Free agent SS Dansby Swanson agrees to 7-year, $177 million deal with Chicago Cubs
The last of this offseason's "big four" free agent shortstops has finally found a home.
Dansby Swanson, who helped lead the Atlanta Braves to a World Series title in 2021, has agreed to a 7-year, $177 million deal with the Chicago Cubs, according to a baseball official with direct knowledge of the deal.
The official spoke to USA TODAY Sports on the condition of anonymity because the deal has not been announced.
ESPN's Jeff Passan was the first to report the deal.
Swanson, 28, hit .277 with 25 home runs last season for Atlanta, and set career highs with 99 runs scored, 96 RBI and 18 stolen bases as the Braves won their fifth consecutive NL East division title.
Swanson's deal caps a lucrative offseason for elite shortstops -- with fellow free agents Carlos Correa, Xander Bogaerts and Trea Turner all signing contracts for at least 11 years and $280 million.
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Swanson was the No. 1 overall pick in the 2015 MLB draft by the Arizona Diamondbacks, but was traded to Atlanta later that year in a five-player swap that sent pitcher Shelby Miller to Arizona.
He made his MLB debut in August of 2016 and instantly became a fixture as the team's starting shortstop.
Swanson became an All-Star for the first time in 2022 and also won his first Gold Glove, while becoming one of only two players in the majors to appear in all 162 games.
Contributing: Bob Nightengale