Powershell

Question

我正在尝试 import/read 散列数组 table（键，值）中的文件中的数据，但无法执行，因为默认分配符号（ =) 不使用，而是使用冒号 (:)。我尝试了不同的方法来读取数据，但 powershell 每次都会给我一个错误。数据（此处截断）采用以下格式（注意等号被替换为：

[{"name":"Alex","IQ":"A+"},{"name":"Steve","IQ":"60"},{"name":"Lisa","IQ":"90"},{"name":"Jackie","IQ":"B"}]

如何将这些数据读入数组或变量，以便我可以使用它？这是错误消息：

> @a = [{"name":"Alex","IQ":"A+"},{"name":"Steve","IQ":"60"},{"name":"Lisa","IQ":"90"},{"name":"Jackie","IQ":"B"}]
At line:1 char:7
+ @a = [{"name":"Alex","IQ":"A+"},{"name":"Steve","IQ":"60"},{"name":"L ...
+       ~
Missing type name after '['.
At line:1 char:14
+ @a = [{"name":"Alex","IQ":"A+"},{"name":"Steve","IQ":"60"},{"name":"L ...
+              ~~~~~~~
Unexpected token ':"Alex"' in expression or statement.
At line:1 char:21
+ @a = [{"name":"Alex","IQ":"A+"},{"name":"Steve","IQ":"60"},{"name":"L ...
+                     ~
Missing argument in parameter list.
At line:1 char:40
+ @a = [{"name":"Alex","IQ":"A+"},{"name":"Steve","IQ":"60"},{"name":"L ...
+                                        ~~~~~~~~
Unexpected token ':"Steve"' in expression or statement.
At line:1 char:48
+ @a = [{"name":"Alex","IQ":"A+"},{"name":"Steve","IQ":"60"},{"name":"L ...
+                                                ~
Missing argument in parameter list.
At line:1 char:67
+ ... ":"Alex","IQ":"A+"},{"name":"Steve","IQ":"60"},{"name":"Lisa","IQ":"9 ...
+                                                           ~~~~~~~
Unexpected token ':"Lisa"' in expression or statement.
At line:1 char:74
+ ... :"Alex","IQ":"A+"},{"name":"Steve","IQ":"60"},{"name":"Lisa","IQ":"90 ...
+                                                                 ~
Missing argument in parameter list.
At line:1 char:93
+ ... "Steve","IQ":"60"},{"name":"Lisa","IQ":"90"},{"name":"Jackie","IQ":"B ...
+                                                         ~~~~~~~~~
Unexpected token ':"Jackie"' in expression or statement.
At line:1 char:102
+ ... Steve","IQ":"60"},{"name":"Lisa","IQ":"90"},{"name":"Jackie","IQ":"B" ...
+                                                                 ~
Missing argument in parameter list.
At line:1 char:1
+ @a = [{"name":"Alex","IQ":"A+"},{"name":"Steve","IQ":"60"},{"name":"L ...
+ ~~
The splatting operator '@' cannot be used to reference variables in an expression. '@a' can be used only as an
 argument to a command. To reference variables in an expression use '$a'.
    + CategoryInfo          : ParserError: (:) [], ParentContainsErrorRecordException
    + FullyQualifiedErrorId : MissingTypename

Answer 1

您的输入似乎是 JSON，因此您需要：

首先使其成为 PowerShell string^[1]，将其包含在 '...' (假设要使用该值 verbatim).
然后使用 ConvertFrom-Json cmdlet 将字符串解析为对象图：

$arrayOfObjects = ConvertFrom-Json '[{"name":"Alex","IQ":"A+"},{"name":"Steve","IQ":"60"},{"name":"Lisa","IQ":"90"},{"name":"Jackie","IQ":"B"}]'

$arrayOfObjects 然后包含代表 JSON 输入的 [pscustomobject] 个实例的数组；如果你输出 $arrayOfObjects，你会看到：

name   IQ
----   --
Alex   A+
Steve  60
Lisa   90
Jackie B

^{[1] 您问题 (=) 中作业的 RHS 以未加引号的 [ 开头，这意味着 PowerShell 在中解析它表达式模式，因此期望 [ 成为类型文字的一部分，例如 [string] - 请参阅概念 about_Parsing帮助主题。

另外，请注意，sigil @ 永远不会在变量 assignments 中使用（其中必须使用 $），仅在 splatting 的上下文中使用；也就是说，使用 $var = ...，从不使用 @var = ...。}

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