Bash : date -d throws "invalid date" when date is parameterized

Question

When I parameterize the date in the code as :

str_last_log_date="2017-07-24"
last_log_date=$(date -d '${str_last_log_date}' +%s)
threshold_days_ago=$(date -d 'now - 2 days' +%s)
echo "last_log_date ${last_log_date}  thres_days_ago ${threshold_days_ago}"

Gives the error :

date: invalid date ‘${str_last_log_date}’ last_log_date thres_days_ago 1500969455

But if I don't parameterize the date and pass directly, it gives the correct result :

last_log_date=$(date -d '2017-07-24' +%s)
threshold_days_ago=$(date -d 'now - 2 days' +%s)
echo "last_log_date ${last_log_date}  thres_days_ago ${threshold_days_ago}"

last_log_date 1500854400 thres_days_ago 1500969511

Any tips?

score 8 · Answer 1 · answered Jul 27 '17 at 08:10

8

Variables are not expanded when put inside single quotes; use double quotes instead:

date -d "${str_last_log_date}"

answered Jul 27 '17 at 08:10

heemayl

54,820
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score 4 · Accepted Answer · answered Jul 27 '17 at 08:09

4

last_log_date=$(date -d '${str_last_log_date}' +%s)

Should be updated to be (remove single quotes):

last_log_date=$(date -d ${str_last_log_date} +%s)

answered Jul 27 '17 at 08:09

Yaron

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1

better yet, replace the single-quotes with double-quotes, so that it still works if $last_log_date is changed to contain a space (like "2017-07-24 21:16:00"). more re-usable too. – cas Jul 27 '17 at 11:18
2

@cas, or `$IFS` is modified to include `-` or `:` or digits... – Stéphane Chazelas Jul 27 '17 at 11:21
and, of course, the many other reasons why it's a [good idea to always quote your variables](https://unix.stackexchange.com/questions/171346/security-implications-of-forgetting-to-quote-a-variable-in-bash-posix-shells) – cas Jul 27 '17 at 12:28

Bash : date -d throws "invalid date" when date is parameterized

2 Answers2