Question

DATE1=`perl -e 'use POSIX qw(strftime); print strftime "%Y-%m-%d",localtime(time()- 3600*72);'`

DATE2=`perl -e 'use POSIX qw(strftime); print strftime "%Y%m%d",localtime(time()- 3600*72);'`

DATE3=`perl -e 'use POSIX qw(strftime); print strftime "%Y%m%d",localtime(time()- 3600*48);'`

DATE4=`perl -e 'use POSIX qw(strftime); print strftime "%Y/%m/%d",localtime(time()- 3600*48);'`

Below is my shell script(test1.sh), in which I need to print four dates and all those four dates should come from above. Meaning I need to pass above those four dates from the command prompt only to the below shell script.

#!/bin/bash

echo Date1
echo Date2
echo Date3
echo Date4

So when I am running the shell script like this- it should get Date1, Date2, Date3, Date4 from the above four dates?

sh -x test1.sh Date1 Date2 Date3 Date4

Is it possible to do in shell script?

Answer 1

If you execute your first code-block in a shell, then you’ll have defined 4 variables with the output of those commands. To access any of those variables, you have to prepend a $ to the variable name – say, $DATE1.

So, for running your script with those parameters, you should run:

 bash -x test1.sh $DATE1 $DATE2 $DATE3 $DATE4

Pay attention to the $ and the case-sensitiveness.

Finally, in your script, arguments are received in a family of variables that are called with the number of order of the parameter. So, your first parameter is retrieved by $1, the second by $2, and so on.

So, your script ends up being:

#!/bin/bash

echo $1
echo $2
echo $3
echo $4

Watch also that you are defining a she-bang that tells the script to be interpreted by bash, but your previous invocation was using sh. They have lots of compatibilities, but they are not the same.