write an xml and read with a condition

Question

0.00/5 (No votes)

See more:

Hai,

I would like to create an xml file with the following structure

XML

<?xml version="1.0"?>
<Category>Mobile Phones
  <Subcategory>Samsung Phones
  <Brand>Samsung
    <Product>
      <title>SAMSUNG B7722I WHITE DUALSIM SAMSUNG B7722I WHITE DUALSIM
      </title>
      <Network>GSM 850 / 900 / 1800 / 1900 - SIM 1 </Network>
  <Dimensions>113.5 x 55.5 x 14.3 mm Greutate : 112 g Weight: 112 g  </Dimensions>
    <Camera>5 MP, 2560x1920 pixels, LED flash </Camera>
    <Radio>Stereo FM radio with RDS, FM recording </Radio><Standby>Up to 420 h </Standby>
    <TalkTime>Up to 12 h 50 min</TalkTime>
    </Product>
    <Product>
      <title>
        SAMSUNG GALAXY S WIFI 4.0 WHITE
      </title>
      <Network>N / A  </Network>
      <Weight> </Weight>
      <Camera>3.15 MP, 2048x1536 pixels, autofocus, LED flash  </Camera>
      <Radio>Stereo FM radio with RDS  </Radio>

    </Product>
  </Brand>

  </Subcategory>
  <Subcategory>
    NOKIA Phones

    <Brand>
     Nokia
      <Product>
        <title>
          NOKIA N8 PINK
        </title>
        <Network>GSM 850 / 900 / 1800 / 1900  </Network>
        <Dimensions>113.5 x 59.1 x 12.9 mm, 86 cc </Dimensions>
        <Camera>5 MP, 2560x1920 pixels, LED flash </Camera>
        <Standby>Up to 420 h </Standby>
        <TalkTime>Up to 12 h 50 min</TalkTime>
        <weight>135 g</weight>
        <Rooms>12 MP, 4000x3000 pixels, Carl Zeiss optics, autofocus, Xenon flash </Rooms>
        <Radio>Stereo FM radio with RDS, FM transmitter </Radio>
      </Product> </Brand>
  </Subcategory>
</Category>

I would like to show the xml file like the following link with a specific condition like subcategory= nokia phone or samsung phone

http://translate.google.com/translate?hl=en&sl=ro&tl=en&u=http%3A%2F%2Fwww.quickmobile.ro%2Ftelefoane-nokia[^]

How its possible
pls help

Thank you,
soumya

Posted 2-Jun-11 0:58am

soumyaraj

Add a Solution

Comments

Sergey Alexandrovich Kryukov 2-Jun-11 14:26pm

Is the XML schema (meta-data) already pre-defined or you can modify it? What you show is not a schema but just one sample. I'm asking because there is much better approach. What if you leave the data structure the same but change the form of XML?
--SA

3 solutions

Solution 1

You can use LINQ to XML[^] to create/modify and delete elements in XML.

With LINQ to XML you can easy create XML like this:

C#

XElement xmlTree1 = new XElement("Root",
    new XElement("Child1", 1),
    new XElement("Child2", 2),
    new XElement("Child3", 3),
    new XElement("Child4", 4),
    new XElement("Child5", 5),
    new XElement("Child6", 6)
);

And output is:

XML

<Root>
  <Child3>3</Child3>
  <Child4>4</Child4>
  <Child5>5</Child5>
</Root>

Hope it can help you.

Posted 4-Jun-11 23:41pm

Kim Togo

Updated 5-Jun-11 15:17pm

Chris Maunder773.5K

v2

Comments

Wonde Tadesse 5-Jun-11 16:37pm

5+

Kim Togo 6-Jun-11 2:47am

Thanks

Solution 3

I would also suggest XSLT as a way of rendering XML conditionally, and converting it to be more human readable. You don't say, but I assume you don't want to show raw XML ?

Posted 5-Jun-11 15:21pm

Christian Graus

Comments

Wonde Tadesse 6-Jun-11 7:27am

Good suggestion. 5+

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Wonde Tadesse · Accepted Answer · 2011-06-05T10:37:00

Solution 2

As Kim said LINQ to XML[^] is good. But if you want to work with OO way, then XML Serialization[^] is way to go.

I hope this helps you well.

Posted 5-Jun-11 10:37am

Wonde Tadesse