Enumerable.All<TSource> Método

public:
generic <typename TSource>
[System::Runtime::CompilerServices::Extension]
 static bool All(System::Collections::Generic::IEnumerable<TSource> ^ source, Func<TSource, bool> ^ predicate);

public static bool All<TSource> (this System.Collections.Generic.IEnumerable<TSource> source, Func<TSource,bool> predicate);

static member All : seq<'Source> * Func<'Source, bool> -> bool

<Extension()>
Public Function All(Of TSource) (source As IEnumerable(Of TSource), predicate As Func(Of TSource, Boolean)) As Boolean

Parâmetros de tipo

Parâmetros

Retornos

true se todos os elementos da sequência de origem passarem no teste no predicado especificado ou se a sequência for vazia; caso contrário, false.

Exceções

Exemplos

O exemplo de código a seguir demonstra como usar All para determinar se todos os elementos em uma sequência atendem a uma condição. A variável allStartWithB será verdadeira se todos os nomes de animais de estimação começarem com "B" ou se a pets matriz estiver vazia.

class Pet
{
    public string Name { get; set; }
    public int Age { get; set; }
}

public static void AllEx()
{
    // Create an array of Pets.
    Pet[] pets = { new Pet { Name="Barley", Age=10 },
                   new Pet { Name="Boots", Age=4 },
                   new Pet { Name="Whiskers", Age=6 } };

    // Determine whether all pet names
    // in the array start with 'B'.
    bool allStartWithB = pets.All(pet =>
                                      pet.Name.StartsWith("B"));

    Console.WriteLine(
        "{0} pet names start with 'B'.",
        allStartWithB ? "All" : "Not all");
}

// This code produces the following output:
//
//  Not all pet names start with 'B'.

Structure Pet
    Public Name As String
    Public Age As Integer
End Structure

Sub AllEx()
    ' Create an array of Pets.
    Dim pets() As Pet =
{New Pet With {.Name = "Barley", .Age = 2},
 New Pet With {.Name = "Boots", .Age = 4},
 New Pet With {.Name = "Whiskers", .Age = 7}}

    ' Determine whether all pet names in the array start with "B".
    Dim allNames As Boolean =
pets.All(Function(ByVal pet) pet.Name.StartsWith("B"))

    ' Display the output.
    Dim text As String = IIf(allNames, "All", "Not all")
    Console.WriteLine($"{text} pet names start with 'B'.")
End Sub

' This code produces the following output:
'
' Not all pet names start with 'B'.

O valor booliano que o All método retorna normalmente é usado no predicado de uma where cláusula (Where cláusula no Visual Basic) ou em uma chamada direta para o Where método . O exemplo a seguir demonstra esse uso do All método .

class Pet
{
    public string Name { get; set; }
    public int Age { get; set; }
}
class Person
{
    public string LastName { get; set; }
    public Pet[] Pets { get; set; }
}

public static void AllEx2()
{
    List<Person> people = new List<Person>
        { new Person { LastName = "Haas",
                       Pets = new Pet[] { new Pet { Name="Barley", Age=10 },
                                          new Pet { Name="Boots", Age=14 },
                                          new Pet { Name="Whiskers", Age=6 }}},
          new Person { LastName = "Fakhouri",
                       Pets = new Pet[] { new Pet { Name = "Snowball", Age = 1}}},
          new Person { LastName = "Antebi",
                       Pets = new Pet[] { new Pet { Name = "Belle", Age = 8} }},
          new Person { LastName = "Philips",
                       Pets = new Pet[] { new Pet { Name = "Sweetie", Age = 2},
                                          new Pet { Name = "Rover", Age = 13}} }
        };

    // Determine which people have pets that are all older than 5.
    IEnumerable<string> names = from person in people
                                where person.Pets.All(pet => pet.Age > 5)
                                select person.LastName;

    foreach (string name in names)
    {
        Console.WriteLine(name);
    }

    /* This code produces the following output:
     *
     * Haas
     * Antebi
     */
}

Structure Pet
    Public Name As String
    Public Age As Integer
End Structure

Structure Person
    Public LastName As String
    Public Pets() As Pet
End Structure

Sub AllEx2()
    Dim people As New List(Of Person)(New Person() _
{New Person With {.LastName = "Haas",
                  .Pets = New Pet() {New Pet With {.Name = "Barley", .Age = 10},
                                     New Pet With {.Name = "Boots", .Age = 14},
                                     New Pet With {.Name = "Whiskers", .Age = 6}}},
  New Person With {.LastName = "Fakhouri",
                   .Pets = New Pet() {New Pet With {.Name = "Snowball", .Age = 1}}},
  New Person With {.LastName = "Antebi",
                   .Pets = New Pet() {New Pet With {.Name = "Belle", .Age = 8}}},
  New Person With {.LastName = "Philips",
                   .Pets = New Pet() {New Pet With {.Name = "Sweetie", .Age = 2},
                                      New Pet With {.Name = "Rover", .Age = 13}}}})

    ' Determine which people have pets that are all older than 5.
    Dim names = From person In people
                Where person.Pets.All(Function(pet) pet.Age > 5)
                Select person.LastName

    For Each name As String In names
        Console.WriteLine(name)
    Next

    ' This code produces the following output:
    '
    ' Haas
    ' Antebi

End Sub

Comentários

A enumeração de source é interrompida assim que o resultado pode ser determinado.

Na sintaxe da expressão de consulta do Visual Basic, uma Aggregate Into All() cláusula é convertida em uma invocação de All.

Aplica-se a

Confira também

• Cláusula Aggregate (Visual Basic)