分享程序员开发的那些事...
更新时间：2022-05-30 22:53:17
我可以请参阅以下三种方式：
I can see three ways around this:
使用设计为使用的模板字符串，而不需要任何格式 function：
Use template strings like they were designed to be used, without any format function:
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，而是使用纯字符串文字：
Don't use a template string, but a plain string literal:
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，替换仍将被处理程序进行评估，而不会使用 p0 之类的标识符，而不使用名为so的变量。 如果不同的替代主体语法提案是，则此行为可能会更改接受（更新：不是）。
Use a tagged template literal. Notice that the substitutions will still be evaluated without interception by the handler, so you cannot use identifiers like p0 without having a variable named so. This behavior may change if a different substitution body syntax proposal is accepted (Update: it was not).
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: ^ ^