Kobe Bryant ties record as longest tenured player with one NBA franchise

When the NBA season officially has its opening tip, Kobe Bryant will break yet another historic league record that may not be rivaled again.

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Drafted by the Los Angeles Lakers in 1996, Kobe Bryant will tie Utah Jazz legend John Stockton as the most tenured player with one franchise in NBA history with 19 consecutive seasons.

It’s just one of many records Bryant has broken and will break (like possibly besting Michael Jordan’s scoring record in Chicago) before he hangs up his sneakers, though it’s very likely that more statistical-based records may eventually be broken by other talent (LeBron James, etc). What may not be broken is how many years he’s spent with the same franchise.

As things currently sit, Bryant ranks second all-time, tied with Indiana Pacers legend Reggie Miller in terms of tenure. He will tie John Stockton of the Utah Jazz come late October, though likely stands to pass him come the 2015-16 season as Bryant believes he has a few good seasons left.

It may not seem like a significant record, but free agency is different than it used to be. More than that, even great talent often finds themselves playing out the twilight of their career on a different roster. Given that, it’s very possible that when Bryant breaks the record come the 2015-16 season, it may stand for quite some time.

Ironically, Tim Duncan of the San Antonio Spurs is just one season behind Kobe Bryant and he himself will move up into a tie with Reggie Miller for second all-time (18) when the season tips.

Dirk Nowitzki of the Dallas Mavericks is the only active player closer to either Bryant or Duncan with 16 straight seasons with one franchise.

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