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Shell script: How can I populate a shell script variable with result of sql statement

I want to populate a shell script variable with the result of a select count(*) sql statement. At the moment I can select the results and write them to a file but now I just want to get the count and populate a shell script varaiable with the value

i.e. I want the variable 'COUNT' to be populated with the result of 'Select count(*) from Table'

I'm a newcomer to shell scripting to so any help would be greatly appreciated

Thanks

mahome

COUNT=3 or export count=3, if you want call other scripts which want to use it. If you post your exisiting script we can give more hints.

$ COUNT=3
$ echo $COUNT
3

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omarfarid

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GlobexCorp

ASKER

Thanks for the reply

I'm not sure of the username/password for the db, I'm working from someone elses code. I'm already accessing the DB so is there anyway of using that code without the inclusion of the user/pass?

Thanks

omarfarid

I am not aware of any other method to access database without username and password

GlobexCorp

ASKER

At the moment I'm successfully outputting a sql select to a file without using username/password:

'CODE=echo "select max(col) from TestTable where col like 'A%'" | ac_bl -qs -| ~/developers/bin/runparallel -l20000 -n30","' >> Test_File.csv

I'm not using the user/pass here so I figured it was possible to read from the db the same way and instead of writing to a file populate my variable

omarfarid

I think either ac_bl or ~/developers/bin/runparallel has the username and password