Dansby Swanson reportedly agrees to seven-year deal with Chicago Cubs
CHICAGO — The Chicago Cubs and Dansby Swanson agreed to a seven-year, $177-million contract Saturday, adding the All-Star shortstop to their rebuilding project.
The deal includes a full no-trade clause, according to a person familiar with the negotiations who spoke to the AP on condition of anonymity because the contract was pending a physical.
Swanson was part of a stellar group of free-agent shortstops that also included Trea Turner, Carlos Correa and Xander Bogaerts. He was the last one of the four to decide on a team.
Jason Heyward thought his career might be over after the Cubs released him. The Dodgers are giving him a chance, signing him to minor league contract.
Swanson was selected by Arizona with the No. 1 overall pick in the 2015 amateur draft, but he was traded to Atlanta in a multiplayer deal that December.
With the Braves, Swanson became part of a young core that led the franchise back to the top of the NL East. He batted .248 with 27 homers and 88 RBIs in 2021, helping Atlanta to its first World Series championship since 1995.
The 28-year-old Swanson is coming off perhaps his best big-league season, hitting .277 with 25 homers and a career-best 96 RBIs in 162 games. The shortstop made the NL All-Star team for the first time and also won his first Gold Glove.
A pitch clock, pickoff limit, the changing of shifts on defense and bigger bases could make for a different game in the upcoming season.
While Atlanta managed to sign a couple of its young stars to team-friendly, long-term deals, it was unable to do the same with Swanson. After the team was eliminated by Philadelphia in the NL Division Series, the Braves gave Swanson a $19.65-million qualifying offer. But the Kennesaw, Ga., native declined the proposal.
Swanson, who played college ball at Vanderbilt, made his big league debut with Atlanta in 2016. He is a .255 hitter with 102 homers and 411 RBIs in 827 career games.
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