && vals, [])) + ');'
  )();
}

render("hello ${ name }", {name:'mo'}); // "hello mo"

注意:Function构造函数总是在全局作用域中创建，这可能会导致全局变量被模板覆盖，例如render("hello ${someGlobalVar = 'some new value'}"， {name:'mo'});

@Mateusz Moska，解决方案很好，但当我在React Native(构建模式)中使用它时，它抛出一个错误:无效字符'”，尽管当我在调试模式下运行它时它是有效的。

所以我用正则表达式写出了我自己的解。

String.prototype.interpolate = function(params) {
  let template = this
  for (let key in params) {
    template = template.replace(new RegExp('\\$\\{' + key + '\\}', 'g'), params[key])
  }
  return template
}

const template = 'Example text: ${text}',
  result = template.interpolate({
    text: 'Foo Boo'
  })

console.log(result)

演示:https://es6console.com/j31pqx1p/

注:因为我不知道问题的根本原因，我在react-native repo中提出了一张票，https://github.com/facebook/react-native/issues/14107，这样一旦他们能够修复/指导我大致相同:)