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get file date as column value
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deesh
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Joined: 08 Oct 2007
Posts: 184

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Post Posted: Mon Apr 10, 2017 9:49 pm Reply with quote    Back to top    
DataStage® Release: 8x
Job Type: Parallel
OS: Unix
Hi,

I have a scenario like source file include date like ABCD10112016.dat . I need to derive that file date value '10112016' for one of the target field.

Could you help how to reach, Moreover will get multiple files same day with different dates.

1job source files:

ABCD09112016.dat
ABCD10112016.dat
ABCD11112016.dat
ABCD12112016.dat
ABCD13112016.dat

2job source files:
EFGH09112016.dat
EFGH10112016.dat
EFGH11112016.dat
EFGH12112016.dat
EFGH13112016.dat
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ray.wurlod

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Location: Sydney, Australia
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Post Posted: Tue Apr 11, 2017 5:13 am Reply with quote    Back to top    
Assuming that the file name is a job parameter, you can manipulate that. For example Code: Convert(Convert("0123456789", "", jpFileName), "", jpFile ...

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deesh
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Post Posted: Wed Apr 19, 2017 11:42 am Reply with quote    Back to top    
Achieved by the file name column property in sequential file stage
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