view IPs from apache access.log for date period
Clash Royale CLAN TAG#URR8PPP
I have seen someone use this command -- and it works where it pulls back all the IPS in a list/group:
cat access.* | awk ' print $1 ' | sort | uniq -c | sort -n
However, I want to choose the past 2 days; how can I do that?
logs apache-httpd ip date
edited Jul 13 '16 at 18:10
Jeff Schaller
41.5k1056132
asked Jul 13 '16 at 18:04
WMK83WMK83
307
add a comment |
I have seen someone use this command -- and it works where it pulls back all the IPS in a list/group:
cat access.* | awk ' print $1 ' | sort | uniq -c | sort -n
However, I want to choose the past 2 days; how can I do that?
logs apache-httpd ip date
edited Jul 13 '16 at 18:10
Jeff Schaller
41.5k1056132
asked Jul 13 '16 at 18:04
WMK83WMK83
307
so before issuing the above coomand u need to filter the access log and get the entries for last two days , and pipe the output to the above command , to get the entries in a timespan see this stackoverflow.com/questions/7706095/…
– Ijaz Ahmad Khan
Jul 13 '16 at 22:56
add a comment |
I have seen someone use this command -- and it works where it pulls back all the IPS in a list/group:
cat access.* | awk ' print $1 ' | sort | uniq -c | sort -n
However, I want to choose the past 2 days; how can I do that?
logs apache-httpd ip date
edited Jul 13 '16 at 18:10
Jeff Schaller
41.5k1056132
asked Jul 13 '16 at 18:04
WMK83WMK83
307
I have seen someone use this command -- and it works where it pulls back all the IPS in a list/group:
cat access.* | awk ' print $1 ' | sort | uniq -c | sort -n
However, I want to choose the past 2 days; how can I do that?
logs apache-httpd ip date
logs apache-httpd ip date
edited Jul 13 '16 at 18:10
Jeff Schaller
41.5k1056132
asked Jul 13 '16 at 18:04
WMK83WMK83
307
edited Jul 13 '16 at 18:10
Jeff Schaller
41.5k1056132
asked Jul 13 '16 at 18:04
WMK83WMK83
307
edited Jul 13 '16 at 18:10
Jeff Schaller
41.5k1056132
edited Jul 13 '16 at 18:10
Jeff Schaller
41.5k1056132
edited Jul 13 '16 at 18:10
Jeff Schaller
41.5k1056132
41.5k1056132
asked Jul 13 '16 at 18:04
WMK83WMK83
307
asked Jul 13 '16 at 18:04
WMK83WMK83
307
asked Jul 13 '16 at 18:04
WMK83WMK83
307
307
so before issuing the above coomand u need to filter the access log and get the entries for last two days , and pipe the output to the above command , to get the entries in a timespan see this stackoverflow.com/questions/7706095/…
– Ijaz Ahmad Khan
Jul 13 '16 at 22:56
add a comment |
so before issuing the above coomand u need to filter the access log and get the entries for last two days , and pipe the output to the above command , to get the entries in a timespan see this stackoverflow.com/questions/7706095/…
– Ijaz Ahmad Khan
Jul 13 '16 at 22:56
so before issuing the above coomand u need to filter the access log and get the entries for last two days , and pipe the output to the above command , to get the entries in a timespan see this stackoverflow.com/questions/7706095/…
– Ijaz Ahmad Khan
Jul 13 '16 at 22:56
so before issuing the above coomand u need to filter the access log and get the entries for last two days , and pipe the output to the above command , to get the entries in a timespan see this stackoverflow.com/questions/7706095/…
– Ijaz Ahmad Khan
Jul 13 '16 at 22:56
add a comment |
1 Answer
1
active
oldest
votes
tail -100 access.* | awk ' print $1 ' | sort | uniq -c | sort -n
or
cat access.* | grep "28/Mar/2016" | awk ' print $1 ' | sort | uniq -c | sort -n
edited Jul 15 '16 at 21:17
clk
1,6091821
answered Jul 15 '16 at 21:15
FabianJimenezFabianJimenez
113
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
tail -100 access.* | awk ' print $1 ' | sort | uniq -c | sort -n
or
cat access.* | grep "28/Mar/2016" | awk ' print $1 ' | sort | uniq -c | sort -n
edited Jul 15 '16 at 21:17
clk
1,6091821
answered Jul 15 '16 at 21:15
FabianJimenezFabianJimenez
113
add a comment |
tail -100 access.* | awk ' print $1 ' | sort | uniq -c | sort -n
or
cat access.* | grep "28/Mar/2016" | awk ' print $1 ' | sort | uniq -c | sort -n
edited Jul 15 '16 at 21:17
clk
1,6091821
answered Jul 15 '16 at 21:15
FabianJimenezFabianJimenez
113
add a comment |
tail -100 access.* | awk ' print $1 ' | sort | uniq -c | sort -n
or
cat access.* | grep "28/Mar/2016" | awk ' print $1 ' | sort | uniq -c | sort -n
edited Jul 15 '16 at 21:17
clk
1,6091821
answered Jul 15 '16 at 21:15
FabianJimenezFabianJimenez
113
tail -100 access.* | awk ' print $1 ' | sort | uniq -c | sort -n
or
cat access.* | grep "28/Mar/2016" | awk ' print $1 ' | sort | uniq -c | sort -n
edited Jul 15 '16 at 21:17
clk
1,6091821
answered Jul 15 '16 at 21:15
FabianJimenezFabianJimenez
113
edited Jul 15 '16 at 21:17
clk
1,6091821
edited Jul 15 '16 at 21:17
clk
1,6091821
edited Jul 15 '16 at 21:17
clk
1,6091821
1,6091821
answered Jul 15 '16 at 21:15
FabianJimenezFabianJimenez
113
answered Jul 15 '16 at 21:15
FabianJimenezFabianJimenez
113
answered Jul 15 '16 at 21:15
FabianJimenezFabianJimenez
113
113
add a comment |
add a comment |
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so before issuing the above coomand u need to filter the access log and get the entries for last two days , and pipe the output to the above command , to get the entries in a timespan see this stackoverflow.com/questions/7706095/…
– Ijaz Ahmad Khan
Jul 13 '16 at 22:56