单击按钮时打开弹出窗口

Question

我在 Flodesk 中制作了一个弹出窗体，并在我的网站的结束 head 标签之前安装了以下 Javascript 片段。

<script>
  (function(w, d, t, h, s, n) {
    w.FlodeskObject = n;
    var fn = function() {
      (w[n].q = w[n].q || []).push(arguments);
    };
    w[n] = w[n] || fn;
    var f = d.getElementsByTagName(t)[0];
    var v = '?v=' + Math.floor(new Date().getTime() / (120 * 1000)) * 60;
    var sm = d.createElement(t);
    sm.async = true;
    sm.type = 'module';
    sm.src = h + s + '.mjs' + v;
    f.parentNode.insertBefore(sm, f);
    var sn = d.createElement(t);
    sn.async = true;
    sn.noModule = true;
    sn.src = h + s + '.js' + v;
    f.parentNode.insertBefore(sn, f);
  })(window, document, 'script', 'https://assets.flodesk.com', '/universal', 'fd');
</script>
<script>
  window.fd('form', {
    formId: '615edb53f88d548e68f5c863'
  });
</script>

目前，我的弹出表单会在页面加载时自动打开。相反，我希望它在单击按钮时打开。有什么办法可以做到这一点吗？谢谢！

Answer 1

好像

window.fd('form', {
    formId: '615edb53f88d548e68f5c863'
  });

打开模式，这样您就可以在单击按钮时使用函数调用它。示例：

<script>
  (function(w, d, t, h, s, n) {
    w.FlodeskObject = n;
    var fn = function() {
      (w[n].q = w[n].q || []).push(arguments);
    };
    w[n] = w[n] || fn;
    var f = d.getElementsByTagName(t)[0];
    var v = '?v=' + Math.floor(new Date().getTime() / (120 * 1000)) * 60;
    var sm = d.createElement(t);
    sm.async = true;
    sm.type = 'module';
    sm.src = h + s + '.mjs' + v;
    f.parentNode.insertBefore(sm, f);
    var sn = d.createElement(t);
    sn.async = true;
    sn.noModule = true;
    sn.src = h + s + '.js' + v;
    f.parentNode.insertBefore(sn, f);
  })(window, document, 'script', 'https://assets.flodesk.com', '/universal', 'fd');
</script>
<script>
  function openModal() {
    window.fd('form', {
      formId: '615edb53f88d548e68f5c863'
    });
  }
</script>
<button onclick="openModal()">Open Modal</button>

单击按钮时打开弹出窗口

Open a popup when a button is clicked

html

javascript

shopify