[英]Send XML Request and receive XML Response from other server using Ajax or Javascript
I have to send an Xml to other server(an web service of a travel company) and receive response in Xml An example of xml request is: .......................................................................................................................................................我必须向其他服务器(旅游公司的 Web 服务)发送一个 Xml 并以 Xml 接收响应 一个 xml 请求示例是:....... ………………………………………………………………………………………………………………………………………………………… ………………………………………………………………………………………………………………………………………………………… ……………………………………………………………………………………………………………………………………………………………………………………………………
<?xml version="1.0" encoding="UTF-8"?>
<Request RequestType="getCountryRequest">
<AuditInfo>
<RequestId>
001
</RequestId>
<RequestUser>
user
</RequestUser>
<RequestPass>
password
</RequestPass>
<RequestTime>
2012-09-04T18:00:46
</RequestTime>
<RequestLang>
RO
</RequestLang>
</AuditInfo>
<RequestDetails>
<getCityRequest CountryCode="RO"/>
</RequestDetails>
</Request>
And the response must look like并且响应必须看起来像
<?xml version="1.0" encoding="utf-8"?>
<Response ResponseType="getCityResponse">
<AuditInfo>
<ResponseId>
12940524
</ResponseId>
<RequestId>
12949986
</RequestId>
<ResponseTime>
2012-09-04T18:10:02
</ResponseTime>
</AuditInfo>
<ResponseDetails>
<getCityResponse>
<City>
<CountryCode>
ENG
</CountryCode>
<CityCode>
ENG
</CityCode>
<CityName>
Madrid
</CityName>
</City>
</getCityResponse>
</ResponseDetails>
</Response>
Please help me!请帮我!
It should be something close to this (using jQuery):它应该与此接近(使用 jQuery):
var query = '<?xml version="1.0" encoding="UTF-8"?><Request RequestType="getCountryRequest"><AuditInfo><RequestId>001</RequestId>'+
'<RequestUser>user</RequestUser><RequestPass>password</RequestPass><RequestTime>2012-09-04T18:00:46</RequestTime><RequestLang>'+
'RO</RequestLang></AuditInfo><RequestDetails><getCityRequest CountryCode="RO"/></RequestDetails></Request>';
$.ajax({
url: 'YOUR_URL_HERE',
data: query,
type: 'POST',
contentType: "text/xml",
dataType: "text",
success : function (xmlResponse){
xmlResponse = $.parseXML( xmlResponse ),
$xml = $( xmlResponse ),
$title = $xml.find( "ResponseId" ); //to get the ResponseId for example
},
});
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