Date: Sun, 28 Nov 2004 13:10:01 +0100 From: =?iso-8859-1?Q?Jimmy_M=E4kel=E4_|_Loopia_Webbhotell_AB?= <jimmy.makela@loopia.se> To: <freebsd-questions@freebsd.org> Subject: Timezone conversion Message-ID: <045701c4d543$2ff2c160$c600a8c0@i11>
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Hi I'm trying to convert a date from one timezone into another using the date-command, but I can't seem to get it to work. The problem can be illustrated as below, though in reality I get the date from another source of course, otherwise I wouldn't have to do this. Take a date in some timezone other than your own, in this example I generate the current time in UTC: > TZ=UTC date +"%d %b %Y %H:%M:%S %z" 28 Nov 2004 12:02:18 +0000 Try to use date -j to convert a date of this format into the default display-format: > TZ=CET date -j -f "%d %b %Y %H:%M:%S %z" "28 Nov 2004 12:02:18 +0000" Warning: Ignoring 5 extraneous characters in date string (+0000) Sun Nov 28 12:02:18 CET 2004 The result is wrong of course because the timezone is ignored even though i specify %z in the format-string. The date in CET should be 13:02:18. Now the question; does anyone know why the %z is ignored? Am I missing something, and if so what? The man-page for the -f parameter states "Parsing is done using strptime(3)." and since strptime allows %z I was assuming that the example above should work. I would be really thankful for some help in getting around this. Thanks in advance. Best regards, Jimmy Mäkelä
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