A homegrown talent will be staying with the San Francisco Giants as shortstop Brandon Crawford was inked to a six-year extension deal that will earn him a guaranteed $75 million.
The Associated Press reported that the pact will cover Crawford's two years of arbitration and four years of free agency, with it also having a "full no-trade clause."
A breakdown of what Crawford will be earning with his new contract was also mentioned in the report.
He will get $5.8 million in the first season (2016); $8 million in the second (2017); and $15 million in each of the four remaining seasons in the contract.
Crawford will also be receiving a $1.2 million signing bonus, which will be due this Dec. 15, according to the report.
Last season, the 28-year-old was named an All-Star for the first time in his career and was batting .256 with 21 home runs and 84 runs batted in (RBI) in 143 regular season games played.
Unfortunately, the Giants missed the playoffs last season despite winning the World Series three times in the past five years (2010, 2012, and 2014).
In their last two World Series victories, Crawford was a huge part of the club's success, proving that he is an efficient contributor.
Crawford's hard work was highlighted last season when he was awarded the Golden Glove and Silver Slugger awards.
"Brandon is an exceptionally talented baseball player who has earned this through his hard work, dedication and competitive spirit," Giants general manager Bobby Evans said, according to the Associated Press.
"He took great strides both offensively and defensively last season, winning his first-ever Gold Glove and Silver Slugger award. This is a great day for Brandon and his family, for the Giants and for our fans," he added.
The deal is great for the Giants as they will be keeping a player who has been a tremendous part of their success and is continually improving, just reaching his prime, especially with his performance last season.
