php - 带有 gtag_report_conversion 的 php mysql 表单

Question

我正在尝试使用表单获取数据并使用 gtag_report_conversion 跟踪 Adwords

我的代码：

                  <form id="form1" method="post" enctype="multipart/form-data" role="form" autocomplete="off">
          <ul class="form-box">
              <li>
                 <input name="adi" type="text" id="adi" class="form-text" required="required" placeholder="Adınız" />
              </li>
              <li>
                 <input name="mesaj" type="text" id="mesaj" class="form-text" required="required" placeholder="Soyadınız" />
              </li>
              <li>
                 <input name="telefon" type="tel"  id="telefon" class="form-text" required="required" placeholder="Telefon" />
              </li>
              <li>
                 <input name="mail" type="email" id="mail" class="form-text" required="required" placeholder="E-Posta" />
              </li>
              <li class="izin">
                 <input type="checkbox" required="required" checked> <a href="javascript:;" data-fancybox="modal" data-src="#bilgilendirme-modal">İletişim sözleşmesini </a> okudum,
                 <br> kabul ediyorum.
              </li>

           </ul>
        </form>
        <button  type="submit" name="submit" form="form1" class="form-submit" value="Submit" onclick="return gtag_report_conversion('http://localhost/thanks.php');">GÖNDER</button>

我将数据保存在 mysql 中。

这是同一页面中的代码。

if(isset($_POST['submit']))
{
$adi = $_POST["adi"];
$telefon = $_POST["telefon"];
$mail = $_POST["mail"];
$mesaj = $_POST["mesaj"];
$tarih = date("d.m.Y");
$ip = $_SERVER['REMOTE_ADDR'];
$stmt = $DB_con->prepare("INSERT INTO iletisim_formu (adi,telefon,mail,mesaj,tarih,ip) VALUES ('" . $adi. "','" . $telefon. "','" . $mail. "','" . $mesaj. "','" . $tarih. "','" . $ip. "')");
$stmt->execute();
}

当我写按钮

onclick="return gtag_report_conversion(' http://localhost/thanks.php ');"

它会感谢页面，但不会保存到数据库。当我删除它时，它会保存到 db 但我无法使用 gtag。

怎么了？

我该怎么做才能正确？

标签： phphtmlmysqlgoogle-ads-api