Need to pass variable in a command and assign value to a variable Post: 302999803

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Top Forums UNIX for Beginners Questions & Answers Need to pass variable in a command and assign value to a variable Post 302999803 by bakunin on Wednesday 28th of June 2017 06:34:42 AM

06-28-2017

Registered User

Quote:

Originally Posted by ektubbe

I tried using below code but it did not work -

Code:

param1 = $inputfile1
param2  = $inputfile2
param3 = $inputfile3
header_date_14=$(m_dump /wload/baot/app/data_abinitio/serial/uk_cust/ukrb_ukba_acnt_bde27_src.dml $'$param1' | head -35)
header_date_15=$(m_dump /wload/baot/app/data_abinitio/serial/uk_cust/ukrb_ukba_acnt_bde27_src.dml $'$param2' | head -35)
header_date_16=$(m_dump /wload/baot/app/data_abinitio/serial/uk_cust/ukrb_ukba_acnt_bde27_src.dml $'$param3' | head -35)

First: "did not work" is NOT an analysis - it is like calling the doctor and then tell him "it hurts". It might be somewhat relevant to know what "it" is...

Second: RudiC already told you that spaces around equal signs are a syntax error. You sould have gotten some diagnostic message to that effect, no?

Instead of

Code:

param1 = $inputfile1

Code:

param1=$inputfile1

and better yet, because even this will fail if $inputfile contains white space, do

Code:

param1="$inputfile1"

Third: I don't know what you are trying to do with:

Code:

header_date_14=$(m_dump /some/file  $'$param1' | head -35)

but it is not gonna work. The single quotes are preventing the expansion of the variable $param1 (actually this is what their purpose is) and therefore you don't have the value of $param1 there, but the literal string "$param1". This is most probably not what you wanted. Instead write:

Code:

header_date_14=$(m_dump /some/file  "$param1" | head -35)

Again, the double quotes prevent the command from failing if a filename contains spaces.

I hope this helps.

bakunin

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