我解决了这个插值模板使用:

function flatKeys(inputObject: any): {[key: string]: any} {
    const response: {[key: string]: any} = {};
  function iterative(currentObject: any, parentKeys: string[]=[]) {
    const llaves = Object.keys(currentObject);
    for (let i=0; i<llaves.length; i++) {
        const llave: string = llaves[i];
      const valor = currentObject[llave];
      const llavesNext = parentKeys.concat(llave);
      if (typeof valor == 'object') {
        iterative(valor, llavesNext);
      } else {
        response[llavesNext.join('.')] = valor;
      }
    }
  }
  iterative(inputObject);
  return response;
}

function interpolate(template: string, values: any, defaultValue='') {
  const flatedValues = flatKeys(values);
  const interpolated = template.replace(/\${(.*?)}/g, function (x,g) {
    const value = flatedValues[g];
    if ([undefined, null].indexOf(value) >= 0) {
      return defaultValue;
    }
    return value;
  });
  return interpolated;
}

const template = "La ${animal.name} tomaba ${alimento.name} con el ${amigos.0.names}";
const values = {
    animal: {
    name:"Iguana"
  },
  alimento: {
    name: "café"
  },
  amigos: [
    { name: "perro" },
    true
  ]
};

const interpolated = interpolate(template, values);

console.log(interpolated);

一般来说，我反对使用邪恶的eval()，但在这种情况下，它是有意义的:

var template = "`${a}.${b}`";
var a = 1, b = 2;
var populated = eval(template);

console.log(populated);         // shows 1.2

然后如果你改变值并再次调用eval()，你会得到新的结果:

a = 3; b = 4;
populated = eval(template);

console.log(populated);         // shows 3.4

如果你想把它写在一个函数中，那么它可以这样写:

function populate(a, b){
  return `${a}.${b}`;
}

如果你正在寻找一些相当简单的东西(只是固定的变量字段，没有计算，条件…)，但这也适用于没有模板字符串支持的浏览器客户端，如IE 8,9,10,11…

开始吧:

fillTemplate = function (templateString, templateVars) {
    var parsed = templateString;
    Object.keys(templateVars).forEach(
        (key) => {
            const value = templateVars[key]
            parsed = parsed.replace('${'+key+'}',value)
        }
    )
    return parsed
}

你可以只使用一行标签模板，比如:

const SERVICE_ADDRESS = (s,tenant) => `http://localhost/${tenant}/api/v0.1/service`;

在客户端代码中，你可以这样使用它:

const myTenant = 'me';
fetch(SERVICE_ADDRESS`${myTenant}`);

你可以像这样使用内联箭头函数， 定义:

const template = (substitute: string) => `[^.?!]*(?<=[.?\s!])${substitute}(?=[\s.?!])[^.?!]*[.?!]`;

用法:

console.log(template('my replaced string'));

所有的道具在这里教我关于javascript的一个功能，我从来不知道-我知道字符串模板文字，但不知道你可以用它们调用函数没有括号!

作为感谢，我在这里分享我的typescript改编，这使得它很容易用typescript知道的命名变量来制作一个可重用的模板——它允许任何类型，因为它们会自动转换为字符串，但如果你不喜欢这个策略，你可以自己调整。

/**
 * Use this with a template literal in order to create reusable string template;
 * use interpolation to add strings for each variable you want to use in the template.
 * 
 * e.g.:
 * 
 *  const reUsableStringTemplate = stringTpl`${'name'} and ${'otherName'} are going to ${'place'}`;
 * 
 * You can then call it with:
 * 
 *  const filled = reUsableStringTemplate({name: 'John', otherName: 'Jane', place: 'Paris'});
 *  // John and Jane are going to Paris
 * 
 * reUsableStringTemplate will have types and know the names of your variables
 * 
 * @returns String template function with full typescript types
 */
export function stringTpl<keys extends string>(parts: TemplateStringsArray, ...keys: keys[]) {
  return (opts: Record<keys, any>) => {
    let outStr = '';
    for (let i = 0; i < parts.length; ++i) {
      outStr += parts[i];
      const key = keys.shift();
      if (key && key in opts) {
        outStr += opts[key];
      } else {
        outStr += key ?? '';
      }
    }
    return outStr;
  };
}