php微信公众号code获取不到 微信公众号调取用户信息 遇到invalid code的问题

使用了php，php部分如下：

$code = $_GET["code"];

$userinfo = getUserInfo($code);

function getUserInfo($code)

{

$appid = "1111111111";

$appsecret = "1111111111111";//appid和appsecret在这里隐去，在源代码中是正确的

$access_token = "";

//Get access_token

$access_token_url = "https://api./sns/oauth2/access_token?appid=$appid&secret=$appsecret&code=$code&grant_type=authorization_code";

$access_token_json = https_request($access_token_url);

$access_token_array = json_decode($access_token_json,true);

$access_token = $access_token_array['access_token'];

$openid = $access_token_array['openid'];

//Get user info

$userinfo_url = "https://api./sns/userinfo?access_token=$access_token&openid=$openid";

$userinfo_json = https_request($access_token_url);

$userinfo_array = json_decode($userinfo_json,ture);

echo $userinfo_json;

return userinfo_array;

}

function https_request($url)

{

$curl = curl_init();

curl_setopt($curl, CURLOPT_URL, $url);

curl_setopt($curl, CURLOPT_SSL_VERIFYPEER, FALSE);

curl_setopt($curl, CURLOPT_SSL_VERIFYHOST, FALSE);

curl_setopt($curl, CURLOPT_RETURNTRANSFER, 1);

$data = curl_exec($curl);

if (curl_errno($curl)){

return 'ERROR'.curl_error($curl);

}

curl_close($curl);

return $data;

}

然后现在出现的问题就是，调userinfo_json时出现了问题，导致了最后结果无法显示。

中间过程把userinfo_json输出时，提示{"errcode":40029,"errmsg":"invalid code"}

而如果我把userinfo_url的链接复制下来在手机打开的话，还是能调用到用户的详细资料

所以问题到底出现在了哪里？有什么解决办法？求大神解答