Grep logs from the last hour

Question

I am trying to grep logs that happened in the last hour, time format is like this: 2020-11-01 18:46:56.036

I tried to run this:

grep "^$(date -d -1 hour +'%Y-%m-%d %H')" postgresql-Mon.log | tail -n 100

but that doesn't work. Also how do I use the tail command without -n? I want to grep from bottom to top but to get all the logs from the last hour not only 100.

Help please :>

I know there are other ways doing it but I need it with grep command

`tail -100` is tail without `n` – armagedescu Nov 02 '20 at 15:03 — armagedescu, Nov 02 '20 at 15:03

Дмитрий Шатов · Answer 1 · 2020-11-02T09:19:56.080

1

grep "$(date "+%Y-%m-%d %H" -d '1 hour ago')"

must work

[root@dell-shatov-dk ~]# cat ./test.log 
2020-11-02 07:21:10.104 t1
2020-11-02 08:21:10.104 t2
2020-11-02 09:21:10.104 t3
2020-11-02 10:21:10.104 t4
2020-11-02 11:21:10.104 t5
[root@dell-shatov-dk ~]# grep "$(date "+%Y-%m-%d %H" -d '1 hour ago')" test.log
2020-11-02 11:21:10.104 t5

edited Nov 02 '20 at 09:19

answered Nov 02 '20 at 08:30

Дмитрий Шатов

129
5

`[root@pg67 log]# cat test.log 2020-11-02 07:21:10.104 2020-11-02 08:21:10.104 2020-11-02 09:21:10.104 2020-11-02 10:21:10.104 You have new mail in /var/spool/mail/root [root@pg67 log]# grep $(date "+%Y-%m-%d %H" -d '1 hour ago') test.log grep: 07: No such file or directory test.log:2020-11-02 07:21:10.104 test.log:2020-11-02 08:21:10.104 test.log:2020-11-02 09:21:10.104 test.log:2020-11-02 10:21:10.104 [root@pg67 log]# date Mon Nov 2 08:55:02 UTC 2020 ` – dddd4 Nov 02 '20 at 08:59
`grep "$(date "+%Y-%m-%d %H" -d '1 hour ago')" test.log`, sorry, missing quotes – Дмитрий Шатов Nov 02 '20 at 09:15
This will fail if the grepped date is not in the file. – kvantour Nov 02 '20 at 09:23

score 0 · Answer 2 · edited Nov 02 '20 at 09:16

0

grep "$(date "+%Y-%m-%d %H" -d '1 hour ago')" test.log

edited Nov 02 '20 at 09:16

RavinderSingh13

130,504
14
57
93

answered Nov 02 '20 at 09:13

user3188140

265
1
5

Grep logs from the last hour

2 Answers2