Bonjour tous,

J'ai juste besoin d'en extraire la partie date [*] à partir d'un fichier log d'Apache
Je suis en mesure de le faire en utilisant la chaîne de commandement --

Logfile contenu - First record --
\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d
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 \u003d cco lang_ja & lr \u003d & ie \u003d utf-8 & oe \u003d
utf-8 & aq \u003d t & rls \u003d org.mozilla: jafficial & client \u003d 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 \u003d 202.226.241.234.1211271614048643; SMIDENTITY \u003d 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 \u003d eNpLTE5OLS72SS1LzbE1tgYAKM8FFg \u003d\u003d; __utma \u003d 152635051.1371569464997210600.1234262319.1234262319.1234865834.2; __utmz \u003d 152635051.1234262319.1.1.utmcsr \u003d (dire
ct) | utmccn \u003d (direct) | utmcmd \u003d (none); s_pers \u003d% 20s_nr% 3D1236048421913% 7C1238640421913% 3B% 20gpv_p20% 3Dtools.cisco.com/search/jsp/search-results.get% 7C12367597
50363% 3B% s_sess \u003d 20s_ria% 3Dflash% 252010% 257C% 3B% 20s_cc% 3Dtrue% 3B% 20s_sq% 3D% 3B "" 202.227.241.239 "

\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d
Comment je suis en train

dt \u003d `head -1 log-date-checker.txt | awk-F"] " '(print $ 1)' | awk-F" [ " '(print $ 2)' | cut-f1-d" "`

echo $ dt
\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d
Sortie

11/Mar/2009: 01:00:00


Qu'est-ce que j'ai besoin? --

\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d\u003d
Tout suppléant et la solution optimale pour prendre la date portion

Toute aide apprécier

Merci

echo '198.133.219.248 - - [11/Mar/2009:01:00:00 -0700] "GET /jp/index.shtml HTTP/1.1" 200 24255' | sed 's#[^[]*[[]\([^]][^]]*\).*#\1#'

Merci Vgersh
It works!

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'[^]][^]]*' - À n'importe quel caractère, MAIS ']"Répété 1 fois ou plus

je vous remercie de votre aide
Je vous remercie de votre temps précieux.