Returning Failure if File does not exist










2















I am running the below script in SQL Server Agent.



All the subsequent steps in job are dependent upon this step. This is pretty simple Looking for a file and exits with failure if file is not found.



 $file = "C:\Data\FileDrop\.done"
$CheckFile = Test-Path -Path $file

if (!($CheckFile)) exit 1


However when the agent job runs, it says the step failed because file not found and existing with code 0 - success.



What am I doing wrong here?










share|improve this question






















  • I'm not sure what the issue is but if you don't need to store the result of test-path then you can just use if(!t(test-path -path $file)). You can also use if((test-path -path $file) -eq $false))

    – emanresu
    Nov 14 '18 at 21:15












  • Thank you for your response. The issue is, it exits with code=0 success. How do I fail this if file is not found?

    – Lucky
    Nov 14 '18 at 21:22






  • 1





    Don't escape backslashes in PowerShell string literals: has no special meaning (` is PowerShell's escape character), so it doesn't need escaping. That said, in file paths double backslashes are usually benign (treated the same as a single backslash). Is the file the job reports as not found really C:DataFileDrop.done? In your invocation scenario, is any exit code from your PowerShell script reflected in the job?

    – mklement0
    Nov 15 '18 at 3:16
















2















I am running the below script in SQL Server Agent.



All the subsequent steps in job are dependent upon this step. This is pretty simple Looking for a file and exits with failure if file is not found.



 $file = "C:\Data\FileDrop\.done"
$CheckFile = Test-Path -Path $file

if (!($CheckFile)) exit 1


However when the agent job runs, it says the step failed because file not found and existing with code 0 - success.



What am I doing wrong here?










share|improve this question






















  • I'm not sure what the issue is but if you don't need to store the result of test-path then you can just use if(!t(test-path -path $file)). You can also use if((test-path -path $file) -eq $false))

    – emanresu
    Nov 14 '18 at 21:15












  • Thank you for your response. The issue is, it exits with code=0 success. How do I fail this if file is not found?

    – Lucky
    Nov 14 '18 at 21:22






  • 1





    Don't escape backslashes in PowerShell string literals: has no special meaning (` is PowerShell's escape character), so it doesn't need escaping. That said, in file paths double backslashes are usually benign (treated the same as a single backslash). Is the file the job reports as not found really C:DataFileDrop.done? In your invocation scenario, is any exit code from your PowerShell script reflected in the job?

    – mklement0
    Nov 15 '18 at 3:16














2












2








2








I am running the below script in SQL Server Agent.



All the subsequent steps in job are dependent upon this step. This is pretty simple Looking for a file and exits with failure if file is not found.



 $file = "C:\Data\FileDrop\.done"
$CheckFile = Test-Path -Path $file

if (!($CheckFile)) exit 1


However when the agent job runs, it says the step failed because file not found and existing with code 0 - success.



What am I doing wrong here?










share|improve this question














I am running the below script in SQL Server Agent.



All the subsequent steps in job are dependent upon this step. This is pretty simple Looking for a file and exits with failure if file is not found.



 $file = "C:\Data\FileDrop\.done"
$CheckFile = Test-Path -Path $file

if (!($CheckFile)) exit 1


However when the agent job runs, it says the step failed because file not found and existing with code 0 - success.



What am I doing wrong here?







powershell






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 14 '18 at 20:59









LuckyLucky

597




597












  • I'm not sure what the issue is but if you don't need to store the result of test-path then you can just use if(!t(test-path -path $file)). You can also use if((test-path -path $file) -eq $false))

    – emanresu
    Nov 14 '18 at 21:15












  • Thank you for your response. The issue is, it exits with code=0 success. How do I fail this if file is not found?

    – Lucky
    Nov 14 '18 at 21:22






  • 1





    Don't escape backslashes in PowerShell string literals: has no special meaning (` is PowerShell's escape character), so it doesn't need escaping. That said, in file paths double backslashes are usually benign (treated the same as a single backslash). Is the file the job reports as not found really C:DataFileDrop.done? In your invocation scenario, is any exit code from your PowerShell script reflected in the job?

    – mklement0
    Nov 15 '18 at 3:16


















  • I'm not sure what the issue is but if you don't need to store the result of test-path then you can just use if(!t(test-path -path $file)). You can also use if((test-path -path $file) -eq $false))

    – emanresu
    Nov 14 '18 at 21:15












  • Thank you for your response. The issue is, it exits with code=0 success. How do I fail this if file is not found?

    – Lucky
    Nov 14 '18 at 21:22






  • 1





    Don't escape backslashes in PowerShell string literals: has no special meaning (` is PowerShell's escape character), so it doesn't need escaping. That said, in file paths double backslashes are usually benign (treated the same as a single backslash). Is the file the job reports as not found really C:DataFileDrop.done? In your invocation scenario, is any exit code from your PowerShell script reflected in the job?

    – mklement0
    Nov 15 '18 at 3:16

















I'm not sure what the issue is but if you don't need to store the result of test-path then you can just use if(!t(test-path -path $file)). You can also use if((test-path -path $file) -eq $false))

– emanresu
Nov 14 '18 at 21:15






I'm not sure what the issue is but if you don't need to store the result of test-path then you can just use if(!t(test-path -path $file)). You can also use if((test-path -path $file) -eq $false))

– emanresu
Nov 14 '18 at 21:15














Thank you for your response. The issue is, it exits with code=0 success. How do I fail this if file is not found?

– Lucky
Nov 14 '18 at 21:22





Thank you for your response. The issue is, it exits with code=0 success. How do I fail this if file is not found?

– Lucky
Nov 14 '18 at 21:22




1




1





Don't escape backslashes in PowerShell string literals: has no special meaning (` is PowerShell's escape character), so it doesn't need escaping. That said, in file paths double backslashes are usually benign (treated the same as a single backslash). Is the file the job reports as not found really C:DataFileDrop.done? In your invocation scenario, is any exit code from your PowerShell script reflected in the job?

– mklement0
Nov 15 '18 at 3:16






Don't escape backslashes in PowerShell string literals: has no special meaning (` is PowerShell's escape character), so it doesn't need escaping. That said, in file paths double backslashes are usually benign (treated the same as a single backslash). Is the file the job reports as not found really C:DataFileDrop.done? In your invocation scenario, is any exit code from your PowerShell script reflected in the job?

– mklement0
Nov 15 '18 at 3:16













3 Answers
3






active

oldest

votes


















1














I don't think the return value / error code of the job has anything to do with whatever value the script returns.



If you want to fail with an error message, try this:



# make sure to stop on errors
$ErrorActionPreference = 'Stop'
$path = 'C:DataFileDrop.done'
# check for file explicitly (in case a directory with that name exists)
if(![System.IO.File]::Exists($path))
# throwing an exception will abort the job
throw (New-Object System.IO.FileNotFoundException("File not found: $path", $path))

# anything after this will not be executed on error...


This should fail the job entirely and show the error message in the job's history.






share|improve this answer

























  • If the OP doesn't actually care about the setting the exit code and just wants to ensure that the job fails in this condition then I think this is the correct answer.

    – emanresu
    Nov 16 '18 at 17:09


















0














Maybe this is what you're looking for:



if(!(Test-Path "C:DataFileDrop.done"))
return 1

else
return 0






share|improve this answer






























    0














    This is not going to be a great answer because I have not found the documentation that explains this but the use of Exit followed by a number seems to not have any effect on the exit code.



    The following example works, with the general idea taken from a blog.
    First create a test script like the following:



    $host.SetShouldExit(12345)
    exit


    Next call this script from the Powershell command line as in the following example:



    $command = "C:YourTestScriptPathAndName.ps1"
    Powershell -NonInteractive -NoProfile -Command $command


    The script will run and exit, and we can then check the exit code from the Powershell prompt by writing out the contents of $LASTEXISTCODE



    $LASTEXITCODE


    In this case $LASTEXISTCODE would contain 12345.



    In the case of using this with Test-Path it would look like this:



    #using a path that does not exist
    if(Test-Path -Path 'C:nope')
    $host.SetShouldExit(1234)

    else
    $host.SetShouldExit(2222)



    This sets the exit code to 2222 when invoked from the command line, but when I invoke that test script from another script instead of from the command line the $LASTEXITCODE variable is set to 0 still.






    share|improve this answer






















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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1














      I don't think the return value / error code of the job has anything to do with whatever value the script returns.



