用 JavaScript 中的所有匹配项替换“标签”（字符串）答案

【问题标题】：Replace a "Tag" ( string ) with all the matches in JavaScript用 JavaScript 中的所有匹配项替换“标签”（字符串）
【发布时间】：2020-01-27 04:04:39
【问题描述】：

我正在尝试将如下所示的字符串置换为其可能的值。

我有一个格式如下的对象

{
  descriptions: ["Here's some <tag> exemple", "Can be something without Tag"],
  headlines: ["<tag> another exemple", "<tag>"]
}

我有一个具有不同排列的数组

["First","Second","Third"]

我正在尝试创建与排列一样多的对象以在最后得到这个结果

[
  {
    descriptions: ["Here's some First exemple", "Can be something without Tag"],
    headlines: ["First another exemple", "First"]
  },
  {
    descriptions: [
      "Here's some Second exemple",
      "Can be something without Tag"
    ],
    headlines: ["Second another exemple", "Second"]
  },
  {
    descriptions: ["Here's some Third exemple", "Can be something without Tag"],
    headlines: ["Third another exemple", "Third"]
  }
];

我被困在这里..

function foo(adCopy: AdCopy[], tag: string, variants: string[]) {
  variants.forEach(variant => {
    adCopy.forEach(adCopy => {
      adCopy.headlines.map(headline => {

      })
    })
  })
}

【问题讨论】：

标签： javascript reactjs typescript logic frontend

【解决方案1】：

const mapObj = (obj, map) => Object.fromEntries(Object.entries(obj).map(map));

const result = variants.map(variant =>
   adCopies.map(adCopy =>
     mapObj(adCopy, ([key, values]) => ([
        key, 
        values.map(value => value.replace(/<tag>/, variant))
     ]))
   )
);

您可以将所有变量映射到它们的排列，并且您可以使用Object.fromEntries 和Object.entries 来映射对象。

【讨论】：

【解决方案2】：

假设您不想修改原始对象，您可以这样做：

let obj = {
	descriptions: ["Here's some <tag> exemple", "Can be something without Tag"],
	headlines: ["<tag> another exemple", "<tag>"]
};
let permutations = ['First', 'Second', 'Third'];

let objMapped = permutations.map(permutation =>
	Object.entries(obj).reduce((acc, [key, values]) => {
		acc[key] = values.map(value => value.replace(/<tag>/g, permutation));
		return acc;
	}, {}));

console.log(objMapped);

【讨论】：