Question

我正在尝试通过Google API获取地址的坐标。这是我的要求：http://maps.googleapis.com/maps/api/geocode/xml?address=＆＃34; +＆＃34;柏林Hbf（tief）＆＃34; +＆＃34;＆amp; sensor = true＆＃34; 这是谷歌的回应：

<?xml version="1.0" encoding="UTF-8"?>
<GeocodeResponse>
    <status>OK</status>
    <result>
        <type>transit_station</type>
        <type>point_of_interest</type>
        <type>establishment</type>
        <formatted_address>Berlin Hbf (tief), 10557 Berlin, Germany</formatted_address>
        <address_component>
            <long_name>Berlin Hbf (tief)</long_name>
            <short_name>Berlin Hbf (tief)</short_name>
            <type>point_of_interest</type>
            <type>establishment</type>
        </address_component>
        <address_component>
            <long_name>Mitte</long_name>
            <short_name>Mitte</short_name>
            <type>sublocality_level_1</type>
            <type>sublocality</type>
            <type>political</type>
        </address_component>
        <address_component>
            <long_name>Berlin</long_name>
            <short_name>Berlin</short_name>
            <type>locality</type>
            <type>political</type>
        </address_component>
        <address_component>
            <long_name>Berlin</long_name>
            <short_name>Berlin</short_name>
            <type>administrative_area_level_1</type>
            <type>political</type>
        </address_component>
        <address_component>
            <long_name>Germany</long_name>
            <short_name>DE</short_name>
            <type>country</type>
            <type>political</type>
        </address_component>
        <address_component>
            <long_name>10557</long_name>
            <short_name>10557</short_name>
            <type>postal_code</type>
        </address_component>
        <geometry>
            <location>
                <lat>52.5253840</lat>
                <lng>13.3692087</lng>
            </location>
            <location_type>APPROXIMATE</location_type>
            <viewport>
                <southwest>
                    <lat>52.5240350</lat>
                    <lng>13.3678597</lng>
                </southwest>
                <northeast>
                    <lat>52.5267330</lat>
                    <lng>13.3705577</lng>
                </northeast>
            </viewport>
        </geometry>
        <place_id>ChIJ0SkfA5ZRqEcRo-ThHeCZJl8</place_id>
    </result>
</GeocodeResponse>

这是我用来反序列化它的代码：

string servResp = await response.Content.ReadAsStringAsync();

try
{
    XmlSerializer serializer = new XmlSerializer(typeof(GoogleAPI.GeocodeResponse));
    geoCodeResponse = ((GoogleAPI.GeocodeResponse)(serializer.Deserialize(XmlReader.Create(servResp))));
}
    catch (Exception ex)
{
    System.Diagnostics.Debug.WriteLine(ex.Message);
}

这些是我的课程：http://codeshare.io/qGChA

出于某种原因，它崩溃了一个例外：

$ exception.InnerException {System.ArgumentException：非法路径中的字符。在System.IO.Path.CheckInvalidPathChars（String 在System.IO.Path.GetFileName（String。的路径，布尔检查附加）在System.IO.FileStream..ctor（String路径，FileMode模式， FileAccess访问，FileShare共享，Int32 bufferSize）at System.Xml.XmlRelativePathResolver.OpenStream（URI uri）} System.Exception {System.ArgumentException}

为什么会这样？

Answer 1

Deserialize方法应该提供StringReader对象，而不是字符串。正确的解决方案是：

string servResp = await response.Content.ReadAsStringAsync();
            StringReader stringReader = new StringReader(servResp);

            try
            {
                XmlSerializer serializer = new XmlSerializer(typeof(GoogleAPI.GeocodeResponse));
                geoCodeResponse = ((GoogleAPI.GeocodeResponse)(serializer.Deserialize(XmlReader.Create(stringReader))));
            }
            catch (Exception ex)
            {
                System.Diagnostics.Debug.WriteLine(ex.Message);
            }

Answer 2

XmlReader.Create(string)从输入参数指定的 URI中读取XML。您传递的是实际的XML而不是引用它的URI。您需要通过StringWriter提供XML字符串，如以下扩展方法：

public static class XmlExtensions { public static T DeserializeXml<T>(this string xmlString) { using (var reader = new StringReader(xmlString)) { return (T)new XmlSerializer(typeof(T)).Deserialize(reader); } } }

然后做：

string servResp = await response.Content.ReadAsStringAsync(); try { geoCodeResponse = servResp.DeserializeXml<GoogleAPI.GeocodeResponse>(); } catch (Exception ex) { System.Diagnostics.Debug.WriteLine(ex.Message); }

为什么无法反序列化此GoogleAPI XML响应？

2 个答案: