Returning Failure if File does not exist
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
asked Nov 14 '18 at 20:59
LuckyLucky
597
add a comment |
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
asked Nov 14 '18 at 20:59
LuckyLucky
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:`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 reallyC: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
add a comment |
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
asked Nov 14 '18 at 20:59
LuckyLucky
597
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
powershell
asked Nov 14 '18 at 20:59
LuckyLucky
597
asked Nov 14 '18 at 20:59
LuckyLucky
597
asked Nov 14 '18 at 20:59
LuckyLucky
597
asked Nov 14 '18 at 20:59
LuckyLucky
597
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:`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 reallyC: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
add a comment |
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:`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 reallyC: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
add a comment |
3 Answers
3
active
oldest
votes
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.
answered Nov 15 '18 at 10:22
marszemarsze
5,16832041
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
add a comment |
Maybe this is what you're looking for:
if(!(Test-Path "C:DataFileDrop.done")){
return 1
}
else {
return 0
}
answered Nov 15 '18 at 9:27
TobyUTobyU
2,3681921
add a comment |
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.
answered Nov 16 '18 at 16:16
emanresuemanresu
40626
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
answered Nov 15 '18 at 10:22
marszemarsze
5,16832041
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
add a comment |
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.
answered Nov 15 '18 at 10:22
marszemarsze
5,16832041
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
add a comment |
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.
answered Nov 15 '18 at 10:22
marszemarsze
5,16832041
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.
answered Nov 15 '18 at 10:22
marszemarsze
5,16832041
edited Nov 15 '18 at 12:36
answered Nov 15 '18 at 10:22
marszemarsze
5,16832041
answered Nov 15 '18 at 10:22
marszemarsze
5,16832041
answered Nov 15 '18 at 10:22
marszemarsze
5,16832041
5,16832041
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
add a comment |
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
add a comment |
Maybe this is what you're looking for:
if(!(Test-Path "C:DataFileDrop.done")){
return 1
}
else {
return 0
}
answered Nov 15 '18 at 9:27
TobyUTobyU
2,3681921
add a comment |
Maybe this is what you're looking for:
if(!(Test-Path "C:DataFileDrop.done")){
return 1
}
else {
return 0
}
answered Nov 15 '18 at 9:27
TobyUTobyU
2,3681921
add a comment |
Maybe this is what you're looking for:
if(!(Test-Path "C:DataFileDrop.done")){
return 1
}
else {
return 0
}
answered Nov 15 '18 at 9:27
TobyUTobyU
2,3681921
Maybe this is what you're looking for:
if(!(Test-Path "C:DataFileDrop.done")){
return 1
}
else {
return 0
}
answered Nov 15 '18 at 9:27
TobyUTobyU
2,3681921
answered Nov 15 '18 at 9:27
TobyUTobyU
2,3681921
answered Nov 15 '18 at 9:27
TobyUTobyU
2,3681921
answered Nov 15 '18 at 9:27
TobyUTobyU
2,3681921
2,3681921
add a comment |
add a comment |
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.
answered Nov 16 '18 at 16:16
emanresuemanresu
40626
add a comment |
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.
answered Nov 16 '18 at 16:16
emanresuemanresu
40626
add a comment |
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.
answered Nov 16 '18 at 16:16
emanresuemanresu
40626
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.
answered Nov 16 '18 at 16:16
emanresuemanresu
40626
answered Nov 16 '18 at 16:16
emanresuemanresu
40626
answered Nov 16 '18 at 16:16
emanresuemanresu
40626
answered Nov 16 '18 at 16:16
emanresuemanresu
40626
40626
add a comment |
add a comment |
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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:
`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 reallyC: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