如何在Tampermonkey中實(shí)現(xiàn)對(duì)多個(gè)鏈接的GET請(qǐng)求并依次判斷條件？

async function processLinks(links, conditionFunc) {   for (const link of links) {     const response = await fetch(link); // 使用fetch代替GM_xmlhttpRequest，更現(xiàn)代化     const data = await response.text(); // 獲取響應(yīng)文本      if (conditionFunc(data)) {       return data; // 滿足條件，返回結(jié)果并結(jié)束     }   }   return null; // 沒(méi)有鏈接滿足條件 }  // 示例用法： const links = [   "https://example.com/link1",   "https://example.com/link2",   "https://example.com/link3" ];  const condition = (data) => data.includes("success"); // 判斷條件函數(shù)  processLinks(links, condition)   .then((result) => {     if (result) {       console.log("條件滿足，結(jié)果：", result);     } else {       console.log("所有鏈接均不滿足條件");     }   })   .catch((error) => {     console.error("請(qǐng)求錯(cuò)誤：", error);   });

function processLinksRecursive(links, conditionFunc, index = 0, result = null) {   if (index >= links.length || result !== null) {     return result; // 結(jié)束遞歸   }    fetch(links[index])     .then(response => response.text())     .then(data => {       if (conditionFunc(data)) {         result = data; // 滿足條件       } else {         result = processLinksRecursive(links, conditionFunc, index + 1, result); // 繼續(xù)遞歸       }     })     .catch(error => console.error("請(qǐng)求錯(cuò)誤:", error));    return result; // 返回結(jié)果 }  // 示例用法 (與方法一相同)： 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("條件滿足，結(jié)果：", finalResult); } else {   console.log("所有鏈接均不滿足條件"); }