Set Bash script variable equal to function with parameters

Question

I have a script that kicks off a database query dependent on dates. Right now, the script defaults to yesterday's date:

function startDate() {
  date --date="yesterday" "+%Y-%m-%d";
}

START= "`startDate`"

What I want to do is pass an argument to the script, so that cron (or whatnot) can have configurable dates. I am unable to get the right syntax for neither the function nor the function call:

function startDate() {
  if [ -z "$1" ]
  then 
    date --date="yesterday" "+%Y-%m-%d";
  else
    "$1"
  fi
}

START= "`startDate \"$1\"`"

$ sh shTest.sh 2014-05-19
shTest.sh: line 6: 2014-05-19: command not found
shTest.sh: line 10: : command not found

What am I missing here?

anubhava · Accepted Answer · 2014-05-20 19:44:36Z

1

Using BASH you can do:

START="$(startDate "$1")"

Also your function needs echo $1:

function startDate() {
  if [ -z "$1" ]
  then 
    date --date="yesterday" "+%Y-%m-%d";
  else
    echo "$1"
  fi
}

answered May 20, 2014 at 19:44

anubhava

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