Powershell：将文件路径指定为变量 [英] Powershell: Specify file path as variable

问题描述

我通过powershell脚本运行以下SQL查询，并需要针对不同的文件多次运行脚本。所以我想弄清楚的是，当我运行脚本时，如何指定一个文件路径作为一个变量？

 更新[$ Db_name]。[dbo]。[$ BatchTable] 
 set [$ Db_name]。[dbo ] [$ BatchTable] .Wave ='Wave1.1'
 from [$ Db_name]。[dbo]。[$ BatchTable] 
 inner join OPENROWSET（BULK'FilePath\file.csv' 
 FORMATFILE ='E：\import.xml'）作为
 on（[$ Db_name]。[dbo]。[$ BatchTable] .Name = a.Name）和
 [$ Db_name]。[dbo]。[$ BatchTable] .Domain = a.Domain）
  

'FilePath \file.csv'是我需要定义为一个变量的文件路径，使我的代码改为如下：

  inner join OPENROWSET（BULK'$ INPUTFILEPATH'，
 FORMATFILE ='E：\import.xml'）AS a 
  

任何帮助或潜在的更好的方法来完成这将非常有帮助。

从命令像我想要能够像这样运行脚本：

CMD：updatescript。 ps1 $ INPUTFILEPATH = C：\Documents\myfile.csv

同样，我不确定这是最好的办法吗？

解决方案

你就在那里。
您需要在脚本的开头添加一个参数块，例如

  Param（
 [Parameter（Mandatory = $ true）] 
 [ValidateScript（{Test-Path $ _ -PathType'leaf'}）] 
 [string] $ InputFilePath 
）
  

这将创建一个强制的（不是可选的）字符串参数 InputFilePath 并且 ValidateScript 是用于验证参数的代码，在这种情况下，使用 Test-Path cmdlet检查文件是否存在和路径类型（如果检查目录使用'container'）。

下面的语法：

  updatescript.ps1 -INPUTFILEPATHC：\Documents\myfile.csv
  

并且在脚本中使用该变量作为路径完全符合您的问题：

  inner join OPENROWSET（BULK'$ INPUTFILEPATH'，
 FORMATFILE ='E：\import.xml'）AS a 
  
 
 
 
 注意：在PowerShell中运行脚本时使用参数，只需使用最少的字符从您的param块中的所有其他参数中唯一标识该参数 - 在这种情况下 -I 工作原理以及 -InputFilePath 。
 
I am running the following SQL query through a powershell script and need to run the script multiple times against different files. So what I am trying to figure out is how to specify a file path as a variable when I run the script? 
update [$Db_name].[dbo].[$BatchTable]
set [$Db_name].[dbo].[$BatchTable].Wave = 'Wave1.1'
from [$Db_name].[dbo].[$BatchTable]
inner join OPENROWSET(BULK 'FilePath\file.csv',    
FORMATFILE= 'E:\import.xml') AS a
on ([$Db_name].[dbo].[$BatchTable].Name= a.Name) and  
([$Db_name].[dbo].[$BatchTable].Domain = a.Domain)
The 'FilePath\file.csv' is the file path I need to define as a variable so that my code would instead look like this: 
inner join OPENROWSET(BULK '$INPUTFILEPATH',    
FORMATFILE= 'E:\import.xml') AS a
Any help or potentially better methods to accomplish this would help very much. 

From the command like I want to be able to run the script like this: 

CMD: updatescript.ps1 $INPUTFILEPATH = C:\Documents\myfile.csv 

Again, I'm not sure this is the best way to go about this? 
解决方案
You're nearly there.
You will need to add a parameter block at the very start of your script e.g.
Param( 
[Parameter(Mandatory=$true)]
[ValidateScript({Test-Path $_ -PathType 'leaf'})]  
[string] $InputFilePath  
)
This creates a mandatory (not optional) string parameter, called InputFilePath, and the ValidateScript is code used to validate the parameter, in this case checking the file exists using the Test-Path cmdlet and pathtype of leaf (if checking existence of a directory use 'container').

When running your script use the syntax below:
updatescript.ps1 -INPUTFILEPATH "C:\Documents\myfile.csv"
and in the script use the variable as the path exactly as in your question:
inner join OPENROWSET(BULK '$INPUTFILEPATH',    
FORMATFILE= 'E:\import.xml') AS a
NOTE: in powershell when using parameters when running a script you only need to use the least amount of characters that uniquely identify that parameter from all the others in your param block - in this case -I works just as well as -InputFilePath.

                        
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