Get xml node with filter

Question

I have next xml:

<Histories>
  <History>
    <Date>2011.11.11 08:45</Date>
    <Action>Add</Action>
  </History>
  <History>
    <Date>2011.11.12 08:45</Date>
    <Action>Modify</Action>
  </History>
  <History>
    <Date>2011.11.13 08:45</Date>
    <Action>Delete</Action>
  </History>
  <History>
    <Date>2011.11.14 08:45</Date>
    <Action>Add</Action>
  </History>
  <History>
    <Date>2011.11.15 08:45</Date>
    <Action>Modify</Action>
  </History>
  <History>
    <Date>2011.11.16 08:45</Date>
    <Action>Delete</Action>
  </History>
  <History>
    <Date>2011.11.17 08:45</Date>
    <Action>Add</Action>
  </History>
  <History>
    <Date>2011.11.18 08:45</Date>
    <Action>Modify</Action>
  </History>
  <History>
    <Date>2011.11.19 08:45</Date>
    <Action>Delete</Action>
  </History>
  <History>
    <Date>2011.12.20 08:45</Date>
    <Action>Modify</Action>
  </History>
</Histories>

I need to get last node with Action (Add/Modify/Delete). How can I do?

Example:

Add 2011.11.17 08:45

Modify 2011.12.20 08:45

Delete 2011.11.19 08:45

I do: /Histories/History/Action[text()='Add']/../../History[position()=last()] - it do not work.

Thanks

Is there a requirement to use Xpath? Could you use LINQ to XMl? — TGH
– TGH, Commented Mar 28, 2012 at 5:42
Need xpath. I found solution: /Histories/History [Action ='Delete'][last()] /Date — ShurikEv
– ShurikEv, Commented Mar 28, 2012 at 6:07

Dimitre Novatchev · Accepted Answer · 2012-03-28 13:21:34Z

1

Use:

/*/History[Action = 'Delete'] [last()]

This selects the last History element that has a child Action with string value "Delete" and that (the History element) is a child of the top element of the XML document.

answered Mar 28, 2012 at 13:21

Dimitre Novatchev

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Comments

inhan · Accepted Answer · 2012-03-28 06:39:37Z

1

$arr = array('Add','Modify','Delete');
foreach ($arr as $action) {
    $res = $xPath->query('/Histories/History[Action="'.$action.'"][last()]/Date');
    echo "Last $action: " . $res->item(0)->nodeValue . "\n";
}

EDIT: Here's the compact response: /Histories/History[Action="Add"][last()]/Date

edited Mar 28, 2012 at 6:39

answered Mar 28, 2012 at 6:33

inhan

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4 Comments

Rody Over a year ago

ehh, topicstarter has his post tagged .NET and you post a PHP sample?

inhan Over a year ago

No, I'm posting the xPath actually. I had the chance to try it in PHP, that's why.

ShurikEv Over a year ago

I found solution :) See top ;-)

inhan Over a year ago

@RodyvanSambeek Thank you :) I was going to post 3 separate xPath strings but then I thought using a loop instead might give an idea about a more foolproof approach.

Simon · Accepted Answer · 2012-03-28 06:50:16Z

Consider using linq to XML

Helper method:

string GetValue(XDocument xDocument, string action)
{
    var xElements = xDocument.Descendants("Action").Where(x => x.Value == action);
    var xElement = xElements.Last().Parent;
    return xElement.Element("Date").Value;
}

Usage:

var xml = @"<Histories>
  <History>
    <Date>2011.11.11 08:45</Date>
    <Action>Add</Action>
  </History>
...
  <History>
    <Date>2011.12.20 08:45</Date>
    <Action>Modify</Action>
  </History>
</Histories>";

var xDocument = XDocument.Parse(xml);

var lastAdd= GetLast(xDocument, "Add");
var lastModify = GetLast(xDocument, "Modify");
var lastDelete = GetLast(xDocument, "Delete"); 

string GetLast(XDocument xDocument, string action)
{
    var xElements = xDocument.Descendants("Action").Where(x => x.Value == action);
    var xElement = xElements.Last().Parent;
    return xElement.Element("Date").Value;
}

Chris · Accepted Answer · 2013-11-19 09:23:00Z

public void insertToDo(string item, string date, string time, string due, string description)
        {
            XElement xEmp = XElement.Load(@"C://Users//Khulu//Documents//Visual Studio 2012//Projects//AMD//Schedule//ToDo.xml");
            //
            xEmp.Add(
                     new XElement("ToDo",
                      new XElement("Item", item),
                       new XElement("date", date),
                        new XElement("time", time),
                        new XElement("due", due),
                        new XElement("description", description))
            );
            xEmp.Save(@"C://Users//Khulu//Documents//Visual Studio 2012//Projects//AMD//Schedule//ToDo.xml");
        }
        public DataSet displayGrid()`enter code here`
        {
            DataSet ds = new DataSet();
            ds.ReadXml("C://Users//Khulu//Documents//Visual Studio 2012//Projects//AMD//Schedule//ToDo.xml");
            return ds;
        }
        public void deleteXml(string item)
        {
            XDocument doc = XDocument.Load("C://Users//Khulu//Documents//Visual Studio 2012//Projects//AMD//Schedule//ToDo.xml");
            doc.Root.Elements("ToDo")
            .Elements("Item")
            .Where(l => l.Value == item)
            .Select(x => x.Parent)
            .Remove();
            doc.Save("C://Users//Khulu//Documents//Visual Studio 2012//Projects//AMD//Schedule//ToDo.xml");
        }
        public void updateTodo(string item, string date, string time, string due, string description)
        {

            XmlDocument xmlDoc = new XmlDocument();
            xmlDoc.Load("C://Users//Khulu//Documents//Visual Studio 2012//Projects//AMD//Schedule//ToDo.xml");

            foreach (XmlNode node in xmlDoc.SelectNodes("toDoList/ToDo"))
            {
                if (node.SelectSingleNode("Item").InnerText == item)
                {
                    node.SelectSingleNode("Item").InnerText = item;
                    node.SelectSingleNode("date").InnerText = date;
                    node.SelectSingleNode("time").InnerText = time;
                    node.SelectSingleNode("due").InnerText = due;
                    node.SelectSingleNode("description").InnerText = description;

                }

            }
           xmlDoc.Save("C://Users//Khulu//Documents//Visual Studio 2012//Projects//AMD//Schedule//ToDo.xml");
        }

    }
}

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