asked on

convert XML to Json C#

Hi Team,

I need to convert following XML results to JSON.also i have used RestSharp
Results

<?xml version="1.0"?>
<table>
 <row id="18" employeeId="20">
  <field id="date">2017-05-15</field>
  <field id="location">test</field>
  <field id="department">Production (Direct)</field>
  <field id="division">abc</field>
  <field id="jobTitle">Stores Material Handler</field>
  <field id="reportsTo">abc cde</field>
 </row>

</table>

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I have used following way but it's not working

 public class table
    {
        public row[] row { get; set; }

    }
    public class row
    {
        public string id { get; set; }
        public string employeeId { get; set; }

    }
    public class field
    {
        public string id { get; set; }
        public string Value { get; set; }

    }

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Rikin Shah

Hi,

Try this code, generated from online tool - https://app.quicktype.io

namespace QuickType
{
    using System;
    using System.Collections.Generic;

    using System.Globalization;
    using Newtonsoft.Json;
    using Newtonsoft.Json.Converters;

    public partial class Temperatures
    {
        [JsonProperty("row", NullValueHandling = NullValueHandling.Ignore)]
        public Row Row { get; set; }
    }

    public partial class Row
    {
        [JsonProperty("@id", NullValueHandling = NullValueHandling.Ignore)]
        public string Id { get; set; }

        [JsonProperty("@employeeId", NullValueHandling = NullValueHandling.Ignore)]
        public string EmployeeId { get; set; }

        [JsonProperty("field", NullValueHandling = NullValueHandling.Ignore)]
        public List<Field> Field { get; set; }
    }

    public partial class Field
    {
        [JsonProperty("@id", NullValueHandling = NullValueHandling.Ignore)]
        public string Id { get; set; }

        [JsonProperty("#text", NullValueHandling = NullValueHandling.Ignore)]
        public string Text { get; set; }
    }
}

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ukerandi

ASKER

@Rikin Shah
It's not working only EmployeeId and Id will get data.
But after calling "Row " class
[code]
public class Row
[/code]
But this part not working
[code]
public partial class Field
[/code]
I have tried different ways like changing
public List<Field> Field { get; set; } to public Field[] Field { get; set; }

but not working any idea

ASKER CERTIFIED SOLUTION

louisfr

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ukerandi

ASKER

@louisfr
it's looks very close to answer. But problem is it's showing one one records
I think if we can add to Table List it will work not sure how to change this coding to List
[code]
public static List<string> Xml2Json(string xml) { var x = new XmlSerializer(typeof(Table)); Table table; using (var r = new StringReader(xml)) { table = (Table)x.Deserialize(r); } var j = new JsonSerializer(); using (var w = new StringWriter()) { j.Serialize(w, table); return w.ToString(); } }

[/code]

request details like that(XML)

<?xml version="1.0"?>
<table>
 <row id="18" employeeId="20">
  <field id="date">2017-05-15</field>
  <field id="location">test</field>
  <field id="department">Production (Direct)</field>
  <field id="division">abc</field>
  <field id="jobTitle">Stores Material Handler</field>
  <field id="reportsTo">abc cde</field>
 </row>
 <row id="19" employeeId="20">
  <field id="date">2017-06-15</field>
  <field id="location">test</field>
  <field id="department">Production (Direct)</field>
  <field id="division">abc</field>
  <field id="jobTitle">Stores Material Handler</field>
  <field id="reportsTo">abc cdettt</field>
 </row>
 <row id="20" employeeId="20">
  <field id="date">2017-07-15</field>
  <field id="location">test</field>
  <field id="department">Production (Direct)</field>
  <field id="division">abc</field>
  <field id="jobTitle">Stores Material Handler</field>
  <field id="reportsTo">abc cde111</field>
 </row>
</table>

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