动态标记 - Google Maps API - 标记不会在循环中使用i值递增

时间:2012-05-04 01:29:44

标签: google-maps dynamic google-maps-markers

我在下面创建了地图api,并且有一个来自数据库的地址数组。 地图加载正确,但我有以下问题。

  1. 动态标记没有以“i”的值递增..所以地图显示所有标记“1”而不是“1,2,3,4,5”

    // Define Vars
    var pinColor = "FE7569";
    var marker, i;
    
    // Set default map center location
    var latlng = new google.maps.LatLng(-27.79014,153.18049);
    
    // Create pinShadow for each marker 
    var pinShadow = new google.maps.MarkerImage("http://chart.apis.google.com        /chart?chst=d_map_pin_shadow",
        new google.maps.Size(40, 37),
        new google.maps.Point(0, 0),
        new google.maps.Point(12, 35));
    
    // Function to create the dynamic marker
    function pinImage(i){
        return image = new google.maps.MarkerImage("http://www.googlemapsmarkers.com/v1/"+i+"/"+pinColor+"/", new google.maps.Size(21, 34), new google.maps.Point(0,0), new google.maps.Point(10, 34));
    }
    
    // Function to run once page has loaded
        function initialize(){      
    // Set Default settigns for google map
    var settings = {
        zoom: 10,
        center: latlng,
        mapTypeControl: true,
        mapTypeControlOptions: {style: google.maps.MapTypeControlStyle.DROPDOWN_MENU},
        navigationControl: true,
        navigationControlOptions: {style: google.maps.NavigationControlStyle.SMALL},
        mapTypeId: google.maps.MapTypeId.ROADMAP};
    
    // Start Google Maps and assign it to div on site
    var map = new google.maps.Map(document.getElementById("map_canvas"), settings);     
    
    // loop though total numner of address in array
    for (i = 1; i < total; i++) {  
        // Use geocoder to grab latlong for user inputed address
        geocoder.geocode( { 'address': address[i]}, function(results, status) {
            if (status == google.maps.GeocoderStatus.OK) {
                // Redefine map center location
                if (i = 1){ map.setCenter(results[0].geometry.location); }
    
                // Create dynamic markers on map as per the address array
                var marker = new google.maps.Marker({
                    map: map, 
                    position: results[0].geometry.location,
                    title:address[i],
                    zIndex: i,
                    // PROBLEM CODE --- Using the function below it creates all Markers with a "1" instead of the "i" value..??  
                    icon: pinImage(i),
                    shadow: pinShadow
                });         
            }
        });
        }
    }
    

1 个答案:

答案 0 :(得分:0)

旧的“函数范围包装器”技巧(function(i) {...})(i)在这里工作。没有它,执行将获取i的最终值。此外,if(i = 1)应为if(i == 1)

玩耍:http://jsfiddle.net/BK8vv/

for (var i = 1; i < total; i++) {  
   (function(i) {

    // Use geocoder to grab latlong for user inputed address
    geocoder.geocode( { 'address': address[i]}, function(results, status) {
        if (status == google.maps.GeocoderStatus.OK) {
            // Redefine map center location
            if (i == 1){ map.setCenter(results[0].geometry.location); }

            // Create dynamic markers on map as per the address array
                    alert(i);
            var marker = new google.maps.Marker({
                map: map, 
                position: results[0].geometry.location,
                title:address[i],
                zIndex: i,
                // PROBLEM CODE --- Using the function below it creates all Markers with a "1" instead of the "i" value..??  
                icon: pinImage(i),
                shadow: pinShadow
            });         
        }
       });
     })(i);
    }
}