Question

是否有正确/最好/任何方法使用PowerShell 5.0中的类构造制作单例类？我尝试过这样的事情：

class ScannerGeometry
{
    [int] $EEprom;              # read eeprom
    # ....

    ScannerGeometry()
    {
        # ... fetch calibration from instrument
    }

    [ScannerGeometry] Instance() {
        if ($this -eq $null)
        {
            $this = [ScannerGeometry]::new()
        }
        return $this
    }
}

并为其分配如下内容：

$scanner = [ScannerGeometry]::Instance()

唉，我收到Method invocation failed because [ScannerGeometry] does not contain a method named 'Instance'.

的运行时错误

另外，可以在PS5类中使构造函数（或任何其他方法）成为私有的吗？

Answer 1

您的方法是可见的，因为它不是静态的：

[ScannerGeometry]::new() | set x
PS C:\Work> $x | gm


   TypeName: ScannerGeometry

Name        MemberType Definition
----        ---------- ----------
Equals      Method     bool Equals(System.Object obj)
GetHashCode Method     int GetHashCode()
GetType     Method     type GetType()
Instance    Method     ScannerGeometry Instance()
ToString    Method     string ToString()
EEprom      Property   int EEprom {get;set;}

您需要使用关键字static。在这种情况下，$this无效，因此您必须稍微更改代码。这是代码：

class ScannerGeometry
{
    [int] $EEprom;              # read eeprom


    static [ScannerGeometry] Instance() {
          return [ScannerGeometry]::new()
    }
}

修改

好的，这是工作代码：

class ScannerGeometry { [int] $EEprom # read eeprom static [ScannerGeometry] $instance static [ScannerGeometry] GetInstance() { if ([ScannerGeometry]::instance -eq $null) { [ScannerGeometry]::instance = [ScannerGeometry]::new() } return [ScannerGeometry]::instance } } $s = [ScannerGeometry]::GetInstance()

Answer 2

这个怎么样？

class ScannerGeometry
{
    [int] $EEprom;              # read eeprom
    # ....

    ScannerGeometry()
    {
        # ... fetch calibration from instrument
    }

    [ScannerGeometry] Instance() {
        if ($this -eq $null)
        {
            $this = [ScannerGeometry]::new()
        }
        return $this
    }
}

$scangeo = [ScannerGeometry]::new().Instance()

PS：代码只是没有错误，我对单例上下文完全无能为力：（

Source : HSGB

Powershell 5班和单身人士

2 个答案: