Can't the parameters of url that Wechat wants to share be escaped?

error https://sf.gg?url=https://baidu.com/%E6%90%9C%E7%B4%A2"

normal https://sf.gg?url=https://baidu.com/ search


depending on what you mean, do you want to add parameters after a link like' https://baidu.com?
if so, you can copy:

function set_url(url, param) {
    //url
    var n_param = urlToString(param);
    var n_url = "";
    if(param != '' && param != null)
        n_url = url + '?' + n_param;
    else
        n_url = url;
    return n_url;
}
//url 'https://baidu.com'
//param 

help you find it, it seems that only one backend method can be found: Portal
String temp = URLEncoder.encode (str," UTF-8 ");

are you using the wrong method? Try this encodeURIComponent (url)

Menu