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What's the syntax for searching a generic collection with more than 1 condition?

Hi,
I need to search and see if there is a match in my generic collection with 4 constraints.
myColleciton.Find(x => x.Name == users.GivenName);

How do I add on match requirements for last name, middle name...?
Thank you.

plusone3055

myColleciton.Find("john","a","smith","whatever4thConstrinatismatch");

lapucca

ASKER

I don't have to use that predicate syntax crap like x=>x......
?

plusone3055

not if you don't want to.

your constraintsahoul be in the colletion so either use strings (if you have strings that have the values you need)
or just use double quotes for the values you wan to match)

lapucca

ASKER

Go error when I put in just my 4 constraints, mouse over to see the error message. Error says "No overload method for Find that takes 4 arguments.

plusone3055

for example if you were looking for metadata in a colletion like a keyword

C.KeywordTypes.Find("Account #")

ASKER CERTIFIED SOLUTION

kaufmed

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kaufmed

P.S.

Just keep in mind that the stuff on the left side of the arrow ( => ) is the variable declarations (with types optionally specified), and the stuff on the right is what will actually be executing.

lapucca

ASKER

I tried both with and without parentheses and both got me the same red underline error message.

users.Find(x => x.GivenName == gridEditUser.GivenName && x.Surname == gridEditUser.Surname && x.MiddleName == gridEditUser.MiddleName && x.VoiceTelephoneNumber = gridEditUser.VoiceTelephoneNumber);

users.Find(x => (x.GivenName == gridEditUser.GivenName) && (x.Surname == gridEditUser.Surname) && (x.MiddleName == gridEditUser.MiddleName) && (x.VoiceTelephoneNumber = gridEditUser.VoiceTelephoneNumber));

I'm using VS2012. According to this link, http://stackoverflow.com/questions/12406750/find-item-in-generic-list-by-specifying-multiple-condition, it should work with either. Can you see what's wrong with my code? Thank you.

lapucca

ASKER

Found my typo, I'm missing a "=" at the last constraint. Should be "==".

Thank you.