如何使用PowerShell将选择查询的输出从DataSet复制到日志文件？

Posted 2021-05-03

我使用下面的代码从SQLServer中的表中选择一个值，代码执行成功并在PowerShell的命令提示符上显示我已分配给变量的输出，但是当我尝试将分配给$MsgBody的输出添加到日志时文件将System.Data.DataRow复制到日志文件中。

如何将输出添加到日志文件任何帮助将不胜感激。

$scriptPath = $PSScriptRoot
$logFilePath = Join-Path $scriptPath "DemoResults.log"

# If log file exists, then clear its contents 
if (Test-Path $logFilePath) {
    Clear-Content -Path $logFilePath
} 

# It displays the date and time of execution of powershell script in log file.
$log = "Date Of Testing: {0} " -f (Get-Date)
$logString = "Process Started." 
Add-Content -Path $logFilePath -Value $log -Force 
Add-Content -Path $logFilePath -Value $logString -Force
$SQLServer = "AP-PON-SRSTEPMSSQLSERVER12" #use ServerInstance for named SQL instances! 
$SQLDBName = "SystemDB"

$SqlQuery2 = "Select * from SystemDB.dbo.Version_Solution WHERE Notes ='9.2.7'"
$sa = "srp"
$pass = "Stayout"

$connectionString = "Data Source=$SQLServer;Initial Catalog=$SQLDBName;User ID=$sa;Password=$pass";

$connection = New-Object System.Data.SqlClient.SqlConnection($connectionString);
$command = New-Object System.Data.SqlClient.SqlCommand($SqlQuery2, $connection);
$connection.Open();
$SqlAdapter = New-Object System.Data.SqlClient.SqlDataAdapter
$SqlAdapter.SelectCommand = $command
$DataSet = New-Object System.Data.DataSet
$SqlAdapter.Fill($DataSet)
$MsgBody = $DataSet.Tables[0] 
#Displays output in Command shell
$MsgBody

$MsgBody | Add-Content $logFilePath
Get-Content $logFilePath

$connection.Close();

虽然我尝试使用它并将输出复制到文件但删除了日志文件中的其他先前输出。

$MsgBody > $logFilePath 

编辑部分： - 使用此>>后

$MsgBody >> $logFilePath 

它以水平方式将输出复制到日志文件中，

S o l u t i o n             :   i n t e l l   C o m p o n e n t           :   S y s t e m D B  M a j o r                   :   9  M i n o r                   :  2 S e r v i c e P a c k       :   1  H o t f i x                 :   0  I n s t a l l e d D a t e   :   1 2 / 1 2 / 2 0 1 7   6 : 5 7 : 4 8   P M  N o t e s                   :   9 . 2 . 1  R o l l U p R e l e a s e   :   0

看起来很难看，我想让它以这种方式垂直复制 -

Solution      : intell
Component     : SystemDB
Major         : 9
Minor         : 2
ServicePack   : 1
Hotfix        : 0
InstalledDate : 12/12/2017 6:57:48 PM
Notes         : 9.2.1
RollUpRelease : 0

答案

虽然我尝试使用它并将输出复制到文件但删除了日志文件中的其他先前输出。

$MsgBody > $logFilePath 

看看Get-Help about_redirection告诉我们：

Operator  Description               Example
--------  ----------------------    ------------------------------
>         Sends output to the       Get-Process > Process.txt
          specified file.

>>        Appends the output to     dir *.ps1 >> Scripts.txt
          the contents of the
          specified file.

[...]

这意味着，只需将>运算符替换为大于>>的双倍运算符即可。

另一答案

您可以尝试以下非常简单的修改：

$MsgBody | Add-Content $logFilePath

至

$MsgBody | Out-File -Path $logFilePath -Append -Force

我认为这就是你要找的东西。

另一答案

你可以试试

$ DataSet.Tables [0] | format-list >> abc.txt

它对我有用

另一答案

我发现上述问题的答案虽然它有点冗长但对我来说效果很好。

$scriptPath = $PSScriptRoot
$logFilePath = Join-Path $scriptPath "DemoResults.log"

# If log file exists, then clear its contents 
if (Test-Path $logFilePath) {
    Clear-Content -Path $logFilePath
} 

# It displays the date and time of execution of powershell script in log file.
$log = "Date Of Testing: {0} " -f (Get-Date)
$logString = "Process Started." 
Add-Content -Path $logFilePath -Value $log -Force 
Add-Content -Path $logFilePath -Value $logString -Force
$SQLServer = "AP-PON-SRSTEPMSSQLSERVER12" #use ServerInstance for named SQL instances! 
$SQLDBName = "SystemDB"

$SqlQuery2 = "Select * from SystemDB.dbo.Version_Solution WHERE Notes ='9.2.1'"
$sa = "srp"
$pass = "Stayout"

$connectionString = "Data Source=$SQLServer;Initial Catalog=$SQLDBName;User ID=$sa;Password=$pass";

$connection = New-Object System.Data.SqlClient.SqlConnection($connectionString);
$command = New-Object System.Data.SqlClient.SqlCommand($SqlQuery2, $connection);
$connection.Open();
$SqlAdapter = New-Object System.Data.SqlClient.SqlDataAdapter
$SqlAdapter.SelectCommand = $command
$DataSet = New-Object System.Data.DataSet
$SqlAdapter.Fill($DataSet)
$MsgBody = $DataSet.Tables[0] 
#Displays output in Command shell
$MsgBody

#Instead of this code, I have used the below code to copy output in Vertical Format in log file. 
<#
$MsgBody | Add-Content $logFilePath
Get-Content $logFilePath
#>

#Code to copy the output of select statement to log file.
    $logString="Version_Solution Table in SystemDB"
    add-content -Path $logFilePath -Value $logString -Force
    add-content -Path $logFilePath -Value "`n" -Force

    $MsgBody = $MsgBody | Select-Object Solution, Component, Major, Minor ,ServicePack,Hotfix,InstalledDate,Notes,RollUpRelease
    $Solution=$MsgBody.Solution
    $Component=$MsgBody.Component
    $Major=$MsgBody.Major
    $Minor=$MsgBody.Minor
    $ServicePack=$MsgBody.ServicePack
    $Hotfix=$MsgBody.Hotfix
    $InstalledDate=$MsgBody.InstalledDate
    $Notes=$MsgBody.Notes
    $RollUpRelease=$MsgBody.RollUpRelease

    add-content -Path $LogFilePath -Value "Solution: $Solution" -Force   
    add-content -Path $LogFilePath -Value "Component: $Component" -Force  
    add-content -Path $LogFilePath -Value "Major: $Major" -Force  
    add-content -Path $LogFilePath -Value "Minor: $Minor" -Force  
    add-content -Path $LogFilePath -Value "ServicePack: $ServicePack" -Force  
    add-content -Path $LogFilePath -Value "Hotfix: $Hotfix" -Force  
    add-content -Path $LogFilePath -Value "InstalledDate: $InstalledDate" -Force  
    add-content -Path $LogFilePath -Value "Notes: $Notes" -Force  
    add-content -Path $LogFilePath -Value "RollUpRelease: $RollUpRelease" -Force  
    add-content -Path $logFilePath -Value "`n" -Force

$connection.Close();