清理字符串列表并仅保留特定内容

Question

我有一个包含许多电子邮件地址的列表。格式类似如下。您会注意到每个“企业”都有以相同方式开头的相似电子邮件地址。

Business Name 01    email@business01.com; info@business01.com; admin@business01.com; other@business01.com; contact@business01.com;`enter code here`
Business Name 02    email@business02.com; info@business02.com; admin@business02.com; other@business02.com; contact@business02.com;
Business Name 03    email@business03.com; info@business03.com; admin@business03.com; other@business03.com; contact@business03.com;

我希望删除除以“信息”和“联系人”开头的电子邮件之外的所有内容。所以最终列表应该是这样的：

Business Name 01    info@business01.com; contact@business01.com;
Business Name 02    info@business02.com; contact@business02.com;
Business Name 03    info@business03.com; contact@business03.com;

有任何想法吗？

Answer 1

不确定只使用记事本++是否可行，但使用一些 JavaScript，您可以分别处理每个业务，然后过滤以仅保留您想要的电子邮件地址，然后重新格式化它们。

const data = `Business Name 01    email@business01.com; info@business01.com; admin@business01.com; other@business01.com; contact@business01.com;
Business Name 02    email@business02.com; info@business02.com; admin@business02.com; other@business02.com; contact@business02.com;
Business Name 03    email@business03.com; info@business03.com; admin@business03.com; other@business03.com; contact@business03.com;`

const output = data.split("
")
  .map((business) => {
    const [name, emails] = business.split("    ");
    const newEmails = emails.slice(0, -1).split("; ").filter((email) => email.startsWith("info") || email.startsWith("contact"));
    return `${name}    ${newEmails.join("; ")};`;
  })
  .join("
");
  
console.log(output);
a.innerText = output;

<pre id=a></pre>