Question

我正在尝试在局部视图中实现谷歌地图，并尝试在页面上调用部分视图这是我放在主视图上的谷歌地图脚本

 <script src="https://maps.googleapis.com/maps/api/js?key=AIzaSyDLhntIiMFhAoHu5XtrxDuh4EeNbm2ZuH0&libraries=places&callback=initAutocomplete" async defer></script>
    <script type="text/javascript">
        function initAutocomplete()
        {

            var map = new google.maps.Map(document.getElementById('map'), 
                {
                center: { lat: -33.8688, lng: 151.2195 },
                zoom: 13,
                mapTypeId: 'roadmap'
            });
            console.log('map', map);
            // Create the search box and link it to the UI element.
            var input = document.getElementById('pac-input');
            var searchBox = new google.maps.places.SearchBox(input);
            map.controls[google.maps.ControlPosition.TOP_LEFT].push(input);

            // Bias the SearchBox results towards current map's viewport.
            map.addListener('bounds_changed', function () {
                searchBox.setBounds(map.getBounds());
            });

            var markers = [];
            // Listen for the event fired when the user selects a prediction and retrieve
            // more details for that place.
            searchBox.addListener('places_changed', function () {
                var places = searchBox.getPlaces();

                if (places.length == 0) {
                    return;
                }

                // Clear out the old markers.
                markers.forEach(function (marker) {
                    marker.setMap(null);
                });
                markers = [];

                // For each place, get the icon, name and location.
                var bounds = new google.maps.LatLngBounds();
                places.forEach(function (place) {
                    if (!place.geometry) {
                        console.log("Returned place contains no geometry");
                        return;
                    }
                    var icon = {
                        url: place.icon,
                        size: new google.maps.Size(71, 71),
                        origin: new google.maps.Point(0, 0),
                        anchor: new google.maps.Point(17, 34),
                        scaledSize: new google.maps.Size(25, 25)
                    };
                    // Create a marker for each place.
                    markers.push(new google.maps.Marker({
                        map: map,
                        icon: icon,
                        title: place.name,
                        position: place.geometry.location
                    }));

                    if (place.geometry.viewport) {
                        // Only geocodes have viewport.
                        bounds.union(place.geometry.viewport);
                    } else {
                        bounds.extend(place.geometry.location);
                    }
                    var address = $('#pac-input').val();
                    //document.getElementById('pac-input');
                    $.ajax({
                        url: "http://maps.googleapis.com/maps/api/geocode/json?address=" + address + "&sensor=false",
                        type: "POST",
                        success: function (res) {
                            latitude = res.results[0].geometry.location.lat;
                            longitude = res.results[0].geometry.location.lng;
                        }
                    });
                });
                map.fitBounds(bounds);
            });
        }

这是我的HTML放置在局部视图中

 <div class="form-group">
            <label>Type a location to Search on Map</label>
            <input type="text" id="pac-input" class="form-control" ng-model="MyLocations.Location" required>
            <div class="selectlocatinonmap" id="map">
            </div>
    </div>

我已将此脚本放在我必须调用部分视图的页面上，但谷歌地图无法在局部视图上工作我也附上了这张照片请看一下

但是控制台没有显示任何错误。

Answer 1

也许是因为当调用生成地图的回调函数时，地图容器会被隐藏。在完成淡入淡出或滑动效果后，您可以在显示表单的按钮上调用init map回调。

修改

$('#my_button_wich_show_form').click(initAutocomplete);

请记住从地图网址中删除回调参数。如果在显示模态之前使用某些效果，请在效果完成后调用initAutocomplete函数。

在initAutocomplete函数中，您必须检查#map是否为空，这样您就可以执行脚本或只触发map resize事件：

<script type="text/javascript"> function initAutocomplete() { if( !$('#map').is(':empty') ){ google.maps.event.trigger(this.map, 'resize'); return false; } this.map = new google.maps.Map(document.getElementById('map'), { center: { lat: -33.8688, lng: 151.2195 }, zoom: 13, mapTypeId: 'roadmap' }); var map = this.map; console.log('map', map); // Create the search box and link it to the UI element. var input = document.getElementById('pac-input'); var searchBox = new google.maps.places.SearchBox(input); map.controls[google.maps.ControlPosition.TOP_LEFT].push(input); // Bias the SearchBox results towards current map's viewport. map.addListener('bounds_changed', function () { searchBox.setBounds(map.getBounds()); }); var markers = []; // Listen for the event fired when the user selects a prediction and retrieve // more details for that place. searchBox.addListener('places_changed', function () { var places = searchBox.getPlaces(); if (places.length == 0) { return; } // Clear out the old markers. markers.forEach(function (marker) { marker.setMap(null); }); markers = []; // For each place, get the icon, name and location. var bounds = new google.maps.LatLngBounds(); places.forEach(function (place) { if (!place.geometry) { console.log("Returned place contains no geometry"); return; } var icon = { url: place.icon, size: new google.maps.Size(71, 71), origin: new google.maps.Point(0, 0), anchor: new google.maps.Point(17, 34), scaledSize: new google.maps.Size(25, 25) }; // Create a marker for each place. markers.push(new google.maps.Marker({ map: map, icon: icon, title: place.name, position: place.geometry.location })); if (place.geometry.viewport) { // Only geocodes have viewport. bounds.union(place.geometry.viewport); } else { bounds.extend(place.geometry.location); } var address = $('#pac-input').val(); //document.getElementById('pac-input'); $.ajax({ url: "http://maps.googleapis.com/maps/api/geocode/json?address=" + address + "&sensor=false", type: "POST", success: function (res) { latitude = res.results[0].geometry.location.lat; longitude = res.results[0].geometry.location.lng; } }); }); map.fitBounds(bounds); }); }

Answer 2

运行Fiddler它是免费的，看看你的请求是否正在进行googgle。根据您的屏幕截图，我没有看到任何xhr到Google地图的API。我想你被封锁了。

Google地图未在部分视图中显示？

2 个答案: