Serialize multiple objects

Question

My serialization code is like this..

public class slab
{
    public int lowerlimit {get; set;}
    public int upperlimit { get; set; }
    public int percentage { get; set; }

}

public class Details
{
    static void Main(string[] args)
    {
        slab s= new slab();
        s.lowerlimit = 0;
        s.upperlimit = 200000;
        s.percentage = 0;
        XmlSerializer serializer = new XmlSerializer(s.GetType());
        StreamWriter writer = new StreamWriter(@"filepath");
        serializer.Serialize(writer.BaseStream, s);
    }
}

it's working fine and I am getting output file as:

<?xml version="1.0"?>
<slab xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
    <lowerlimit>0</lowerlimit>
    <upperlimit>200000</upperlimit>
    <percentage>0</percentage>
</slab>

But how can I serialize more than one objects? I would like to get an output file as

<slabs>
    <slab>
        <lowerlimit>0</lowerlimit>
        <upperlimit>200000</upperlimit>
        <percentage>0</percentage>
    </slab>
    <slab>
        <lowerlimit>200000</lowerlimit>
        <upperlimit>500000</upperlimit>
        <percentage>10</percentage>
    </slab>
    <slab>
        <lowerlimit>500000</lowerlimit>
        <upperlimit>1000000</upperlimit>
        <percentage>20</percentage>
    </slab>
    <slab>
        <lowerlimit>1000000</lowerlimit>
        <upperlimit>0</upperlimit>
        <percentage>30</percentage>
    </slab>
</slabs>

Put them in an array and serialize the array...

Adriano Repetti
– Adriano Repetti

2013-03-06 12:12:25 +00:00
Commented Mar 6, 2013 at 12:12 — Adriano Repetti
– Adriano Repetti, Commented Mar 6, 2013 at 12:12

igorushi · Accepted Answer · 2016-12-01 11:16:26Z

9

Actually the desired output format is not valid XML, as an XML file always requires a single root element. You could put your slabs into a list (List<Slab> slabs = new List<Slab>();) and serialize that, but you'll probably get output like that:

<slabs>
    <slab>
    <lowerlimit>0</lowerlimit>
    <upperlimit>200000</upperlimit>
    <percentage>0</percentage>
    </slab>

    <slab>
    <lowerlimit>200000</lowerlimit>
    <upperlimit>500000</upperlimit>
    <percentage>10</percentage>
    </slab>

    <slab>
    <lowerlimit>500000</lowerlimit>
    <upperlimit>1000000</upperlimit>
    <percentage>20</percentage>
    </slab>

    <slab>
    <lowerlimit>1000000</lowerlimit>
    <upperlimit>0</upperlimit>
    <percentage>30</percentage>
    </slab>
</slabs>

EDIT
Another way of serializing could be this, telling the serializer more about the root element:

List<Slab> slabs = new List<Slab>();
slabs.Add(...);

XmlSerializer serializer = new XmlSerializer(slabs.GetType(), new XmlRootAttribute("slabs"));
StreamWriter writer = new StreamWriter(@"filepath");
serializer.Serialize(writer.BaseStream, slabs);

edited Dec 1, 2016 at 11:16

igorushi

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answered Mar 6, 2013 at 12:18

Thorsten Dittmar

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3 Comments

Ranadheer Reddy Over a year ago

working fine :-). but i am not getting parent tag as <slabs> </slabs>. i am getting like <arrayOfSlabs></arrayOfSlabs>. how can i change it ??

Ranadheer Reddy Over a year ago

if i write it in the another class my main class(where am adding items to Slabs) is unable to find "Slabs".

Thorsten Dittmar Over a year ago

I changed my answer a bit more so that it is closed to your existing code.

Evan Knowles · Accepted Answer · 2013-03-06 12:36:03Z

2

To encapsulate nicely, and ensure the type name, you could create a new object called slabs, that contains just a List<Slab>. Add the slabs to this new object and serialize it.

answered Mar 6, 2013 at 12:36

Evan Knowles

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Comments

Rajesh Subramanian · Accepted Answer · 2013-03-06 12:17:37Z

1

You can make use of following code.

List<Slab> listSlabs = new List<Slab>();
//add Slab to listSlabs

You can serialize the list.

answered Mar 6, 2013 at 12:17

Rajesh Subramanian

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1 Comment

Ranadheer Reddy Over a year ago

slab s=new slab(); slab s2=new slab(); listSlabs.add(s); listSlabs.add(s2); in this way?

demonplus · Accepted Answer · 2017-11-22 09:33:02Z

0

Use

XmlSerializer serializer = new XmlSerializer(**slabs**.GetType(), new XmlRootAttribute("slabs"));

Instead of

XmlSerializer serializer = new XmlSerializer(s.GetType(), new XmlRootAttribute("slabs"));

edited Nov 22, 2017 at 9:33

demonplus

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answered Jul 24, 2014 at 16:02

upkar

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