i want to create some xml with this format
<root>
<columns>
<column name="name1" value="value1" />
<column name="name2" value="value2" />
<column name="name3" value="value3" />
...
</columns>
<rows>
<row name="name1" value="value1" />
<row name="name2" value="value2" />
<row name="name3" value="value3" />
...
</rows>
</root>
so i create a
XElement tree = new XElement("root", new XElement("columns", from c in columns select new XElement("column", c) ...
And the result is
<root>
<columns>
<column>
<column>
<column>
...
</column>
</column>
</columns>
...
</root>
how can i close a node children or something to have what i want??
Following code:
List<KeyValuePair<string, string>> columns = new Dictionary<string,string> {
{"name1", "value1"},
{"name2", "value2"},
{"name3", "value3"}
}.ToList();
XElement tree =
new XElement("root",
new XElement("columns",
from c in columns
select new XElement("column",
new XAttribute("name", c.Key),
new XAttribute("value", c.Value)
)
),
new XElement("rows",
from r in rows // same dictionary as columns
select new XElement("row",
new XAttribute("name", r.Key),
new XAttribute("value", r.Value)
)
)
);
Produces required xml:
<root>
<columns>
<column name="name1" value="value1" />
<column name="name2" value="value2" />
<column name="name3" value="value3" />
</columns>
<rows>
<row name="name1" value="value1" />
<row name="name2" value="value2" />
<row name="name3" value="value3" />
</rows>
</root>
XElement tree = new XElement("root",
new XElement("columns",
from c in columns
select new XElement("column", new XAttribute("name", c.Name,), new XAttribute("value", c.Value))),
new XElement("rows",
from r in rows
select new XElement("row", new XAttribute("name", r.Name,), new XAttribute("value", r.Value))));
Related
I am trying to add new attribute to my xml files in c#. my xml file format is shown below:
<Root MessageOfRoot="Welcome to Xml">
<Header Size="36">
<Parameter Name="ID" Index="0" Value="23" />
<Parameter Name="Name" Index="4" Value="Uncle Bob" />
<Parameter Name="Number" Index="8" Value="4" />
</Header>
<Body Size="0">
<Parameter Index="0" UnitNumber="0" Name="UnitBarcode" Type="Integer" />
<Parameter Index="4" PromotionId="0" Name="PromotionalUnit" Type="Integer" />
</Body>
</Root>
I want to add new attribute my xml file which should be like:
<Root MessageOfRoot="Welcome to Xml">
<Header Size="36" NewAttr="1">
<Parameter Name="ID" Index="0" Value="23" NewAttr="1"/>
<Parameter Name="Name" Index="4" Value="Uncle Bob" NewAttr="1"/>
<Parameter Name="Number" Index="8" Value="4" NewAttr="1"/>
</Header>
<Body Size="0" NewAttr="1">
<Parameter Index="0" UnitNumber="0" Name="UnitBarcode" Type="Integer" NewAttr="1"/>
<Parameter Index="4" PromotionId="0" Name="PromotionalUnit" Type="Integer" NewAttr="1"/>
</Body>
</Root>
To do that i write the following code but i am having problem with adding newAttr to all nodes. How can i add NewAttr to my new xml file?
XmlDocument doc = new XmlDocument();
doc.Load("Path of xml");
XmlAttribute NewAttr = doc.CreateAttribute("NewAttr ");
countAttr.Value = "1";
XmlWriter writer = XmlWriter.Create("output.xml", settings);
You can use the following command to load the XML file:
XDocument doc = XDocument.Load(#"C:\Users\myUser\myFile.xml");
Then you can invoke a function that recursively accesses all nodes of the XML starting from the children nodes of the Root element:
AddNewAttribute(doc.Root.Elements());
The function can be like so:
public static void AddNewAttribute(IEnumerable<XElement> elements)
{
foreach (XElement elm in elements)
{
elm.Add(new XAttribute("newAttr", 1));
AddNewAttribute(elm.Elements());
}
}
Finally, you can save the XML back to the original file using:
doc.Save(#"C:\Users\myUser\myFile.xml");
Hi I need to insert some lines into an xml file and save it how I should do it?
