Enumerable.All<TSource> 方法
定义
重要
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确定序列的所有元素是否满足条件。
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
类型参数
- TSource
的元素 source的类型。
参数
- source
- IEnumerable<TSource>
一个 IEnumerable<T> 包含要向其应用谓词的元素。
返回
true 如果源序列的每个元素都通过指定谓词中的测试,或者序列为空,则为 ;否则,为 false.
例外
source 或 predicate 为 null.
示例
下面的代码示例演示如何用于 All 确定序列中的所有元素是否满足条件。 如果所有宠物名称以“B”开头或数组为空,则pets变量allStartWithB为 true。
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'.
方法返回的布尔值All通常用于子句(Visual Basic 中的子句)where的谓词或对方法的Where直接调用Where。 下面的示例演示了此方法的 All 这种用法。
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
注解
注释
此方法不返回集合的所有元素。 而是确定集合的所有元素是否满足条件。
一旦可以确定结果,就会停止枚举 source 。
在Visual Basic查询表达式语法中,Aggregate Into All() 子句转换为调用 All。