js带参数get提交到php

const ShowAppListRequest = new XMLHttpRequest();
const url = `getApps.php?uid=${encodeURIComponent(uid)}`;
ShowAppListRequest.open("GET", url);
ShowAppListRequest.onload = function()
{
if(ShowAppListRequest.status === 200) 
{
 console.log(ShowAppListRequest.responseText)
}
}
ShowAppListRequest.send();

 

posted on 2025-08-15 12:59  小沙盒工作室  阅读(10)  评论(0)    收藏  举报