Question

我正在使用新的ES6 Template Literals功能，而我头脑中的第一件事就是JavaScript的String.format，所以我开始实现原型：

String.prototype.format = function() {
  var self = this;
  arguments.forEach(function(val,idx) {
    self["p"+idx] = val;
  });
  return this.toString();
};
console.log(`Hello, ${p0}. This is a ${p1}`.format("world", "test"));

ES6Fiddle

但是，在传递给我的prototype方法之前，评估了Template Literal。有没有什么办法可以编写上面的代码来推迟结果，直到我动态创建元素？

Answer 1

我可以看到三种解决方法：

使用类似设计使用的模板字符串，没有任何format函数：

console.log(`Hello, ${"world"}. This is a ${"test"}`);
// might make more sense with variables:
var p0 = "world", p1 = "test";
console.log(`Hello, ${p0}. This is a ${p1}`);
// or even function parameters for actual deferral of the evaluation:
const welcome = (p0, p1) => `Hello, ${p0}. This is a ${p1}`;
console.log(welcome("world", "test"));

不要使用模板字符串，而是使用普通字符串文字：

String.prototype.format = function() {
    var args = arguments;
    return this.replace(/\$\{p(\d)\}/g, function(match, id) {
        return args[id];
    });
};
console.log("Hello, ${p0}. This is a ${p1}".format("world", "test"));

使用标记的模板文字。请注意，仍然会在没有处理程序拦截的情况下评估替换，因此如果没有名为so的变量，则不能使用p0之类的标识符。 ~~如果different substitution body syntax proposal is accepted（更新：它不是），此行为可能会更改。~~

function formatter(literals, ...substitutions) {
    return {
        format: function() {
            var out = [];
            for(var i=0, k=0; i < literals.length; i++) {
                out[k++] = literals[i];
                out[k++] = arguments[substitutions[i]];
            }
            out[k] = literals[i];
            return out.join("");
        }
    };
}
console.log(formatter`Hello, ${0}. This is a ${1}`.format("world", "test"));
// Notice the number literals: ^               ^

Answer 2

我也喜欢String.format函数的想法，并且能够显式定义用于解析的变量。

这就是我想出的...基本上是一种String.replace查找的deepObject方法。

const isUndefined = o => typeof o === 'undefined'

const nvl = (o, valueIfUndefined) => isUndefined(o) ? valueIfUndefined : o

// gets a deep value from an object, given a 'path'.
const getDeepValue = (obj, path) =>
  path
    .replace(/\[|\]\.?/g, '.')
    .split('.')
    .filter(s => s)
    .reduce((acc, val) => acc && acc[val], obj)

// given a string, resolves all template variables.
const resolveTemplate = (str, variables) => {
  return str.replace(/\$\{([^\}]+)\}/g, (m, g1) =>
            nvl(getDeepValue(variables, g1), m))
}

// add a 'format' method to the String prototype.
String.prototype.format = function(variables) {
  return resolveTemplate(this, variables)
}

// setup variables for resolution...
var variables = {}
variables['top level'] = 'Foo'
variables['deep object'] = {text:'Bar'}
var aGlobalVariable = 'Dog'

// ==> Foo Bar <==
console.log('==> ${top level} ${deep object.text} <=='.format(variables))

// ==> Dog Dog <==
console.log('==> ${aGlobalVariable} ${aGlobalVariable} <=='.format(this))

// ==> ${not an object.text} <==
console.log('==> ${not an object.text} <=='.format(variables))

或者，如果您不仅想要可变分辨率（例如，模板文字的行为），还可以使用以下内容。

N.B。。eval被认为是“邪恶的”-考虑使用safe-eval替代方案。

// evalutes with a provided 'this' context.
const evalWithContext = (string, context) => function(s){
    return eval(s);
  }.call(context, string)

// given a string, resolves all template variables.
const resolveTemplate = function(str, variables) {
  return str.replace(/\$\{([^\}]+)\}/g, (m, g1) => evalWithContext(g1, variables))
}

// add a 'format' method to the String prototype.
String.prototype.format = function(variables) {
  return resolveTemplate(this, variables)
}

// ==> 5Foobar <==
console.log('==> ${1 + 4 + this.someVal} <=='.format({someVal: 'Foobar'}))

Answer 3

我发布了一个类似问题的答案，该问题提供了两种延迟执行模板文字的方法。当模板文字在函数中时，模板文字仅在调用函数时进行评估，并使用函数的范围进行评估。

https://stackoverflow.com/a/49539260/188963

Answer 4

扩展@Bergi的答案，当您意识到可以返回任何结果（不仅是纯字符串）时，带标签的模板字符串的力量就会彰显出来。在他的示例中，标记构造并返回一个带有闭包和函数属性format的对象。

在我最喜欢的方法中，我自己返回一个函数值，您以后可以调用该函数值并传递新参数来填充模板。像这样：

function fmt([fisrt, ...rest], ...tags) {
  return values => rest.reduce((acc, curr, i) => {
    return acc + values[tags[i]] + curr;
  }, fisrt);
}