the xml file is
<?xml version="1.0" encoding="utf-8"?>
<Dashboard CurrencyCulture="en-US">
<Title Text="Dashboard" />
<DataConnections>
<DataConnection Name="Database1Connection" ProviderKey="Access2007" ConnectionString="XpoProvider=MSAccess;Provider=Microsoft.ACE.OLEDB.12.0;Mode=Share Deny None;data source=D:\Sina\Desktop\Omid\Database1.accdb;Jet OLEDB:Database Password=;">
<Parameters>
<Parameter Name="database" Value="D:\Sina\Desktop\Omid\Database1.accdb" />
<Parameter Name="read only" Value="1" />
<Parameter Name="generateConnectionHelper" Value="false" />
</Parameters>
</DataConnection>
</DataConnections>
<DataSources>
<DataSource Name="Data Source 1">
<DataProvider DataConnection="Database1Connection" SupportSql="true" />
</DataSource>
<DataSource Name="Query 2" />
</DataSources>
and I need to insert these lines
<Selection>
<Table Name="Query2">
<Columns>
<Column Name="PName" />
<Column Name="Prog" />
<Column Name="RDate" />
</Columns>
</Table>
</Selection>
between
<DataProvider DataConnection="Database1Connection" SupportSql="true">
.
.(here)
</DataProvider>
Here is a complete console application that will take the file you provided as input and build a new file with the nodes added.
NOTE: you will clearly have to make the code more dynamic as this is very static. For example, you would build the Column elements with another list and a loop likely.
class Program
{
static void Main(string[] args)
{
var doc = XDocument.Load("XMLFile1.xml");
var selection = new XElement("Selection");
var table = new XElement("Table");
table.Add(new XAttribute("Name", "Query2"));
var columns = new XElement("Columns");
var column = new XElement("Column");
column.Add(new XAttribute("Name", "PName"));
columns.Add(column);
column = new XElement("Column");
column.Add(new XAttribute("Name", "Prog"));
columns.Add(column);
column = new XElement("Column");
column.Add(new XAttribute("Name", "RDate"));
columns.Add(column);
table.Add(columns);
selection.Add(table);
var dataProvider = doc.Root.Descendants("DataProvider").First();
dataProvider.Add(selection);
doc.Save("XMLFile2.xml");
}
}
The output of the new file looks like this:
<DataSources>
<DataSource Name="Data Source 1">
<DataProvider DataConnection="Database1Connection" SupportSql="true">
<Selection>
<Table Name="Query2">
<Columns>
<Column Name="PName" />
<Column Name="Prog" />
<Column Name="RDate" />
</Columns>
</Table>
</Selection>
</DataProvider>
</DataSource>
<DataSource Name="Query 2" />
</DataSources>
Try this,
XmlDocument document = new XmlDocument();
document.Load(filename);
XmlElement childElement = document.CreateElement("child");
XmlNode parentNode = document.SelectSingleNode("root/firstLevel/parent");
parentNode.AppendChild(childElement);
A quick approach would be:
string file = "XMLFile1.xml";
string text = File.ReadAllText(file);
text = text.Replace("<DataProvider DataConnection=\"Database1Connection\" SupportSql=\"true\" />",
"<DataProvider DataConnection=\"Database1Connection\" SupportSql=\"true\">" +
"<Selection>" +
"<Table Name=\"Query2\">" +
"<Columns>" +
" <Column Name=\"PName\" />" +
"<Column Name=\"Prog\" />" +
"<Column Name=\"RDate\" />" +
"</Columns>" +
"</Table>" +
"</Selection>" +
"</DataProvider> ");
File.WriteAllText(file, text);
Assuming having the following xml
<test>
<step>
<var name="name1" />
<var name="name2" />
</step>
<step>
<var name="name3" />
<var name="name4" />
</step>
<step>
<var name="name5" />
<var name="name6" />
</step>
</test>
I m using XmlNodeList, seperated by "step". Is there a way to swap or replace a step directly in the xmlnodelist?
