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Major League Baseball announced its Players of the Week on Monday, and the National League winner was a familiar name.

Dodgers outfielder Mookie Betts won the award for the N.L. after a week that saw him bat .346 with four home runs and seven RBIs. The big game, of course, was Mookie's masterpiece against the Marlins on Friday night, when he hit two home runs and a double, all in clutch situations.

On the season, Betts is now at 31 home runs, one shy of his career high, which came in his American League MVP season in 2018. It's his third season with at least 30 homers, having knocked 31 in 2016 when he finished second behind Mike Trout in the AL MVP voting.

This is Mookie's fourth career Player of the Week Award, but his first with the Dodgers. He won three with the Red Sox, most recently in 2017.

Betts is the third Dodger to win the award this season, joining Freddie Freeman, who won it in July, and Cody Bellinger, who took home the award in April.

Mookie had a scheduled day off on Monday for the series finale in Miami, with the Dodgers not having a break in the schedule over the next week and a half. Manager Dave Roberts said to goal is to "keep that flame going" by giving the red-hot Betts ample rest.

With the Dodgers a lock for a first-round bye in October, individual awards are the only thing going on right now, so congratulations to Mookie on bringing home a little bit of hardware, or a certificate, or whatever it is.