Compare value to array of strings using StartsWith

Question

I have an array:

string[] exceptions = new string[] { "one", two", "one_1", "three" };

.. I want to be able to say:

var result = from c in myCollection
             where not c.Property[3].Value.StartWith(exceptions)
             select c;

So I want myCollection to be filtered to only show those records whose Property[3].Value does not StartWith a value in the exceptions array. I know StartsWith doesn't take a collection so I'm unsure if this is possible via LINQ or not.

Is this possible in LINQ?! Or am I trying to shoehorn my problem into a LINQ solution?

EDIT: I should say, Contains is not an option since I only want to exclude elements whose property startswith the exception string.

Solution 1

var result =  myCollection.Where(c =>  
                           exceptions.All(e => 
                                       !c.Property[3].Value.StartsWith(e));

Solution 2

You could use IndexOfAny (and check result is index position zero) as that takes a collection.

Solution 3

Try this:

string[] exceptions = new string[] { "one", "two", "one_1", "three" };

var result = from c in myCollection
            where !exceptions.Any(exception =>
                c.Property[3].Value.StartsWith(exception))
            select c;

Comments

[pierre]

This one worked for me first time and was a one-liner. Thank you.

[sll]

The query returns collections with property does not start with ANY exception string, !Any() return true even a property does not start single exception but we need to ensure this for all exceptions, I believe there is some confusion with word ANY

[sll]

@pierre : nice to hear, I would suggest covering such logic by unit tests looking forward

[Enigmativity]

With All it won't ever return any results. It should be Any.

[sll]

looks like there is some confusion, we have to ask OP what is work for him

[sll]

From my point of view you guys right so it should be Any(), I will update my answer when @pierre confirm this since answer already acepted

[sll]

@pierre: can you confirm that All() won't work and Any() gives you right results so I will update my answer

[sll]

I've found what I've missed, NOT should be before StartsWith() not before the (All),

[Enigmativity]

The logic of "All Not Starts With" is the same as "Not Any Starts With". Quite different from "Not All Starts With". Well done on fixing the answer.