我想在XML文件中显示多条折线:
<lines>
<line>
<linePoint lat="47.608940" lng="-122.340141"/>
<linePoint lat="47.613590" lng="-122.344391"/>
<linePoint lat="47.624561" lng="-122.356445"/>
</line>
<line>
<linePoint lat="47.616600" lng="-122.344491"/>
<linePoint lat="47.627661" lng="-122.356545"/>
</line>
</lines>
我相信脚本应该迭代两次 - 外部循环遍历line
元素,内部循环遍历每个linePoint
内的line
元素。
但似乎我的脚本将所有5 linePoint
转储到一条折线而不是创建2条折线。
downloadUrl("createXML.php", function(data) {
var xml = data.responseXML;
var line = xml.documentElement.getElementsByTagName("line");
// Read each line
for (var a = 0; a < line.length; a++) {
var linePoint = xml.documentElement.getElementsByTagName("linePoint");
var path = [];
// Read markers on each line
for (var i = 0; i < linePoint.length; i++) {
var lat = parseFloat(linePoint[i].getAttribute("lat"));
var lng = parseFloat(linePoint[i].getAttribute("lng"));
var point = new google.maps.LatLng(lat,lng);
path.push(point);
}
var polyline = new google.maps.Polyline({
path: path
});
polyline.setMap(map);
}
});
有人可以指出我忽略的内容吗?
答案 0 :(得分:2)
是的,这是因为您只是通过all
(复数元素)在一个块中提取<linePoint>
getElementsByTagName("linePoint")
。迭代<line>
以及<line>
遍历<linePoint>
次迭代:
function showPolyLines() {
var parser = new DOMParser();
var xml = parser.parseFromString(responseXML, "application/xml");
var line = xml.querySelectorAll("line");
for (var l = 0; l < line.length; l++) {
var path = [],
linePoints = line[l].querySelectorAll('linePoint');
for (var p = 0; p < linePoints.length; p++) {
var linePoint = linePoints[p],
lat = parseFloat(linePoint.getAttribute("lat")),
lng = parseFloat(linePoint.getAttribute("lng")),
point = new google.maps.LatLng(lat, lng);
path.push(point);
}
var polyline = new google.maps.Polyline({
path: path,
//just to be sure the lines actually is shown on the map :)
strokeColor: '#FF0000',
strokeOpacity: 1.0,
strokeWeight: 2
});
polyline.setMap(map);
}
}
我使用DOMParser替换xml.documentElement
,因为我的小提琴测试中没有真正的responseXML
,但它们实际上是相同的。
演示 - &gt;的 http://jsfiddle.net/9sqastj1/ 强>