Question

我正在添加此type并调用它。它在脚本/功能中工作正常。一旦我试图从powershell类调用它 - 它会给出错误＆＃34;类型未定义＆＃34;。

我设法用一些丑陋的黑客来称呼它。使用Add-Type调用-PassThru，结果将保存在$global:myType中。然后我使用$global:myType.GetMethod('GetSymbolicLinkTarget').invoke($null,$dir)进行间接呼叫。

周围有更好的解决方案吗？

PS：在每次运行之前重置运行空间（在ISE中为Ctrl-T，在PowerGUI和Powershell Studio中为自动）。

PS2：下面的工作样本（简化）。

#works - returns 2
$Source = ' public class BasicTest { public static int Test(int a) { return (a + 1); } } '
Add-Type -TypeDefinition $Source
[BasicTest]::Test(1)


#gives error: Unable to find type [BasicTest]
$Source = ' public class BasicTest { public static int Test(int a) { return (a + 1); } } '
Add-Type -TypeDefinition $Source
class foo { Static [int]bar() { return [BasicTest]::Test(1) } }
[foo]::bar()


#workaround 1 - indirect call with GetMethod(...).Invoke(...)
# returns 4
$Source = ' public class BasicTest1 { public static int Test(int a) { return (a + 1); } } '
$global:BasicTestType1 = (Add-Type -TypeDefinition $Source -PassThru)
class foo {
    static $BasicTestType2 = (Add-Type -TypeDefinition ' public class BasicTest2 { public static int Test(int a) { return (a + 1); } } ' -PassThru)
    Static [int]bar() {
        $ret = $global:BasicTestType1.GetMethod('Test').Invoke($null, [int]1)
        $ret += [foo]::BasicTestType2.GetMethod('Test').Invoke($null, [int]1)
        return $ret
    }
}
[foo]::bar()


#workaround 2 - invoke-expression; has problems passing parameters
# returns 2
$Source = ' public class BasicTest { public static int Test(int a) { return (a + 1); } } '
Add-Type -TypeDefinition $Source
class foo { Static [int]bar() { return invoke-expression '[BasicTest]::Test(1)' } }
[foo]::bar()

PS3：PetSerAl提供了另外两个解决方法here。

Answer 1

问题是在编译时解析了类，并且在运行时之前不会添加对添加的类型的引用。所以，引用无法解决。

我看到的最佳解决方案是强制编译它们的顺序。

# Add the type
$Source = ' public class BasicTest { public static int Test(int a) { return (a + 1); } } '
Add-Type -TypeDefinition $Source

$scriptBlockText = @"
  class foo {
    Static foo() {
    }
   Static [int]bar() { return [BasicTest]::Test(1) } 
  }
"@

# Create a script block from the text
# the class will be compiled at this point
# but is not in scope yet
$scriptBlock = [scriptblock]::Create($scriptBlockText)

# dot source the class
# this brings the class into scope
.([scriptblock]::Create($scriptBlock))

# it would be better if the class was in a separate file
# but I'm using a string to simplify the sample.
# for a file the syntax would be `. .\myClass.ps1`
# this would replace both creating the script block and 
# the dot sourcing statement

# Now you can use the class and the type you addded earlier
[foo]::bar()

Answer 2

另外两种解决方法：

#workaround 3 - type variable
# returns 4
$Source = ' public class BasicTest1 { public static int Test(int a) { return (a + 1); } } '
$global:BasicTestType1 = (Add-Type -TypeDefinition $Source -PassThru)
class foo {
    static $BasicTestType2 = (Add-Type -TypeDefinition ' public class BasicTest2 { public static int Test(int a) { return (a + 1); } } ' -PassThru)
    Static [int]bar() {
        $ret = ($global:BasicTestType1)::Test(1)
        $ret += ([foo]::BasicTestType2)::Test(1)
        return $ret
    }
}
[foo]::bar()


#workaround 4 - most elegant so far
# returns 2
class foo {
    static foo() {
        $Source = ' namespace foo { public class BasicTest1 { public static int Test(int a) { return (a + 1); } } } '
        Add-Type -TypeDefinition $Source 
    }
    Static [int]bar() {
        $ret = ([type]'foo.BasicTest1')::Test(1)
        return $ret
    }
}
[foo]::bar()

解决方法4是最短且最干净的。

也可以使用类型加速器。

如何在Powershell类中正确调用Add-Type添加的类型的静态方法？

2 个答案: