passing timestamp variable to another script as a paramter

Question

I have script that populates tables with data. I need to record in another table STATS, when it started loading and when it completed the load.

So in the first script I stored timestamp into variable:

**script1.sh**
    dtMODEL=$(date +"%d-%b-%y %H.%M.%S.%N %p")
    ./load_table.sh source1
    ./load_table.sh source2
    ./record_table_load_stats.sh MODEL $dtMODEL

**record_table_load_stats.sh**
   #! /bin/bash
   #
   . /etc/profile.d/oracle.sh

   MODEL=$1
   START_DATE=$2
   echo $MODEL
   echo $START_DATE
   sqlplus -s username/password<< !
   /* this is where I wanna use START_DATE variable and populate table*/

when I do:

echo $dtMODEL

15-Oct-13 13.56.46.677879674 PM

but when I pass it to record_table_load_stats.sh, it echoes

15-Oct-13

why?

I need to pass the entire timestamp: 15-Oct-13 13.56.46.677879674 PM — Angelina
– Angelina, Commented Oct 15, 2013 at 17:55

Lars Brinkhoff · Accepted Answer · 2013-10-15 17:57:07Z

1

The way shell evaluation works, it will first substitute the contents of a variable, and then parse the command line into space-separated arguments. To override this, you have to quote the spaces in the variable contents.

Most convenient is to add " quotes around $dtMODEL, i.e.

./record_table_load_stats.sh MODEL "$dtMODEL"

answered Oct 15, 2013 at 17:57

Lars Brinkhoff

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