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Getting elements with default namespace (no namespace prefix) using XPath

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In this SOAP XML file, how can I get the 7 on a using a XPath query? <soap:Envelope xmlns:soap="http://www.w3.org/2003/05/soap-envelope" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema"> <soap:Body> <HelloWorldResponse xmlns="http://tempuri.org/"> <HelloWorldResult>7</HelloWorldResult> </HelloWorldResponse> </soap:Body> </soap:Envelope> This XPath query is not working //*[name () ='soap:Body'] . If you have a namespace prefix set, you could use it, like: //soap:Body But since the nodes you are trying to get use a default namespace , without a

parsing xml containing default namespace to get an element value using lxml

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I have a xml string like this str1 = """<sitemapindex xmlns="http://www.sitemaps.org/schemas/sitemap/0.9"> <sitemap> <loc> http://www.example.org/sitemap_1.xml.gz </loc> <lastmod>2015-07-01</lastmod> </sitemap> </sitemapindex> """ I want to extract all the urls present inside <loc> node i.e http://www.example.org/sitemap_1.xml.gz I tried this code but it didn't word from lxml import etree root = etree.fromstring(str1) urls = root.xpath("//loc/text()") print urls [] I tried to check if my root node is formed correctly. I tried this and get back the same string as str1 etree.tostring(root) '

XPath and namespace specification for XML documents with an explicit default namespace

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问题 I'm struggling to get the correct combination of an XPath expression and the namespace specification as required by package XML (argument namespaces ) for a XML document that has an explicit xmlns namespace defined at the top element. UPDATE Thanks to har07 I was able to put it together: Once you query the namespaces, the first entry of ns has no name yet and that's the problem: nsDefs <- xmlNamespaceDefinitions(doc) ns <- structure(sapply(nsDefs, function(x) x$uri), names = names(nsDefs)) >

Getting elements with default namespace (no namespace prefix) using XPath

阅读更多关于 Getting elements with default namespace (no namespace prefix) using XPath

问题 In this SOAP XML file, how can I get the 7 on a using a XPath query? <soap:Envelope xmlns:soap="http://www.w3.org/2003/05/soap-envelope" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema"> <soap:Body> <HelloWorldResponse xmlns="http://tempuri.org/"> <HelloWorldResult>7</HelloWorldResult> </HelloWorldResponse> </soap:Body> </soap:Envelope> This XPath query is not working //*[name () ='soap:Body'] . 回答1: If you have a namespace prefix set, you

parsing xml containing default namespace to get an element value using lxml

阅读更多关于 parsing xml containing default namespace to get an element value using lxml

问题 I have a xml string like this str1 = \"\"\"<sitemapindex xmlns=\"http://www.sitemaps.org/schemas/sitemap/0.9\"> <sitemap> <loc> http://www.example.org/sitemap_1.xml.gz </loc> <lastmod>2015-07-01</lastmod> </sitemap> </sitemapindex> \"\"\" I want to extract all the urls present inside <loc> node i.e http://www.example.org/sitemap_1.xml.gz I tried this code but it didn\'t word from lxml import etree root = etree.fromstring(str1) urls = root.xpath(\"//loc/text()\") print urls [] I tried to check