如何使用 javascript 中的两个请求与 mailchimp 连接两个数组?

How to concatenate two array by using two request in javascript with mailchimp?

标题可能不是很清楚。我要解释我的问题。

这是我的代码:

mailchimpMarketing = require("@mailchimp/mailchimp_marketing");

mailchimpMarketing.setConfig({
apiKey: "MY API KEY",
server: "MY SERVER",
});

const getEmailMembersFromList = async (listID, offset , count) => {
  const response = await mailchimpMarketing.lists.getListMembersInfo(listID, {offset, count});
    emails = response.members.map(member => {
      return member.email_address
    })
  console.log(emails)
}

getEmailMembersFromList("MY LIST ID", 0, 1000)
getEmailMembersFromList("MY LIST ID", 1000, 58)

它在终端中显示两个电子邮件地址数组。我必须打两次电话,因为有1000多个电子邮件地址(总共1058个),最多1000个。

现在我想连接这两个数组,也许可以使用 .concat() 方法,但我不知道在我的情况下该怎么做。

编辑 1

mailchimpMarketing = require("@mailchimp/mailchimp_marketing");

mailchimpMarketing.setConfig({
apiKey: "MY API KEY",
server: "MY SERVER",
});

const getEmailMembersFromList = async (listID, offset , count) => {
  const response = await mailchimpMarketing.lists.getListMembersInfo(listID, {offset, count});
    emails = response.members.map(member => {
      return member.email_address
    })
  //console.log(emails)
  return emails
}

getEmailMembersFromList("MY LIST ID", 0, 1000)
getEmailMembersFromList("MY LIST ID", 1000, 58)

由于 async 始终发挥作用 return 您可以将所有请求添加到数组中并使用 Promise.all() 获取所有数据的承诺。

仅供参考:您没有从 getEmailMembersFromList 函数中 return 获取使其正常工作所必需的任何内容。

示例:

const firstGroup = getEmailMembersFromList("MY LIST ID", 0, 1000);
const secondGroup = getEmailMembersFromList("MY LIST ID", 1000, 58);

Promise.all([firstGroup, secondGroup]).then((values) => {
  console.log(values);
});

试试这个代码:

let result = [];

const getEmailMembersFromList = async (listID, offset, count) => {
  await mailchimpMarketing.lists
    .getListMembersInfo(listID, { offset, count })
    .then((response) => {
      emails = response.members.map((member) => {
        return member.email_address;
      });
      // try using spread operator, hopefully it should work
      result = [...result, ...emails];
    });
};

getEmailMembersFromList("MY LIST ID", 0, 1000);
getEmailMembersFromList("MY LIST ID", 1000, 58);

console.log(result);