谷歌插件,获取数据库信息,返回到background.js,然后怎么在当前页面右下角新建一个小页面来展示查询到的信息呢
点击小页面上的按钮后,跳转到新页面进行增删,这是代码该怎么继续往下写
// 监听来自content-script的消息
var phpresponse;
chrome.runtime.onMessage.addListener(
function(request, sender, sendResponse){
console.log("收到来自content-script的消息:");
var pageinfo=request.msg;
console.log(pageinfo);
$.ajax({
//type:"post",
url:"http://127.0.0.1/mysqlhelper.php",
//dateType:"json",
//async:true,
type:"post",
data:{
pageinfo:pageinfo,
user:"测试",
type:"inquire",
id:"ceshi1"
},
error: function(request) {
console.log("Connection error");
},
success: function(res) {
console.log(res);
phpresponse=res;
}
});
});