UNIX for Dummies Questions & Answers

Hello All,
How do i overwrite a sourced script variable value.

Sourced Script:

Code:

GEN_PARAM_LIST4=""$LOG_DIR"/dwh_GenerateXMLFile.lst"
GEN_PARAM_LIST4_v2=""$LOG_DIR"/dwh_GenerateXMLFile.v2.lst"

I am using below statement for replacing.
Script2:

Code:

[ ! -z ${UNINUM} ] && "${GEN_PARAM_LIST4}"="${GEN_PARAM_LIST4_v2}"

Error:

Code:

/home/infrmtca/bin/dwh_GenerateXMLFile.sh: line 204: /home/infrmtca/bin/log/dwh_GenerateXMLFile.lst=/home/infrmtca/bin/log/dwh_GenerateXMLFile.v2.lst: No such file or directory

---------- Post updated at 05:11 PM ---------- Previous update was at 05:06 PM ----------

I think i got it but not sure if this is right approach.

Code:

[ ! -z ${UNINUM} ] && GEN_PARAM_LIST4="${GEN_PARAM_LIST4_v2}"

It is a correct approach, but you're missing some quotes.

You don't show us how UNINUM is set. As long as you know that UNINUM will either expand to an empty string or to a single word, what you have will probably work. But, it is poor programming practice to have an unquoted string that sometimes expands to an empty string. If $UNINUM happens to expand to more than one word, the test command:

Code:

[ ! -z ${UNINUM} ]

could yield a syntax error, true, or false depending on the expansion of $UNINUM. The following replacement seems to do what you want and is a lot safer:

Code:

[ ! -z "$UNINUM" ] && GEN_PARAM_LIST4="${GEN_PARAM_LIST4_v2}"