Lineker Expected To Draw Sandhagen For UFC’s Debut On ESPN+

UFC’s debut on ESPN+ next month may be getting another tantalizing matchup, as surging bantamweight contender John Lineker is expected to take on rising prospect Cory Sandhagen, per a recent report by ESPN’s Ariel Helwani.
Lineker was expe…

UFC’s debut on ESPN+ next month may be getting another tantalizing matchup, as surging bantamweight contender John Lineker is expected to take on rising prospect Cory Sandhagen, per a recent report by ESPN’s Ariel Helwani.

Lineker was expected to challenge former UFC bantamweight champion Dominick Cruz at UFC 233 on Jan. 26, but Cruz was forced out of that bout with yet another injury.

Lineker, 28, is coming off a smashing knockout victory over hard-nosed veteran Brian Kelleher this past May. The finish helped push Lineker’s bantamweight record to 6-1 since making his divisional return over three years ago, losing to only T.J. Dillashaw along the way. As one of the most exciting fighters at 135 pounds, UFC is hoping that Lineker can parlay his recent efforts into a title shot sometime next year. The heavy-handed Brazilian will now have to get past one of the most dangerous prospects in the division to get there.

Sandhagen, 26, burst onto the UFC scene earlier this year with a memorable TKO win over veteran Austin Arnett. The young bantamweight then followed his efforts up with a second-round knockout of Brazilian staple Iuri Alcantara. Sandhagen may be biting off a little more than he can chew against a bruising puncher like Lineker, but he’ll have a significant reach advantage heading into the fight. If he can keep Lineker at bay with long punches and kicks, Sandhagen may have a chance to secure the upset and push his UFC record to 3-0.

UFC on ESPN+ 1 will take place on Jan. 19 from inside Barclays Center in Brooklyn, New York, and feature a main even flyweight title fight between current champion Henry Cejudo and current UFC bantamweight titleholder T.J. Dillashaw.

For more UFC on ESPN+ 1 fight card news click here.