Need to make it like this:
<test>
<step>
<var name="name3" />
<var name="name4" />
</step>
<step>
<var name="name1" />
<var name="name2" />
</step>
<step>
<var name="name5" />
<var name="name6" />
</step>
</test>
You can use XDocument class instead of XMLDocument. This will swap the var nodes name3 with name6.
using System.Linq;
using System.Xml.Linq;
class Test
{
static void Main()
{
XDocument document = XDocument.Load("test.xml");
Swap("name3", "name6", document);
document.Save("test.xml");
}
static void Swap(string nameOne, string nameTwo, XDocument document)
{
var nameOneNode = document.Descendants("var").FirstOrDefault(p => p.Attribute("name").Value == nameOne);
var nameTwoNode = document.Descendants("var").FirstOrDefault(p => p.Attribute("name").Value == nameTwo);
nameOneNode.Attribute("name").Value = nameTwo;
nameTwoNode.Attribute("name").Value = nameOne;
}
}
The order of the nodes in an XML file does not necessarily have to be kept when the XML file is read. For example, if your file looks like this:
<xmlcontent>
<node value="Hello" />
<node value="World" />
</xmlcontent>
The XML read may return the nodes like this:
<xmlcontent>
<node value="World" />
<node value="Hello" />
</xmlcontent>
To apply something like an "order" to XML nodes, you need to add an attribute you can sort by, like
<xmlcontent>
<node index="1" value="Hello" />
<node index="2" value="World" />
</xmlcontent>
In that case, "swapping" two elements would come down to swapping the index values.
Finally managed to do it, here is the code:
XmlDocument xml;
XmlNodeList xmlList;
xml = new XmlDocument();
xml.Load(path);
xmlList = xml.GetElementsByTagName("step");
XmlNode refNode = xmlList[1];
XmlNode newNode = xmlList[0];
xml.DocumentElement.InsertAfter(newNode, refNode);
I try to get some specific values from an xml config. See example below.
<?xml version="1.0" encoding="utf-8"?>
<ExcelConfig>
<ExcelDocument name="Customer" type="flat">
<IdentityColumn>
<Column name="Id" />
</IdentityColumn>
<Validate>
<Column name="Name" mandatory="true" />
<Column name="FirstName" mandatory="true" />
<OrColumns mandatory="true">
<Column name="PostalCode" mandatory="false" />
<Column name="PostalCode2" mandatory="false" />
</OrColumns>
</Validate>
</ExcelDocument>
<ExcelDocument name="Company" type="flat">
<IdentityColumn>
<Column name="Id" />
</IdentityColumn>
<Validate>
<Column name="Name" mandatory="true" />
<Column name="FirstName" mandatory="true" />
<OrColumns mandatory="true">
<Column name="PostalCode" mandatory="false" />
<Column name="PostalCode2" mandatory="false" />
</OrColumns>
</Validate>
</ExcelDocument>
<ExcelDocument name="SomeOtherType" type="block">
<IdentityBlock>
<Column name="Period" col="A" />
<Column name="Period2" col="B" />
</IdentityBlock>
<Validate>
<Column name="Name" mandatory="true" />
<Column name="FirstName" mandatory="true" />
</Validate>
</ExcelDocument>
</ExcelConfig>
I use the following code to get some information from the excel file.
"ValidationConfiguration" is the string with the previous configuration.
//Get Different NodeTypes of Excel documents
List<XPathNavigator> types = XmlHelper.GetNodeTypes(validationConfiguration, "/ExcelConfig/ExcelDocument");
List<XPathNavigator> flatTypes = XmlHelper.GetNodeTypes(validationConfiguration,
"//ExcelConfig/ExcelDocument[#type='flat']");
List<XPathNavigator> blockTypes = XmlHelper.GetNodeTypes(validationConfiguration,
"//ExcelConfig/ExcelDocument[#type='block']");
//First we check if the file is from the flat type and get the IdentityColumns
List<XPathNavigator> identityColumnsNode = XmlHelper.GetNodeTypes(validationConfiguration, "//ExcelConfig/ExcelDocument[#type='flat']/IdentityColumn");
You can find the XmlHelper class below.