然后，您构建模板并推迟替换：

> fmt`Test with ${0}, ${1}, ${2} and ${0} again`(['A', 'B', 'C']);
// 'Test with A, B, C and A again'
> template = fmt`Test with ${'foo'}, ${'bar'}, ${'baz'} and ${'foo'} again`
> template({ foo:'FOO', bar:'BAR' })
// 'Test with FOO, BAR, undefined and FOO again'

另一个更接近您编写内容的选项是返回一个从字符串扩展的对象，以开箱即用并尊重接口。 String.prototype的扩展名将不起作用，因为您需要关闭模板标记才能稍后解析参数。

class FormatString extends String {
  // Some other custom extensions that don't need the template closure
}

function fmt([fisrt, ...rest], ...tags) {
  const str = new FormatString(rest.reduce((acc, curr, i) => `${acc}\${${tags[i]}}${curr}`, fisrt));
  str.format = values => rest.reduce((acc, curr, i) => {
    return acc + values[tags[i]] + curr;
  }, fisrt);
  return str;
}

然后，在呼叫站点中：

> console.log(fmt`Hello, ${0}. This is a ${1}.`.format(["world", "test"]));
// Hello, world. This is a test.
> template = fmt`Hello, ${'foo'}. This is a ${'bar'}.`
> console.log(template)
// { [String: 'Hello, ${foo}. This is a ${bar}.'] format: [Function] }
> console.log(template.format({ foo: true, bar: null }))
// Hello, true. This is a null.

您可以在this other answer中引用更多信息和应用程序。

Answer 5

AFAIS，有用的功能“延迟执行字符串模板”仍然不可用。但是，使用lambda是一种表达力强，易读且简短的解决方案：

var greetingTmpl = (...p)=>`Hello, ${p[0]}. This is a ${p[1]}`;

console.log( greetingTmpl("world","test") );
console.log( greetingTmpl("@CodingIntrigue","try") );

Answer 6

您可以使用以下函数将值插入字符串

let inject = (str, obj) => str.replace(/\${(.*?)}/g, (x,g)=> obj[g]);


// --- test ---

// parameters in object
let t1 = 'My name is ${name}, I am ${age}. My brother name is also ${name}.';
let r1 = inject(t1, {name: 'JOHN',age: 23} );
console.log("OBJECT:", r1);


// parameters in array
let t2 = "Values ${0} are in ${2} array with ${1} values of ${0}."
let r2 = inject(t2, {...['A,B,C', 666, 'BIG']} );
console.log("ARRAY :", r2);

My_List = ["adopt", "bake", "beam"]

Answer 7

尽管这个问题已经回答了，但是这里有一个简单的实现，我在加载配置文件时会使用（代码是打字稿，但是很容易转换成JS，只需删除输入内容即可）：

/**
 * This approach has many limitations:
 *   - it does not accept variable names with numbers or other symbols (relatively easy to fix)
 *   - it does not accept arbitrary expressions (quite difficult to fix)
 */
function deferredTemplateLiteral(template: string, env: { [key: string]: string | undefined }): string {
  const varsMatcher = /\${([a-zA-Z_]+)}/
  const globalVarsmatcher = /\${[a-zA-Z_]+}/g

  const varMatches: string[] = template.match(globalVarsmatcher) ?? []
  const templateVarNames = varMatches.map(v => v.match(varsMatcher)?.[1] ?? '')
  const templateValues: (string | undefined)[] = templateVarNames.map(v => env[v])

  const templateInterpolator = new Function(...[...templateVarNames, `return \`${template}\`;`])

  return templateInterpolator(...templateValues)
}

// Usage:
deferredTemplateLiteral("hello ${thing}", {thing: "world"}) === "hello world"

尽管可以使这些功能更强大，更灵活，但它却带来了太多的复杂性和风险，而没有太大的好处。

这里是要点的链接：https://gist.github.com/castarco/94c5385539cf4d7104cc4d3513c14f55

Answer 8

（请参阅上面@Bergi的非常相似的答案）

function interpolate(strings, ...positions) {
  var errors = positions.filter(pos=>~~pos!==pos);
  if (errors.length) {
    throw "Invalid Interpolation Positions: " + errors.join(', ');
  }
  return function $(...vals) {
    var output = '';
    for (let i = 0; i < positions.length; i ++) {
      output += (strings[i] || '') + (vals[positions[i] - 1] || '');
    }
    output += strings[strings.length - 1];
    return output;
  };
}

var iString = interpolate`This is ${1}, which is pretty ${2} and ${3}. Just to reiterate, ${1} is ${2}! (nothing ${0} ${100} here)`;
// Sets iString to an interpolation function

console.log(iString('interpolation', 'cool', 'useful', 'extra'));
// Substitutes the values into the iString and returns:
//   'This is interpolation, which is pretty cool and useful.
//   Just to reiterate, interpolation is cool! (nothing  here)'

此答案与@Bergi的答案之间的主要区别是错误的处理方式（静默与否）。

将这个想法扩展为接受命名参数的语法应该足够容易了：

interpolate`This is ${'foo'}, which is pretty ${'bar'}.`({foo: 'interpolation', bar: 'cool'});

https://github.com/spikesagal/es6interpolate/blob/main/src/interpolate.js

推迟执行ES6模板文字

8 个答案: