php - 这是一个正确的 utm.gif 查询吗？

Question

我想知道我是否以正确的方式构建了这个谷歌分析查询。我不确定“utmcc”部分。

  0 => string 'http://www.google-analytics.com/__utm.gif?utmwv=4.5.7' (length=53)
  1 => string 'utmn=738025229' (length=14)
  2 => string 'utmhn=www.mydomain.de' (length=15)
  3 => string 'utmcs=ISO-8859-1' (length=16)
  4 => string 'utmul=de-de' (length=11)
  5 => string 'utme=8%28myvariable%299%28myvalue%2911%283%29' (length=45)
  6 => string 'utmdt=Homepage' (length=14)
  7 => string 'utmhid=985698389' (length=16)
  8 => string 'utmp=home%2F' (length=12)
  9 => string 'utmac=UA-XXX-6' (length=17)
  10 => string 'utmcc=__utma%253D53343838.1730805896.1287423837.1287423837.1287423837.1%253B%252B__utmz%253D53343838.1287423837.1.1.utmcsr%3D%28direct%29%7Cutmccn%3D%28direct%29%7Cutmcmd%3D%28none%29%253B' (length=188)

Answer 1

对我来说它看起来是正确的，你是为每个用户生成的吗？如果您对每个人都使用相同的 cookie 字符串，您会在分析中得到一些意想不到的结果。

仅供引用，这是我在生产环境中用于生成字符串的静态方法:

public static function generate()
{
    $cookie    = rand(1000000000,9999999999);
    $random    = rand(1000000000,2147483647);
    $visitorId = mt_rand(1147483647,2147483647);
    $today     = time();
    return '__utma=' . $cookie . '.' . $visitorId . '.' . $random . '.' . $today . '.' . $today . '.1;+__utmz=1.'
           . $today . '.1.1.utmcsr=(direct)|utmccn=(direct)|utmcmd=(none);';
}