XML Sorting used LINQ in ASP.NET MVC

Question

I am trying to get sorting working for certain columns from an xml file. Here is what I have.

    public ActionResult Index(string sortOrder)
    {
        IEnumerable<Personal> personals;
        XmlDocument doc = new XmlDocument();
        List<Personal> persons = new List<Personal>();
        doc.Load("C:\\Users\\Matt.Dodson\\Desktop\\SampleWork\\PersonsApplicationFromXMLFile\\PersonsApplicationFromXMLFile\\DAL\\Personal.xml");
        foreach (XmlNode node in doc.SelectNodes("/Persons/record"))
        {
            persons.Add(new Personal
            {
                ID = node["ID"].InnerText,
                Name = node["Name"].InnerText,
                Email = node["Email"].InnerText,
                DateOfBirth = node["DateOfBirth"].InnerText,
                Gender = node["Gender"].InnerText,
                City = node["City"].InnerText
            });
        }
        switch (sortOrder)
        {
            case "ID":
                personals = persons.OrderBy(Personal => Personal.ID);
                break;
            case "Name":
                personals = persons.OrderBy(Personal => Personal.Name);
                break;
            case "City":
                personals = persons.OrderBy(Personal => Personal.City);
                break;
            default:
                break;
        }
        return View(persons);
    }

Things I've tried include .getElementByTagName("ID").toString().OrderBy(...), to no avail. I'm pretty sure I've got syntax all wrong. Please help and thank you.

personals != persons

Henk Holterman
– Henk Holterman

2018-09-26 21:51:20 +00:00
Commented Sep 26, 2018 at 21:51 — Henk Holterman
– Henk Holterman, Commented Sep 26, 2018 at 21:51

D-Shih · Accepted Answer · 2018-09-26 21:49:18Z

4

I think you need to let the was sorted collection result set to persons otherwise, the collection wasn't ordered by.

switch (sortOrder)
{
    case "ID":
        persons = persons.OrderBy(Personal => Personal.ID).ToList();
        break;
    case "Name":
        persons = persons.OrderBy(Personal => Personal.Name).ToList();
        break;
    case "City":
        persons = persons.OrderBy(Personal => Personal.City).ToList();
        break;
    default:
        break;
}
return View(persons);

I would use linq instead of foreach let the code more clear.

SelectNodes("/Persons/record").Cast<XmlNode>() use Cast<XmlNode>() let XmlNodeList to IEnumerable<XmlNode> collection then use linq select.

public ActionResult Index(string sortOrder)
{
    XmlDocument doc = new XmlDocument();
    doc.Load("C:\\Users\\Matt.Dodson\\Desktop\\SampleWork\\PersonsApplicationFromXMLFile\\PersonsApplicationFromXMLFile\\DAL\\Personal.xml");
    IEnumerable<Personal> persons = doc.SelectNodes("/Persons/record")
        .Cast<XmlNode>()
        .Select(node => new Personal()
        {
            ID = node["ID"].InnerText,
            Name = node["Name"].InnerText,
            Email = node["Email"].InnerText,
            DateOfBirth = node["DateOfBirth"].InnerText,
            Gender = node["Gender"].InnerText,
            City = node["City"].InnerText
        });
    switch (sortOrder)
    {
        case "ID":
            persons = persons.OrderBy(Personal => Personal.ID);
            break;
        case "Name":
            persons = persons.OrderBy(Personal => Personal.Name);
            break;
        case "City":
            persons = persons.OrderBy(Personal => Personal.City);
            break;
        default:
            break;
    }
    return View(persons.ToList());
}

edited Sep 26, 2018 at 21:49

answered Sep 26, 2018 at 21:18

D-Shih

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6 Comments

Mathew Dodson Over a year ago

That just gave me another error, taking it out of context(List)

D-Shih Over a year ago

@MathewDodson you just add .ToList(), because I think your view need to use List<Personal> type

Mathew Dodson Over a year ago

Now it's giving me 'use of unassigned local variable, personals'. Trying everything.

D-Shih Over a year ago

@MathewDodson I edit my answer and add another solution I think is better way.

Henk Holterman Over a year ago

Or just return View(personals)

|

m4a · Accepted Answer · 2018-09-27 18:30:31Z

0

Firstly you need return personals

...
return View(personals);

answered Sep 27, 2018 at 18:30

m4a

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