mjhjmtu

I Tried these following, but failed. I need a date format in output.

--------1st try--------

#!/bin/bash
echo "read date yyyymmdd"

read $temp

dd=$(date -j -f '%Y%m%d' "$temp" +'%Y%m%d')

echo $dd

Output: invalid option -- 'j'

--------2nd try--------

#!/bin/bash

echo "read date yyyymmdd"

read $temp

dd=$(date -d "$temp" +'%Y%m%d')

echo $dd

Output: it prints today's date.

The main issue is read $temp. To read something into a variable, don't prefix it with $ (this would access the variable's value):

read temp

The second issue is that you don't seem to know what implementation of the date command you are using. GNU date does not have -j flag. This flag is however available with other implementations of date (e.g. on BSD systems, where it causes the utility to parse the given datestamp and do output, but not to set the date).

The third issue is that you seem to want to convert $temp into YYYYMMDD format using date, but you are already asking the user to input the date on this format, so the conversion is a no-op.

A corrected version of your script (written for GNU date, and not doing anything about the third issue as it's unclear what you actually want to do):

#!/bin/bash

read -p 'Enter date (YYYYMMDD): ' indate
date -d "$indate" +'%Y%m%d'

If all you want to do is to get the date into a variable and then output it, then there is no reason to ask the user for the date:

#!/bin/sh

thedate=$( date +'%Y%m%d' )
printf 'The date is %sn' "$thedate"

or, if you don't need to store it in a variable for later,

#!/bin/sh

date +'The date is %Y%m%d'

So you're trying to read some input into the $temp variable, then print it next to todays date?

You don't want the $ in front of the temp variable for the read command:

#!/bin/bash
echo "read date yyyymmdd"
read temp
echo "$temp + $(date '+%Y%m%d')"