| Tournament information | |
|---|---|
| Dates | 1–7 April 2019 |
| Venue | Olympic Sports Center Gymnasium |
| City | Beijing |
| Country | China |
| Organisation | World Snooker |
| Format | Ranking event |
| Total prize fund | £1,000,000 |
| Winner's share | £225,000 |
| Highest break |  Stuart Bingham (ENG) (147) |
| Final | |
| Champion |  Neil Robertson (AUS) |
| Runner-up | Jack Lisowski (ENG) |
| Score | 11–4 |
← 2018 | |
The 2019 XingPai China Open was a professional ranking snooker tournament, that took place between 1–7 April 2019 in Beijing, China.[1] It was the nineteenth and penultimate ranking event of the 2018/2019 season.
Neil Robertson won his second China Open title, and the 16th ranking title of his career, defeating Jack Lisowski 11–4 in the final.[2]
Mark Selby was the two-time reigning champion, having defended his 2017 title with an 11–3 win against Barry Hawkins in the 2018 final.[3] However, he lost 3–6 to Craig Steadman in qualifying.
Stuart Bingham made the highest break of the event, with his fifth 147 break of his career in his second round match with Peter Ebdon. It was the 151st 147 in snooker history.
- ^ "2018/19 Calendar" (PDF). World Professional Billiards and Snooker Association. 18 July 2018. Retrieved 3 October 2018.
- ^ "Robertson Storms to Beijing Victory". World Snooker. 7 April 2019. Retrieved 7 April 2019.
- ^ "Selby Back On Form With China Open Victory". World Snooker. 8 April 2018. Retrieved 3 October 2018.