In Xml remove node value not the entire node

In Xml remove node value not the entire node

By : beile1968
Date : November 14 2020, 05:16 PM
hop of those help? In Xml How to remove node value in flex4. , Yes it is possible.
Considering your xml is declared like this
code :
var myXML:XML = new XML(<item label="">R1</item>);
resultXML..item(@label == "").setChildren("");

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How to remove a entire node from xml based on content of xml using c#

How to remove a entire node from xml based on content of xml using c#

By : Wurmatron
Date : March 29 2020, 07:55 AM
With these it helps Here's an example that removes the site element containing any element with an href attribute containing www.google.com
code :
using System.Diagnostics;
using System.Linq;
using System.Xml.Linq;

namespace ConsoleApplication6
    class Program
        static void Main(string[] args)
            const string frag = @" <websites>
 <a xmlns=""http://www.w3.org/1999/xhtml"" href=""www.google.com""> Google </a>
 <a xmlns=""http://www.w3.org/1999/xhtml"" href=""www.hotmail.com""> Hotmail </a>

            var doc = XDocument.Parse(frag);

            //Locate all the elements that contain the attribute you're looking for
            var invalidEntries = doc.Document.Descendants().Where(x =>
                //Get the href attribute from the element
                var hrefAttribute = x.Attribute("href");
                //Check to see if the attribute existed, and, if it did, if it has the value you're looking for
                return hrefAttribute != null && hrefAttribute.Value.Contains("www.google.com");

            //Find the site elements that are the parents of the elements that contain bad entries
            var toRemove = invalidEntries.Select(x => x.Ancestors("site").First()).ToList();

            //For each of the site elements that should be removed, remove them
            foreach(var entry in toRemove)

Remove entire xml node

Remove entire xml node

By : Govind Venugopalan
Date : March 29 2020, 07:55 AM
wish help you to fix your issue I try to remove a node from my xml with jdom2 but i don't find the answer
code :
What you are trying to achieve is complex(multi-step process) in nature,
I can share with you the steps(algorithm) that you need to follow:

You need to genearte the schema first:
Generate the schema from this online tool: 

Using these schema generate the Java classes:
using api tools like XMLBeans or JAXB
a) scomp -out employeeschema.jar employeeschema.xsd (Ref: http://xmlbeans.apache.org/)
b) xjc -d out customer.xsd (Ref: http://blog.bdoughan.com/2011/10/jaxb-xjc-imported-schemas-and-xml.html)

Then follow below tutorial for your reference for in-depth CRUD operation that you want to perform on your XML:
Jaxb: http://www.mkyong.com/java/jaxb-hello-world-example/
Xstream: http://x-stream.github.io/tutorial.html
XMLBeans: http://xmlbeans.apache.org/documentation/tutorial_getstarted.html
xslt to move a node under second node into first node and remove second node

xslt to move a node under second node into first node and remove second node

By : Talha Hafeez Qureshi
Date : March 29 2020, 07:55 AM
I wish this helpful for you
move all the remarks tags under the second release tag to first release tag, then delete the second release tag.
code :
<xsl:stylesheet version="1.0" 
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:strip-space elements="*"/>

<!-- identity transform -->
<xsl:template match="@*|node()">
        <xsl:apply-templates select="@*|node()"/>

<xsl:template match="/root">
            <xsl:apply-templates select="release[1]/*"/>
            <xsl:apply-templates select="release[2]/remarks"/>
            <xsl:apply-templates select="*[not(self::release)]"/>

Firebase - How to remove individual data from a node without removing entire node

Firebase - How to remove individual data from a node without removing entire node

By : Meheboob Nadaf
Date : March 29 2020, 07:55 AM
Hope this helps You almost got it. Your query is fine, but when you execute a query against the Firebase Database, there will potentially be multiple results. So the snapshot contains a list of those results. Even if there is only a single result, the snapshot will contain a list of one result.
So you need to loop over those results in the callback:
code :
blockedQuery.once('value').then(function(blockedSnapshot) {
    blockedSnapshot.forEach(function(childSnapshot) {  
        childSnapshot.ref.remove(function (error) {
            if (!error) {
                console.log(friendToUnblock+" should now be unblocked.");
                console.log("There's been an error unblocking the user: "+error);
Boost property tree: Remove a node using pointers to the node and its parent node

Boost property tree: Remove a node using pointers to the node and its parent node

By : SumitS
Date : March 29 2020, 07:55 AM
I hope this helps you . Indeed, there is no direct function to get an iterator from a value reference. So you'll have to write it yourself:
C++: boost ptree relative key
code :
#include <iostream>
#include <boost/property_tree/ptree.hpp>

using namespace boost::property_tree;

ptree::iterator child_iterator(ptree& within, ptree const& child) {
    for (auto it = within.begin(); it != within.end(); ++it)
        if (std::addressof(it->second) == std::addressof(child))
            return it;

    return within.end();

ptree* findParentPTree(std::string const, std::string const&, ptree const&);
ptree* findRemovePTree(std::string const, std::string const&, ptree const&);

void removeElement(const std::string& addr, const std::string& criteria, ptree &ptSource)
    ptree *ptParent = findParentPTree(addr, criteria, ptSource); // Points to "library.booklist"
    ptree *ptRemove = findRemovePTree(addr, criteria, ptSource); // eg the third <book> which contains the <data title="t3"/>

    auto it = child_iterator(*ptParent, *ptRemove);
    if (it != ptParent->end())
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