in job of 30 one job is very slow due to this entire job is
very slow how can u know which job is slow?
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cust id,cust quty like 1,101;1,102;1,103 i want output like cust id,cust quty 1 101,102,103 in oracle please write a query in oracle
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How to create a file using vi editor? 2)how to delete a file in vi editor? 3)How to connect the server datastage to unix? what r the command lines we r using? 4)30 jobs r runnig in unix i want to find out my job. how to do this? give me command?
create a job that splits the data in the Jobs.txt file into four output files. You will direct the data to the different output files using constraints. • Job name: JobLevels • Source file: Jobs.txt • Target file 1: LowLevelJobs.txt − min_lvl between 0 and 25 inclusive. − Same column types and headings as Jobs.txt. − Include column names in the first line of the output file. − Job description column should be preceded by the string “Job Title:” and embedded within square brackets. For example, if the job description is “Designer”, the derived value is: “Job Title: [Designer]”. • Target file 2: MidLevelJobs.txt − min_lvl between 26 and 100 inclusive. − Same format and derivations as Target file 1. • Target file 3: HighLevelJobs.txt − min_lvl between 101 and 500 inclusive. − Same format and derivations as Target file 1. • Rejects file: JobRejects.txt − min_lvl is out of range, i.e., below 0 or above 500. − This file has only two columns: job_id and reject_desc. − reject_desc is a variable-length text field, maximum length 100. It should contain a string of the form: “Level out of range: <min_lvl>”, where <min_lvl> is the value in the min_lvl field. My Question is how do you write the stage variable for reject rows.
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