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2019 Jacksonville mayoral election
From Wikipedia, the free encyclopedia
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The 2019 Jacksonville mayoral election was held on March 19, 2019, to elect the mayor of Jacksonville. Incumbent mayor Lenny Curry, a Republican, won a majority of votes to win a second term in office.[2] No Democratic candidate qualified for the mayoral election.[3]
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Candidates
Republican Party
Declared
- Lenny Curry, incumbent mayor[4]
- Anna Brosche, member of Jacksonville City Council[5]
- Jimmy Hill, small business owner and president of International Association of Fire Fighters Local 2622[6]
Democratic Party
While Democratic candidates did declare their candidacy, no Democratic candidates qualified for the mayoral election in 2019.
Declared
Declined
- Alvin Brown, former mayor of Jacksonville[9]
- Garrett Dennis, member of Jacksonville City Council[6][10]
Independents
Declared
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Campaign and results
Democrats did not field a candidate for Mayor of Jacksonville in the 2019 election. Curry faced Anna Lopez Brosche, described by The Florida Times-Union as a moderate Republican. Brosche's campaign received support from some Democratic politicians, including city councilman Garrett Dennis and former Duval County Democratic Party chair Lisa King.[3]
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References
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