powershell - 构造方法调用

Question

尝试遵循本指南:https://github.com/Readify/Neo4jClient/wiki/cypher-examples#get-all-users-by-label
我需要创建一个lambda表达式，以便将其提供给Return方法。在C#中，它看起来像这样:

.Return(n => n.As<Project>())

在Powershell中，我采用了这种方式(按照@PetSerAl的建议: Return overload fails):

$exp = [System.Linq.Expressions.Expression]
$param = $exp::Parameter([Neo4jClient.Cyper.ICypherResultItem], "n")
$body = $exp::TypeAs($p, (new-object Project).GetType())
$lambda = $exp::Lambda([Func[Project]], $body, $p)

这样，传递给lambda表达式的参数将被键入以接收Neo4j表达式将传递的内容，并且该方法的主体将其转换为 Project(本地定义的类)。现在我可以将其传递给我的方法:

$something.Return($lambda)

但是，我得到这个错误

Exception calling "Return" with "1" argument(s): "The expression must be constructed as either an object initializer (for example: n => new MyResultType { Foo = n.Bar }), an anonymous type initializer (for example: n => new { Foo = n.Bar }), a method call (for example: n => n.Count()), or a member accessor (for example: n => n.As().Bar). You cannot supply blocks of code (for example: n => { var a = n + 1; return a; }) or use constructors with arguments (for example: n => new Foo(n)). If you're in F#, tuples are also supported. Parameter name: expression" At line:1 char:1 + $neo.Cypher.Match("n").Return($return) + ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ + CategoryInfo : NotSpecified: (:) [], MethodInvocationException + FullyQualifiedErrorId : ArgumentException

这清楚表明我没有正确制定lambda表达式的主体。谁能建议应该怎么做？

Answer 1

在C#中，您有:

.Return(n => n.As<Project>())

如果我们将其取出并查看类型， n => n.As<Project>()为:

Expression<Func<ICypherResultItem, Project>>

为了在 Expression Trees中使用 C#来创建该代码，我们最终要做类似的事情:

ParameterExpression parameter = Expression.Parameter(typeof (ICypherResultItem), "n");
MethodCallExpression right = Expression.Call(parameter, typeof (ICypherResultItem).GetMethod("As").MakeGenericMethod(typeof(Project)));
Expression<Func<ICypherResultItem, Project>> expression = Expression.Lambda<Func<ICypherResultItem, Project>>(right, parameter);

因此，将其转换为PowerShell，我认为它类似于:

$exp = [System.Linq.Expressions.Expression]
$param = $exp::Parameter([Neo4jClient.Cypher.ICypherResultItem], "n")
$body = $exp::Call($param, [Neo4jClient.Cypher.ICypherResultItem].GetMethod("As").MakeGenericMethod[Project])
$lambda = $exp::Lambda([Func[ICypherResultItem, Project]], $body, $param)

我不是，而是一个强力的人，我怀疑您将能够更好地翻译 C#，但希望这能使您走上正确的路...

*更新I *
一个小巧的解决方案，使其可以正常工作。声明一个变量以保存 MakeGenericMethod期望的类型数组，并将其传递给:

$PrjType = @((new-object Project).GetType())
$body = $exp::Call($param, [Neo4jClient.Cypher.ICypherResultItem].GetMethod("As").MakeGenericMethod($PrjType))