Using variables with GREP in Bash script-CodePudding

I've searched high and low for a working answer but here I am, still stuck. I'm new to bash scripting and have spent the past few days trying to achieve my goal, but I'm losing my mind instead.

GOAL: I want to run a script that checks for directories that contains yesterday's date (date appears in between other text in the directory name). Sounds simple!

What I have so far:

DATE=$(date -d '1 day'  %y%m%d)
ls /path/to/folders > ~/listofdirs.txt
GREPDIR=$(grep $DATE ~/listofdirs.txt)
if [ -d /path/to/folders/$GREPDIR ]; then
  echo "Dir exists!"
  echo "(cat $GREPDIR)"
  exit 1
else
  echo "Nothing found."
fi

Grep isn't finding any results as I am sure the $DATE isn't working as I expect. If I substitute $DATE with eg: 2022, I get a result. Thanks for any help, direction, advice.

CodePudding user response：

I suggest with bash:

d=$(date -d '1 day'  %y%m%d)

cd /path/to/folders || exit 1
if grep -q "$d" <<< *$d*; then
  echo "found";
else
  echo "not found";
fi

CodePudding user response：

You may simply use ls -d startsWith* and check if the output is empty as follows:

#!/bin/bash

dirsStartingWith="$(date -d '1 day ago'  %y%m%d)*"

if [[ $(ls -d "$dirsStartingWith" 2>/dev/null) ]]; then
  echo "there are folders starting with $dirsStartingWith"
  #ls -d $dirsStartingWith    # to test output
else
    echo "no folders starting with $dirsStartingWith found"
fi

P.S. You may also use find, but I think ls should be sufficient as date is contained in the name of folders.