      If you want to fail with an error message, try this:



      # make sure to stop on errors
      $ErrorActionPreference = 'Stop'
      $path = 'C:DataFileDrop.done'
      # check for file explicitly (in case a directory with that name exists)
      if(![System.IO.File]::Exists($path))
      # throwing an exception will abort the job
      throw (New-Object System.IO.FileNotFoundException("File not found: $path", $path))

      # anything after this will not be executed on error...


      This should fail the job entirely and show the error message in the job's history.






      share|improve this answer

























      • If the OP doesn't actually care about the setting the exit code and just wants to ensure that the job fails in this condition then I think this is the correct answer.

        – emanresu
        Nov 16 '18 at 17:09















      1














      I don't think the return value / error code of the job has anything to do with whatever value the script returns.



      If you want to fail with an error message, try this:



      # make sure to stop on errors
      $ErrorActionPreference = 'Stop'
      $path = 'C:DataFileDrop.done'
      # check for file explicitly (in case a directory with that name exists)
      if(![System.IO.File]::Exists($path))
      # throwing an exception will abort the job
      throw (New-Object System.IO.FileNotFoundException("File not found: $path", $path))

      # anything after this will not be executed on error...


      This should fail the job entirely and show the error message in the job's history.






      share|improve this answer

























      • If the OP doesn't actually care about the setting the exit code and just wants to ensure that the job fails in this condition then I think this is the correct answer.

        – emanresu
        Nov 16 '18 at 17:09













      1












      1








      1







      I don't think the return value / error code of the job has anything to do with whatever value the script returns.



      If you want to fail with an error message, try this:



      # make sure to stop on errors
      $ErrorActionPreference = 'Stop'
      $path = 'C:DataFileDrop.done'
      # check for file explicitly (in case a directory with that name exists)
      if(![System.IO.File]::Exists($path))
      # throwing an exception will abort the job
      throw (New-Object System.IO.FileNotFoundException("File not found: $path", $path))

      # anything after this will not be executed on error...


      This should fail the job entirely and show the error message in the job's history.






      share|improve this answer















      I don't think the return value / error code of the job has anything to do with whatever value the script returns.



      If you want to fail with an error message, try this:



      # make sure to stop on errors
      $ErrorActionPreference = 'Stop'
      $path = 'C:DataFileDrop.done'
      # check for file explicitly (in case a directory with that name exists)
      if(![System.IO.File]::Exists($path))
      # throwing an exception will abort the job
      throw (New-Object System.IO.FileNotFoundException("File not found: $path", $path))

      # anything after this will not be executed on error...


      This should fail the job entirely and show the error message in the job's history.







      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited Nov 15 '18 at 12:36

























      answered Nov 15 '18 at 10:22









      marszemarsze

      5,87132042




      5,87132042












      • If the OP doesn't actually care about the setting the exit code and just wants to ensure that the job fails in this condition then I think this is the correct answer.

        – emanresu
        Nov 16 '18 at 17:09

















      • If the OP doesn't actually care about the setting the exit code and just wants to ensure that the job fails in this condition then I think this is the correct answer.

        – emanresu
        Nov 16 '18 at 17:09
















      If the OP doesn't actually care about the setting the exit code and just wants to ensure that the job fails in this condition then I think this is the correct answer.

      – emanresu
      Nov 16 '18 at 17:09





      If the OP doesn't actually care about the setting the exit code and just wants to ensure that the job fails in this condition then I think this is the correct answer.

      – emanresu
      Nov 16 '18 at 17:09













      0














      Maybe this is what you're looking for:



      if(!(Test-Path "C:DataFileDrop.done"))
      return 1

      else
      return 0






      share|improve this answer



























        0














        Maybe this is what you're looking for:



        if(!(Test-Path "C:DataFileDrop.done"))
        return 1

        else
        return 0






        share|improve this answer

























          0












          0








          0







          Maybe this is what you're looking for:



          if(!(Test-Path "C:DataFileDrop.done"))
          return 1

          else
          return 0






          share|improve this answer













          Maybe this is what you're looking for:



          if(!(Test-Path "C:DataFileDrop.done"))
          return 1

          else
          return 0







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 15 '18 at 9:27









          TobyUTobyU

          2,50421122




          2,50421122





















              0














              This is not going to be a great answer because I have not found the documentation that explains this but the use of Exit followed by a number seems to not have any effect on the exit code.



