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| Bastad 2019 / Sweden, clay , ATP Bastad 2019 Tennis score live qualification |
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| Jul 15, 2019 11:00 |
quarter-final | Arguello Facundo (ARG) (178) | 57.28% | 2 | 6 | 3 | 6 | I H2H O AI | |||
| Diez Steven (CAN) (226) | 42.72% | 1 | 4 | 6 | 3 | ||||||
| Jul 15, 2019 11:00 |
quarter-final | Carreno-Busta Pablo (ESP) (50) | 55.26% | 2 | 7 | 7 | I H2H O AI | ||||
| Quiroz Roberto (ECU) (232) | 44.74% | 0 | 6 (1) | 6 (5) | |||||||
| Jul 15, 2019 11:00 |
quarter-final | Zapata Miralles Bernabe (ESP) (235) | 32.26% | 2 | 6 | 6 | I H2H O AI | ||||
| Vatutin Alexey (RUS) (202) | 67.74% | 0 | 2 | 1 | |||||||
| Jul 15, 2019 11:00 |
quarter-final | Lestienne Constant (FRA) (199) | 50.00% | 2 | 1 | 6 | 6 | I H2H O AI | |||
| Grigelis Laurynas (LTU) (295) | 50.00% | 1 | 6 | 2 | 1 | ||||||
| Jul 14, 2019 10:30 |
eight-final | Vatutin Alexey (RUS) (202) | 65.56% | 2 | 6 | 6 | I H2H O AI | ||||
| Ocleppo Julian (ITA) (414) | 34.44% | 0 | 2 | 2 | |||||||
| Jul 14, 2019 10:30 |
eight-final | Zapata Miralles Bernabe (ESP) (235) | 50.96% | 2 | 6 | 6 | I H2H O AI | ||||
| Horansky Filip (SVK) (181) | 49.04% | 0 | 3 | 4 | |||||||
| Jul 14, 2019 13:00 |
eight-final | Diez Steven (CAN) (226) | 52.52% | 2 | 7 | 6 | I H2H O AI | ||||
| Huesler Marc-Andrea (CHE) (263) | 47.48% | 0 | 5 | 3 | |||||||
| Jul 14, 2019 12:05 |
eight-final | Arguello Facundo (ARG) (178) | 65.12% | 2 | 7 | 6 | I H2H O AI | ||||
| Eriksson Markus (SWE) (445) | 34.88% | 0 | 6 (4) | 1 | |||||||
| Jul 14, 2019 10:30 |
eight-final | Lestienne Constant (FRA) (199) | 50.26% | 2 | 6 | 7 | I H2H O AI | ||||
| Hallquist Lithen John (SWE) | 49.74% | 0 | 0 | 6 (1) | |||||||
| Jul 14, 2019 10:30 |
eight-final | Grigelis Laurynas (LTU) (295) | 38.76% | 2 | 7 | 2 | 6 | I H2H O AI | |||
| Domingues Joao (PRT) (170) | 61.24% | 1 | 6 (4) | 6 | 2 | ||||||
| Jul 14, 2019 12:05 |
eight-final | Quiroz Roberto (ECU) (232) | 63.23% | 2 | 6 | 6 | I H2H O AI | ||||
| Mertens Yannick (BEL) (439) | 36.77% | 0 | 1 | 2 | |||||||
| Jul 14, 2019 13:15 |
eight-final | Carreno-Busta Pablo (ESP) (50) | 55.61% | 2 | 3 | 6 | 6 | I H2H O AI | |||
| Soderlund Carl (SWE) (1746) | 44.39% | 1 | 6 | 4 | 3 | ||||||
