Getting the previous and next record from list using linq

Generic lists, you just gotta love them, especially if using linq! Well to the point. Lets say you got a list with Guid's of which you know one and based on that knowledge want to get either the one before or after.

Guid a = new Guid("709DBFA9-88B3-4518-A32F-C21B6AAE2917");
Guid b = new Guid("B9956FC6-FD9B-4F52-B96E-C1D0C29E1E55");
Guid c = new Guid("BFB9054A-AB18-4349-BCB4-19979D3486BB"); // <-- you know this one, the c guid
Guid d = new Guid("1A8D6CAB-FD42-4517-A0EA-08D3F64DCD69");
Guid e = new Guid("02A8CA8A-5BC8-41BE-8AB8-332EF0B5407B");
List<Guid> guidList = new List<Guid> {a, b, c, d, e};

Solution

Oh wow, now I need to start building indexes pushing and popping yada yada. Well no, a better way to solve this is by using linq. Getting the record after the one you search for

private static Guid GetNext(IEnumerable<Guid> guidList, Guid current)
{
    return guidList.SkipWhile(i => !x.Equals(current)).Skip(1).First();
}

And to get the record before...

private static Guid GetPrevious(IEnumerable<Guid> guidList, Guid current)
{
    return guidList.TakeWhile(i => !x.Equals(current)).Last();
}

Using Generics (you should)

In example

private static T GetNext<T>(IEnumerable<T> list, T current)
{
    try
    {
        return list.SkipWhile(x => !x.Equals(current)).Skip(1).First();
    }
    catch
    {
        return default(T);
    }
}

private static T GetPrevious<T>(IEnumerable<T> list, T current)
{
    try
    {
        return list.TakeWhile(x => !x.Equals(current)).Last();
    }
    catch
    {
        return default(T);
    }
}

Some usage

String example

string f = "first";
string g = "second";
string h = "third";
string i = "fourth";
string j = "fifth";
List<string> stringList = new List<string>{f,g,h,i,j};

var prev = GetPrevious<string>(stringList, j);
var next = GetNext<string>(stringList, j);

Guid example

Guid a = new Guid("709DBFA9-88B3-4518-A32F-C21B6AAE2917");
Guid b = new Guid("B9956FC6-FD9B-4F52-B96E-C1D0C29E1E55");
Guid c = new Guid("BFB9054A-AB18-4349-BCB4-19979D3486BB");
Guid d = new Guid("1A8D6CAB-FD42-4517-A0EA-08D3F64DCD69");
Guid e = new Guid("02A8CA8A-5BC8-41BE-8AB8-332EF0B5407B");
List<Guid> guidList = new List<Guid> {a, b, c, d, e};

var prev = GetPrevious<Guid>(guidList, c);
var next = GetNext<Guid>(guidList, c);