Question

我正在尝试创建一个保持打开的弹出窗口，直到我点击右上角的x将其关闭。做这个的最好方式是什么？我的代码在下面，谢谢！

      //pop up code
     //create custom icon
      var newicon = L.icon({
        iconUrl: 'logo.png',
      })

      // creating marker
      var marker = L.marker(new L.LatLng(41.77, -87.6), {
      icon: L.mapbox.marker.icon({
          'marker-color': 'ff8888'
      }),
      icon: newicon,
      draggable: true,
  }).addTo(map);

      // bind popup to marker
      marker.bindPopup("I am a text that will stay open until closed").openPopup();

Answer 1

使用：

foreach (range(1, 100) as $x) {
    foreach (range(1, 9) as $y) {
        $html = get_data('http://mysite.info/screenshots/'.$x.'-'.$y.'.jpeg' );
    }
}

Answer 2

新解决方案，旧解决方案似乎无效：

var new_popup = L.popup({"autoClose": false, "closeOnClick": null});

marker.bindPopup(new_popup);

Answer 3

要打开弹出窗口，您需要在 L.map() 中指定 closePopupsOnClick: false 以便默认情况下不会关闭弹出窗口

L.map('地图', { ... closePopupOnClick: 假 }

然后，对于您想要关闭的单个弹出窗口，您可以在 click() 中关闭它们

    map.on('popupopen', function (e) {
           //save e.popup to an array or something
        }

    map.on('click', () => {
            //close the popups you want to by calling map.closePopup(popup);
        });

宣传单：保持PopUp Open

3 个答案: