arrays - Powershell - 无法以哈希表数组的形式将数据从文件导入到变量中

Question

我正在尝试从哈希表(键，值)数组中的文件导入/读取数据，但无法执行，因为未使用默认分配符号( = )代替冒号( : ) .我尝试了不同的方法来读取数据，但是 powershell 每次都会给我一个错误。
数据(在此处被截断)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中，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 字符串 [1]，将其包含在 '...' 中(假设要逐字使用该值)。

然后使用 ConvertFrom-Json 将字符串解析为对象图小命令:

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

$arrayOfObjects然后包含一个数组 [pscustomobject]表示 JSON 输入的实例；如果你输出 $arrayOfObjects ， 你会看到的:

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 以表达式模式解析它，因此期望 [成为类型文字的一部分，例如 [string] - 参见概念 about_Parsing帮助主题。
另外，请注意印记 @从不用于变量赋值(必须使用 $)，仅在 splatting 的上下文中使用;也就是说，使用 $var = ... , 从不 @var = ... .