如何在Tampermonkey中实现对多个链接的GET请求并依次判断条件？

async function processLinks(links, conditionFunc) {
  for (const link of links) {
    const response = await fetch(link); // 使用fetch代替GM_xmlhttpRequest，更现代化
    const data = await response.text(); // 获取响应文本

    if (conditionFunc(data)) {
      return data; // 满足条件，返回结果并结束
    }
  }
  return null; // 没有链接满足条件
}

// 示例用法：
const links = [
  "https://example.com/link1",
  "https://example.com/link2",
  "https://example.com/link3"
];

const condition = (data) => data.includes("success"); // 判断条件函数

processLinks(links, condition)
  .then((result) => {
    if (result) {
      console.log("条件满足，结果：", result);
    } else {
      console.log("所有链接均不满足条件");
    }
  })
  .catch((error) => {
    console.error("请求错误：", error);
  });

function processLinksRecursive(links, conditionFunc, index = 0, result = null) {
  if (index >= links.length || result !== null) {
    return result; // 结束递归
  }

  fetch(links[index])
    .then(response => response.text())
    .then(data => {
      if (conditionFunc(data)) {
        result = data; // 满足条件
      } else {
        result = processLinksRecursive(links, conditionFunc, index + 1, result); // 继续递归
      }
    })
    .catch(error => console.error("请求错误:", error));

  return result; // 返回结果
}

// 示例用法 (与方法一相同)：
const links = [
  "https://example.com/link1",
  "https://example.com/link2",
  "https://example.com/link3"
];

const condition = (data) => data.includes("success");

const finalResult = processLinksRecursive(links, condition);
if (finalResult) {
  console.log("条件满足，结果：", finalResult);
} else {
  console.log("所有链接均不满足条件");
}