Hammamet 22 ITF 2017 (Tunisia)
(15,000 $, clay, women)
| Start | Round | S | 1 | 2 | 3 | 4 | 5 | H | A | ||
| 23.09. 15:00 |
F | Eklund / Zhu (3) | 2 | 6 | 6 | info | |||||
| Aponasenk / Lombardo | 0 | 1 | 4 | ||||||||
| 22.09. 15:40 |
SF | Aponasenk / Lombardo | 2 | 2 | 7 | 10 | 2.25 | 1.57 | info | ||
| Bullani / Napolitan | 1 | 6 | 5 | 8 | |||||||
| 22.09. 15:20 |
SF | Eklund / Zhu (3) | 2 | 65 | 7 | 10 | 3.25 | 1.33 | info | ||
| Moratelli / Shinikova (1) | 1 | 7 | 5 | 7 | |||||||
| 21.09. 15:50 |
QF | Moratelli / Shinikova (1) | 2 | 6 | 6 | 1.44 | 2.62 | info | |||
| Ibbou / Sauvant | 0 | 3 | 4 | ||||||||
| 21.09. 15:45 |
QF | Eklund / Zhu (3) | 2 | 6 | 3 | 10 | 2.20 | 1.61 | info | ||
| Salden / Vanhoutte | 1 | 3 | 6 | 5 | |||||||
| 21.09. 14:25 |
QF | Bullani / Napolitan | 2 | 6 | 6 | 1.72 | 2.00 | info | |||
| Pipa / Rosnowska | 0 | 4 | 3 | ||||||||
| 21.09. 14:20 |
QF | Aponasenk / Lombardo | 2 | 6 | 6 | 1.66 | 2.10 | info | |||
| Njoze / Prati | 0 | 4 | 2 | ||||||||
| 20.09. 16:55 |
R16 | Bullani / Napolitan | 2 | 6 | 6 | 1.36 | 3.00 | info | |||
| Cheli / Lene | 0 | 2 | 2 | ||||||||
| 20.09. 15:10 |
R16 | Pipa / Rosnowska | 2 | 6 | 4 | 10 | 2.62 | 1.44 | info | ||
| Remy / Temin (4) | 1 | 2 | 6 | 7 | |||||||
| 20.09. 15:05 |
R16 | Salden / Vanhoutte | 2 | 6 | 6 | 1.05 | 9.00 | info | |||
| Boukholda / Gerald Go | 0 | 2 | 1 | ||||||||
| 19.09. 16:30 |
R16 | Ibbou / Sauvant | 2 | 6 | 6 | 1.09 | 6.08 | info | |||
| Gaelle De / Karunarat | 0 | 3 | 3 | ||||||||
| 19.09. 16:20 |
R16 | Njoze / Prati | 2 | 6 | 6 | 1.05 | 8.14 | info | |||
| Akimova / Malinovic | 0 | 1 | 0 | ||||||||
| 19.09. 15:55 |
R16 | Eklund / Zhu (3) | 2 | 6 | 6 | 1.56 | 2.23 | info | |||
| Kommer / Niedmers | 0 | 3 | 3 | ||||||||
| 19.09. 15:10 |
R16 | Moratelli / Shinikova (1) | 2 | 6 | 6 | 1.03 | 9.64 | info | |||
| Avvantagg / Sonnekus | 0 | 3 | 1 | ||||||||
| 19.09. --:-- |
R16 | Aponasenk / Lombardo | 1 | info | |||||||
| Bosio / Herazo Go (2) | 0 |

round of 16
Eklund / Zhu [3]
Remy / Temin [4]
quarterfinal
Eklund / Zhu [3]
semifinal
Eklund / Zhu [3]
final
Eklund / Zhu [3]
Eklund / Zhu [3]
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