vty: show subscriber: change format of 'last LU seen'
So far, the time string format comes from ctime_r, and we manually add "UTC" to it. The ctime_r format is wildly chaotic IMHO, mixing weekday, day-of-month and hour and year in very unsorted ways. Adding "UTC" to it is non-standard. Instead use an ISO-8601 standardized time string via strftime(). Change-Id: I6731968f05050399f4dd43b241290186e0c59e1a
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@ -36,25 +36,21 @@ struct vty;
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#define hexdump_buf(buf) osmo_hexdump_nospc((void*)buf, sizeof(buf))
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static char *
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get_datestr(const time_t *t, char *datebuf)
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static char *get_datestr(const time_t *t, char *buf, size_t bufsize)
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{
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char *p, *s = ctime_r(t, datebuf);
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/* Strip trailing newline. */
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p = strchr(s, '\n');
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if (p)
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*p = '\0';
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return s;
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struct tm tm;
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gmtime_r(t, &tm);
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strftime(buf, bufsize, "%FT%T+00:00", &tm);
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return buf;
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}
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static void dump_last_lu_seen(struct vty *vty, const char *domain_label, time_t last_lu_seen)
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{
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uint32_t age;
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char datebuf[26]; /* for ctime_r(3) */
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char datebuf[32];
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if (!last_lu_seen)
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return;
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vty_out(vty, " last LU seen on %s: %s UTC", domain_label, get_datestr(&last_lu_seen, datebuf));
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vty_out(vty, " last LU seen on %s: %s", domain_label, get_datestr(&last_lu_seen, datebuf, sizeof(datebuf)));
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if (!timestamp_age(&last_lu_seen, &age))
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vty_out(vty, " (invalid timestamp)%s", VTY_NEWLINE);
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else
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