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?
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.
You only need to reference the global argument variables if passing via perl.
#!/bin/bash
echo "$1" # Date1
echo "$2" # Date2
echo "$3" # Date3
echo "$4" # Date4
else you can open a sub-shell using $() to return the result
> echo $(date "+%Y-%m-%d")
2012-11-20
