首页 \ 问答 \ 使用jQuery过滤Google地图(Filtering Google Maps with jQuery)

使用jQuery过滤Google地图(Filtering Google Maps with jQuery)

有没有办法用jQuery过滤谷歌地图的位置变量?

例如:

  • 汽车
  • 点击带有ID car的链接,过滤ID car的所有字符串。 点击身份证汽车会在地图上隐藏带有ID船的所有字符串,并在地图中仅显示带有ID汽车的字符串。

    可能吗?

    function initialize() {
    
        var locations = [
    
    
      ['<strong>Firm 1</strong><br />1111 Country<br />Street<br /><br />', 47.6801806,8.7499505, 1, id car here?? ],
    
      ['<strong>Firm 2</strong><br />2222 Country<br />Street<br /><br />', 47.6801806,8.7499505, 2, id boat here?? ],
    
    
    
    ];
    
        var map = new google.maps.Map(document.getElementById('map_canvas'), {
          zoom: 10,
          center: new google.maps.LatLng(47.603786, 9.055737),
          mapTypeId: google.maps.MapTypeId.ROADMAP
        });
    
    
    
    
    
    
        var infowindow = new google.maps.InfoWindow();
    
        var marker, i;
    
        for (i = 0; i < locations.length; i++) {  
          marker = new google.maps.Marker({
            position: new google.maps.LatLng(locations[i][1], locations[i][2]),
            map: map,
           icon: new google.maps.MarkerImage('/assets/templates/style/images/iconKarte.png')
          });
    
          google.maps.event.addListener(marker, 'click', (function(marker, i) {
            return function() {
              infowindow.setContent(locations[i][0]);
              infowindow.open(map, marker);
            }
          })(marker, i));
        }
    
     }
    
    </script>
    
    
    
    
     <div id="map_canvas" style="width:750px;height:500px;"></div>
    

    Is there a way to filter Google Maps location variables with jQuery?

    Ex:

  • Car
  • Boat
  • Clicking on the link with ID car, filters all strings with ID car. Clicking by the ID car hides all strings with ID boat on the Maps and show only strings with ID car in the Maps.

    Is it possible?

    function initialize() {
    
        var locations = [
    
    
      ['<strong>Firm 1</strong><br />1111 Country<br />Street<br /><br />', 47.6801806,8.7499505, 1, id car here?? ],
    
      ['<strong>Firm 2</strong><br />2222 Country<br />Street<br /><br />', 47.6801806,8.7499505, 2, id boat here?? ],
    
    
    
    ];
    
        var map = new google.maps.Map(document.getElementById('map_canvas'), {
          zoom: 10,
          center: new google.maps.LatLng(47.603786, 9.055737),
          mapTypeId: google.maps.MapTypeId.ROADMAP
        });
    
    
    
    
    
    
        var infowindow = new google.maps.InfoWindow();
    
        var marker, i;
    
        for (i = 0; i < locations.length; i++) {  
          marker = new google.maps.Marker({
            position: new google.maps.LatLng(locations[i][1], locations[i][2]),
            map: map,
           icon: new google.maps.MarkerImage('/assets/templates/style/images/iconKarte.png')
          });
    
          google.maps.event.addListener(marker, 'click', (function(marker, i) {
            return function() {
              infowindow.setContent(locations[i][0]);
              infowindow.open(map, marker);
            }
          })(marker, i));
        }
    
     }
    
    </script>
    
    
    
    
     <div id="map_canvas" style="width:750px;height:500px;"></div>
    

    原文:https://stackoverflow.com/questions/14585719
    更新时间:2023-05-10 18:05

    最满意答案

    我只想创建新的Java项目(File / New Project / Java), 从模板中选择Create project并选择Command line app 。 这将创建一个包含具有空main方法的类的项目,您可以通过按SHIFT + F10或使用SHIFT + F9进行调试来运行该方法。

    整个过程只需20秒。

    就目录结构的复杂性而言,你只需要关注一个目录--src /。 如果需要,可以将所有类保留在默认包中,因此所有代码都将位于src /中,而不包含任何子目录。

    我不认为这比这更容易。


    I would just create new Java project (File/New Project/Java), select Create project from template and choose Command line app. This creates a project containing a class with empty main method, which you can run by hitting SHIFT + F10 or debug using SHIFT + F9.

    This whole process takes only like 20 seconds.

    And as far as the directory structure complexity goes, you have just one directory that you need to be concerned about - src/. You can keep all your classes in the default package if you want and therefore all your code will be in src/ without any subdirectories for packages.

    I don't think it gets any easier than this.

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