Whilst we can’t say for sure without the XML that’s being parsed, the usual reason for ‘getting blanks’ from childNodes[0] (firstChild) is that there is a whitespace Text node between the parent’s start-tag and the node you are looking for: <data> <![CDATA[ foo ]]> </data> On an XML parser that retains CDATA sections, that’ll give … Read more
Yes, you can do that. You can have one layout XML file and use the same layout in different activities. All you have to do is to use the same layout id in setContentView() method. The same applies to fragments. solved can single xml file be linked to multiple java classes? [duplicate]
I couldn’t quite work out your logic but I think you may benefit from using a key here to look up the SecDef element using its MktSegGrp attribute value <xsl:key name=”MktSeg” match=”SecDef” use=”MktSegGrp/@MktSegID” /> So, for a given MktDef, you would get the SecDef elements for it like so <xsl:variable name=”secDef” select=”key(‘MktSeg’, @MktSegID)” /> Try … Read more
As the error says: structure xmlNode has no member “d”. If you want to compare the content of the node, use the member “content”: xmlStrcmp(cur->content, (const xmlChar *)”d”)) 2 solved C++ program error Libxml2 [duplicate]
You can use the xml_parse_into_struct() function for an easy to parse XML. <?php $simple = “<para><note>simple note</note></para>”; $p = xml_parser_create(); xml_parse_into_struct($p, $simple, $vals, $index); xml_parser_free($p); echo “Index array\n”; print_r($index); echo “\nVals array\n”; print_r($vals); ?> source solved how to Retrieve date from xml file [closed]
Nice question…This problem taken me back to basics of java.. Here we go for the solution.. Car.java package com.sof.test; public class Car { private String model; private String version; public String getModel() { return model; } public void setModel(String model) { this.model = model; } public String getVersion() { return version; } public void setVersion(String … Read more
Yes you can. This question is a duplicate to Is it possible to show an One2many field in a kanban view in Odoo? but here is a link to a module from Serpent Consulting which will be able to do what you are looking for. https://apps.openerp.com/apps/modules/8.0/web_one2many_kanban/ Here is a little more info. <kanban> <field name=”one2manyFieldname”/> … Read more
I solved this problem in this way: @Root(name = “images”) public class Images { @ElementList(entry = “image”, required=false, inline = true) private List<String> imageList; public List<String> getImageList() { return imageList; } } solved Java, Simple XML parsing list
This XSLT will do the thing: <xsl:stylesheet xmlns:xsl=”http://www.w3.org/1999/XSL/Transform” xmlns:x=”http://ws.wso2.org/dataservice” xmlns=”http://ws.wso2.org/dataservice” exclude-result-prefixes=”x” version=”1.0″> <xsl:output indent=”yes” method=”xml” /> <xsl:template match=”x:Standards”> <Standards namespace=”{namespace-uri()}”> <xsl:apply-templates select=”.//x:Standard” /> </Standards> </xsl:template> <xsl:template match=”x:Standard”> <Standard> <xsl:copy-of select=”x:ProductID” /> <xsl:copy-of select=”x:Prefix”/> <xsl:copy-of select=”x:SNumber”/> <RelatedProducts> <xsl:apply-templates select=”.//x:RelatedProduct”/> </RelatedProducts> <xsl:copy-of select=”x:S1″/> <xsl:copy-of select=”x:S2″/> </Standard> </xsl:template> <xsl:template match=”x:RelatedProduct”> <xsl:element name=”RelatedProduct”> <xsl:element name=”RelationType”> <xsl:value-of select=”name(..)” /> </xsl:element> … Read more
Before anybody gives me negative points for this highly unusual solution, please understand the problems. 1) The multiple Folder nested tags 2) The html text I using a xml template file to read intial xml. This xml is complicated so I don’t think it is wise to create the xml from scratch. Use the following … Read more
HTTP basic auth is dead simple. You just need to add the credentials to every request. There is no session, each request requires authentication. solved Creating a generalized HTTP Authentication function [closed]
You can’t. You have to implement it yourself by code. Today I’m in a good mood and I’m bored, so I will write some code for you, even if you made like no effort for it. In shorts, add the switch and the TextView/EditText in your XML, then implement in your code something like this: … Read more
I’d use LINQ to XML, with a helper method: var movies = from element in XDocument.Parse(xml).Descendants(“Movie”) select new Class1 { Id = (int) element.Attribute(“ID”), Subject = (string) element.Element(“Name”), OtherName = (string) element.Element(“OtherName”), Duration = (int) element.Element(“Duration”) .Attribute(“Duration”), Property1 = (string) element.Element(“Properties”) .Elements(“Property”) .Where(x => (string) x.Attribute(“Name”) == “Property1”) .Single(), Property2 = (string) element.Element(“Properties”) .Elements(“Property”) .Where(x … Read more
Document containing unclosed tag is not XML at all. As others suggested in comments, ideally the effort to fix this problem is done by the party that generate the document. Regarding the original question, detecting unclosed tag in general isn’t a trivial task. I would suggest to try HtmlAgilityPack (HAP). It has built in functionality … Read more
So if I get it correct, you want to append new data to your existing xml. For that you can chose to temporarily store the current xml and and add new data to it using Linq Xml. To do this, now you need to modify your current code with an additional check for xml file … Read more