              The following example works, with the general idea taken from a blog.
              First create a test script like the following:



              $host.SetShouldExit(12345)
              exit


              Next call this script from the Powershell command line as in the following example:



              $command = "C:YourTestScriptPathAndName.ps1"
              Powershell -NonInteractive -NoProfile -Command $command


              The script will run and exit, and we can then check the exit code from the Powershell prompt by writing out the contents of $LASTEXISTCODE



              $LASTEXITCODE


              In this case $LASTEXISTCODE would contain 12345.



              In the case of using this with Test-Path it would look like this:



              #using a path that does not exist
              if(Test-Path -Path 'C:nope')
              $host.SetShouldExit(1234)

              else
              $host.SetShouldExit(2222)



              This sets the exit code to 2222 when invoked from the command line, but when I invoke that test script from another script instead of from the command line the $LASTEXITCODE variable is set to 0 still.






              share|improve this answer



























                0














                This is not going to be a great answer because I have not found the documentation that explains this but the use of Exit followed by a number seems to not have any effect on the exit code.



                The following example works, with the general idea taken from a blog.
                First create a test script like the following:



                $host.SetShouldExit(12345)
                exit


                Next call this script from the Powershell command line as in the following example:



                $command = "C:YourTestScriptPathAndName.ps1"
                Powershell -NonInteractive -NoProfile -Command $command


                The script will run and exit, and we can then check the exit code from the Powershell prompt by writing out the contents of $LASTEXISTCODE



                $LASTEXITCODE


                In this case $LASTEXISTCODE would contain 12345.



                In the case of using this with Test-Path it would look like this:



                #using a path that does not exist
                if(Test-Path -Path 'C:nope')
                $host.SetShouldExit(1234)

                else
                $host.SetShouldExit(2222)



                This sets the exit code to 2222 when invoked from the command line, but when I invoke that test script from another script instead of from the command line the $LASTEXITCODE variable is set to 0 still.






                share|improve this answer

























                  0












                  0








                  0







                  This is not going to be a great answer because I have not found the documentation that explains this but the use of Exit followed by a number seems to not have any effect on the exit code.



                  The following example works, with the general idea taken from a blog.
                  First create a test script like the following:



                  $host.SetShouldExit(12345)
                  exit


                  Next call this script from the Powershell command line as in the following example:



                  $command = "C:YourTestScriptPathAndName.ps1"
                  Powershell -NonInteractive -NoProfile -Command $command


                  The script will run and exit, and we can then check the exit code from the Powershell prompt by writing out the contents of $LASTEXISTCODE



                  $LASTEXITCODE


                  In this case $LASTEXISTCODE would contain 12345.



                  In the case of using this with Test-Path it would look like this:



                  #using a path that does not exist
                  if(Test-Path -Path 'C:nope')
                  $host.SetShouldExit(1234)

                  else
                  $host.SetShouldExit(2222)



                  This sets the exit code to 2222 when invoked from the command line, but when I invoke that test script from another script instead of from the command line the $LASTEXITCODE variable is set to 0 still.






                  share|improve this answer













                  This is not going to be a great answer because I have not found the documentation that explains this but the use of Exit followed by a number seems to not have any effect on the exit code.



                  The following example works, with the general idea taken from a blog.
                  First create a test script like the following:



                  $host.SetShouldExit(12345)
                  exit


                  Next call this script from the Powershell command line as in the following example:



                  $command = "C:YourTestScriptPathAndName.ps1"
                  Powershell -NonInteractive -NoProfile -Command $command


                  The script will run and exit, and we can then check the exit code from the Powershell prompt by writing out the contents of $LASTEXISTCODE



                  $LASTEXITCODE


                  In this case $LASTEXISTCODE would contain 12345.



                  In the case of using this with Test-Path it would look like this:



                  #using a path that does not exist
                  if(Test-Path -Path 'C:nope')
                  $host.SetShouldExit(1234)

                  else
                  $host.SetShouldExit(2222)



                  This sets the exit code to 2222 when invoked from the command line, but when I invoke that test script from another script instead of from the command line the $LASTEXITCODE variable is set to 0 still.







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Nov 16 '18 at 16:16









                  emanresuemanresu

                  40626




                  40626



























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