Shell Scripts Integer Expression Expected

Question

I am trying to run the following script in bash

#greetings
set `date`
if [ "$4" -lt 12 ]
then 
  echo "Good Morning"
elif [ "$4" -lt 18 ]
  echo "Good Afternoon"
else
  echo "Good evening"
fi
exit

I keep geeting the error ./greetings: line 3: [: 06:19:20: integer expression expected

What should i do?

What do you pass as `$4` when you're running the script? Because `-lt` require standard numbers. I.E. 12, 13 or 12.5. If you pass 12:30, bash can't parse `:`. So it throws that error. — Prav, Apr 08 '18 at 13:41
this may useful https://stackoverflow.com/questions/1842634/parse-date-in-bash — Achal, Apr 08 '18 at 13:54

Maroun · Answer 1 · 2018-04-08T13:59:51.483

1

If you try to echo $4 at the very beginning of the script, you'll get a result of the following format:

%H:%M:%S

For example,

$ date
16:41:22

This result can't be parsed as integer.

One solution would be setting only the hours instead of the whole date, and use $1 (instead of $4):

# greetings
set `date +%H`
if [ "$1" -lt 12 ]
then
  echo "Good Morning"
elif [ "$1" -lt 18 ]
then
  echo "Good Afternoon"
else
  echo "Good evening"
fi
exit

Also, please note that you're missing then keyword after elif.

edited Apr 08 '18 at 13:59

answered Apr 08 '18 at 13:53

Maroun

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The command set `date` sets the fourth parameter to the value of the time – Yousif Dafalla Apr 08 '18 at 13:54
@YousifShibeika Updated answer. – Maroun Apr 08 '18 at 13:59

score 0 · Answer 2 · answered Apr 08 '18 at 13:57

0

TIME=`date +%H:%M:%S | awk -F: '{ print ($1 * 3600) + ($2 * 60) + $3 }'`

if [ "$TIME" -lt 43200 ]; then 
  echo "Good Morning"
elif [ "$TIME" -lt 64800 ]; then
  echo "Good Afternoon"
else
  echo "Good evening"
fi
exit

All times are converted to seconds

answered Apr 08 '18 at 13:57

Prav

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@Maroun You're absolutely right. But his question had `H:m:s` format and I wasn't sure I can do that with `date` approach. – Prav Apr 08 '18 at 14:03

Shell Scripts Integer Expression Expected

2 Answers2