public static class XmlHelper
{
public static List<XPathNavigator> GetNodeTypes(string xmlConfiguration,string xPath)
{
XPathDocument doc = new XPathDocument(new StringReader(xmlConfiguration));
XPathNavigator nav = doc.CreateNavigator();
XPathExpression expr = nav.Compile(xPath);
List<XPathNavigator> elements = new List<XPathNavigator>();
foreach (XPathNavigator node in nav.Select(expr))
{
elements.Add(node);
}
return elements;
}
public static List<string> GetIdentityColumnNames(List<XPathNavigator> xPathNavigators)
{
List<string> identityColumns = new List<string>();
foreach (XPathNavigator xPathNavigator in xPathNavigators)
{
foreach (XPathNavigator test in xPathNavigator.Select("//Column"))
{
identityColumns.Add(test.GetAttribute("name", ""));
}
}
return identityColumns;
}
}
Now i want to do the following. I selected the identityColumnsNodes(they contains the IdentityColumn from the exceldocuments that have the flat type).
The i get for al that types the colums. But when i try that, i get all columns back from the whole file. He don't only the items from the node that i use.
foreach (XPathNavigator identityColumNode in identityColumnsNode)
{
List<string> identityColumns = XmlHelper.GetIdentityColumnNames(identityColumnsNode);
}
The second problem/thing i want to do --> the best way to select the right validate node from the specific file. With the identityColumns (that i get back and my list of HeaderRow Cells i know what file it is. But how can i select that validate node?
Or are their better methods to do this stuff?
I'm trying to build a datastructure based on xml content.
The structure looks like that:
Dictionary<string, List<Dictionary<string, string>>>
The XML looks like that:
<Table name="testTable">
<Row>
<Column name="test01" value="2029" />
<Column name="test02" value="2029" />
</Row>
<Row>
<Column name="test01" value="2029" />
<Column name="test02" value="2029" />
</Row>
</Table>
<Table name="testTable01">
<Row>
<Column name="test01" value="2029" />
<Column name="test02" value="2029" />
</Row>
<Row>
<Column name="test01" value="2029" />
<Column name="test02" value="2029" />
</Row>
</Table>
It should result in something like that:
Dictionary<tableName, List<Dictionary<columnName, columnValue>>>
It's no problem to do that with some nested foreach-loops, but I'm searching for a way to do that "in one line" with the LINQ extension methods. How could I do that?
Sounds like you want something like:
var tables = doc.Descendants("Table")
.ToDictionary(t => (string) t.Attribute("name"),
t => ExtractRowsFromTable(t));
...
private static List<Dictionary<string, string>> ExtractRowsFromTable(XElement table)
{
return table.Elements("Row")
.Select(row => row.Elements("Column")
.ToDictionary(c => (string) c.Attribute("name"),
c => (string) c.Attribute("value"))
.ToList();
}
You could do all of this in one line, basically inlining ExtractRowsFromTable - but I really wouldn't.
string xml =
#"<Root>
<Table name=""testTable"">
<Row>
<Column name=""test01"" value=""2029"" />
<Column name=""test02"" value=""2029"" />
</Row>
<Row>
<Column name=""test01"" value=""2029"" />
<Column name=""test02"" value=""2029"" />
</Row>
</Table>
<Table name=""testTable2"">
<Row>
<Column name=""test01"" value=""2029"" />
<Column name=""test02"" value=""2029"" />
</Row>
<Row>
<Column name=""test01"" value=""2029"" />
<Column name=""test02"" value=""2029"" />
</Row>
</Table>
</Root>";
XDocument xDoc = XDocument.Parse(xml);
var table = xDoc
.Descendants("Table")
.Select(t => new
{
Name = t.Attribute("name").Value,
Rows = t.Descendants("Row")
.Select(r=> r.Descendants("Column")
.ToDictionary(c=>c.Attribute("name").Value,
c=>c.Attribute("value").Value))
.ToList()
})
.ToList();