enumeration

Get Path To “Links” (AKA Favorites) Folder

情到浓时终转凉″ 提交于 2019-11-29 17:34:27
UPDATE: So I just found out in doing all this leg work that it seems the "Links" folder in your user folder can be named ANYTHING and you can still access it by going to "C:\Users(username)\Links" Ex: Rename "C:\Users\(username)\Links" to "C:\Users\(username)\MyNewLinksFolder" Then try to browse to... "C:\Users\(username)\Links" (forehead to palm) There is a "Favorites" node in the Windows 7 file explorer tree: The path to those favorites (by default) is here: I want to be able to get this path via c# code. I was hoping to use the Environment.SpecialFolder Enum but that seems to be a wash! I

Is there a way to iterate through all enum values? [duplicate]

删除回忆录丶 提交于 2019-11-29 16:04:54
问题 Possible Duplicate: C#: How to enumerate an enum? The subject says all. I want to use that to add the values of an enum in a combobox. Thanks vIceBerg 回答1: string[] names = Enum.GetNames (typeof(MyEnum)); Then just populate the dropdown withe the array 回答2: I know others have already answered with a correct answer, however, if you're wanting to use the enumerations in a combo box, you may want to go the extra yard and associate strings to the enum so that you can provide more detail in the

How to recursively Iterate over properties of an Entity

社会主义新天地 提交于 2019-11-29 15:33:58
Suppose this is what I have in my database table Ancestor ( idAncestor int not null, name varchar(20) not null, ) table Descendant ( idDescendant int not null, name varchar(20) not null, Ancestor_idAncestor int not null ) When ADO.NET generates the entity object for the above 2 tables, I can access Descendant of Ancestor through Ancestors.First().Descendants . If I were to recursively iterate over an ancestor's descendant(s) or descendant's descendant(s) and print out its id and name , the following is my attempt public void Iterate(Ancestor a) { Type t = a.GetType(); PropertyInfo[] props = t

Enumeration and mapping with Scala 2.10

|▌冷眼眸甩不掉的悲伤 提交于 2019-11-29 14:30:50
I'm trying to port my application to Scala 2.10.0-M2. I'm seeing some nice improvements with better warnings from compiler. But I also got bunch of errors, all related to me mapping from Enumeration.values . I'll give you a simple example. I'd like to have an enumeration and then pre-create bunch of objects and build a map that uses enumeration values as keys and then some matching objects as values. For example: object Phrase extends Enumeration { type Phrase = Value val PHRASE1 = Value("My phrase 1") val PHRASE2 = Value("My phrase 2") } class Entity(text:String) object Test { val

How solve compiler enum redeclaration conflict

冷暖自知 提交于 2019-11-29 12:37:54
问题 Consider the following C++ enumerations: enum Identity { UNKNOWN = 1, CHECKED = 2, UNCHECKED = 3 }; enum Status { UNKNOWN = 0, PENDING = 1, APPROVED = 2, UNAPPROVED = 3 }; The Compiler conflicted the both UNKNOWN items and threw this error: error: redeclaration of 'UNKNOWN' I am able to solve this error changing one of the UNKNOWN to UNKNOWN_a , but I would like to do not change the names. How can I solve this conflict without changing the enum items name? 回答1: You can use scoped enumerations

C# - AsEnumerable Example

左心房为你撑大大i 提交于 2019-11-29 11:59:38
What is the exact use of AsEnumerable? Will it change non-enumerable collection to enumerable collection?.Please give me a simple example. After reading the answers, i guess you are still missing a practical example. I use this to enable me to use linq on a datatable var mySelect = from table in myDataSet.Tables[0].AsEnumerable() where table["myColumn"].ToString() == "Some text" select table; From the "Remarks" section of the MSDN documentation : The AsEnumerable<TSource> method has no effect other than to change the compile-time type of source from a type that implements IEnumerable<T> to

How do I convert an integer to an enumerated type?

倖福魔咒の 提交于 2019-11-29 10:59:13
问题 I know how to convert an enumerated type to an integer. type TMyType = (mtFirst, mtSecond, mtThird); var ordValue:integer; enumValue:TMyType; ... ordValue:= Ord(mtSecond); // result is 1 But how do I do the inverse operation and convert an integer to an enumerated type? 回答1: As Ken answered, you just cast it. But to make sure you have correct value you can use code like: if (ordValue >= Ord(Low(TMyType))) and (ordValue <= Ord(High(TMyType))) then enunValue := TMyType(ordValue) else raise

Enumeration of combinations of N balls in A boxes?

好久不见. 提交于 2019-11-29 10:26:21
问题 I want to enumerate all possible combinations of N balls in A boxes. example: I have 8 balls to deal in 3 boxes : box_1 box_2 box_3 case-1 8 0 0 case-2 0 8 0 case-3 0 0 8 case-4 7 1 0 case-5 7 0 1 case-6 6 2 0 ... My first problem is that I need A loops to perform this but I want that A and N to be user's inputs. So how to do without writing all possible number of loops users could need? a and N will be value between 2 and ~800, so it will be strongly demanding in computation time so. How to

Entity Framework enumerating SqlQuery result

强颜欢笑 提交于 2019-11-29 09:12:00
I have strange error while I am trying to view results of SqlQuery: var sql = "SELECT @someParam"; var someParamSqlParameter = new SqlParameter("someParam", "Some Value"); var result = _dbContext.SqlQuery<string>(sql, someParamSqlParameter); var containsAnyElements = result.Any(); So when debugger is at last line and when I try to expand Results View of result it shows me expected result("Some Value") but on invoking last line I got an exception "The SqlParameter is already contained by another SqlParameterCollection.". It looks like when I try to open Result View of result it invokes this

What is the difference between for..in and for each..in in javascript?

≯℡__Kan透↙ 提交于 2019-11-29 07:21:26
What is the difference between for..in and for each..in statements in javascript? Are there subtle difference that I don't know of or is it the same and every browser has a different name for it? Brian R. Bondy "for each...in" iterates a specified variable over all values of the specified object's properties. Example: var sum = 0; var obj = {prop1: 5, prop2: 13, prop3: 8}; for each (var item in obj) { sum += item; } print(sum); // prints "26", which is 5+13+8 Source "for...in" iterates a specified variable over all properties of an object, in arbitrary order. Example: function show_props(obj,