Hello All,
I just need to extract the date portion [ * ] from a apache log file
I am able to do it using the chain of command -
Logfile contents - First record -
==========================
197.130.211.240 - - [11/Mar/2009:01:00:00 -0700] "GET /jp/index.shtml HTTP/1.1" 200 24255 "http://www.google.co.jp/search?q=cco&lr=lang_ja&ie=utf-8&oe=
utf-8&aq=t&rls=org.mozilla:ja
fficial&client=firefox-a" "Mozilla/5.0 (Windows; U; Windows NT 5.1; ja; rv:1.9.0.7) Gecko/2009021910 Firefox/3.0.7" "C
TC=202.226.241.234.1211271614048643; SMIDENTITY=8dvPlQzpqd5+avTXCpx7AWCRc3IbGqY7Mm6ECeAu/lwojS/RC4qmMwC+zQCHmNJekImecfJ9e8qJQVgI8Nb7hBe4rNRzPVXPRqUxTh69a
Bp4UA5oRd5SceTwPkA+RLB/ZvSST0Bvxk1dVnmo8R50VJ+IUDqB+N2854r7x/pihcXYObu7HLRFLdAhwVNYaamwWZ8qsTwgUEfUTZLlEGTlqsJuqo+DiO1nZnUmJ1PIgfIpEHeMpzxTiSpFI7FtTWYivH
GE9WeHfJBPV6EcuWONc9pvTfB4hprbm+sW5HxlzwP6wa2RdhpdO5rrLVX8DEP/I4fKYKTu0LYm0MIrVY1/xpNsHa3EkCgQL9F6r1Ns40qry9RRfCywEGhnDTk4jCsPCGF4s7bnwypVmpraunUq/jQLaBm
zLjB/Jr/uPcNR4zpodSuGbbc/8X3INPMx/LeXxAwau79aeBC40u7yGVZXYcyhdLJrsKOqwOcMcXJbOw9aW0SmQJrzCEtnEvc0nysGttJ5PwYdDO0Mrt5CIFs/IfTSKhDUB/Oa123IVoGO15TPHjE4XoFI
Jxk6FmNojzR1I2iSLgGKF4wwR44i9sPqXWwFfTUOUuRknOu614p368dorLg6gjhOdPvyX0apnP4RZA1iGR8f8FWuyDFQfvO3Mg2f QBoTQYIYSXBu372t8HNbvspnFElLuGqtHOyuC13a; USER_PROFIL
E=eNpLTE5OLS72SS1LzbE1tgYAKM8FFg==; __utma=152635051.1371569464997210600.1234262319.1234262319.1234865834.2; __utmz=152635051.1234262319.1.1.utmcsr=(dire
ct)|utmccn=(direct)|utmcmd=(none); s_pers=%20s_nr%3D1236048421913%7C1238640421913%3B%20gpv_p20%3Dtools.cisco.com/search/jsp/search-results.get%7C12367597
50363%3B; s_sess=%20s_ria%3Dflash%252010%257C%3B%20s_cc%3Dtrue%3B%20s_sq%3D%3B" "202.227.241.239"
============================
How i am trying
dt=`head -1 log-date-checker.txt | awk -F "]" '{print $1}' | awk -F "[" '{print $2}' | cut -f1 -d" "`
echo $dt
========
Output
11/Mar/2009:01:00:00
What i need ?-
===============
Any alternate and optimal solution to pick the date portion
Any help appreciate
Thanks
