Ajax - Sparta Rotterdam — score en direct, compositions, stats, face-à-face et où regarder
Score final : Ajax 2–1 Sparta Rotterdam.
2020-01-19
Gravenberch Ryan60’
van de Beek Donny15’
74’Piroe Joel
Ajax
STATS CLÉS
Sparta Rotterdam
0Possession %0
23
Tirs
19
8
Tirs cadrés
7
9
Corners
4
83% (532/639)
Passes
62% (145/233)
1
Cartons jaunes
1
Faits de match
But
Carton jaune
Carton rouge
MTMi-temps
- Promes Quincy90'+3
- 90'+2Auassar Adil
- 84'Duarte DeroyRigo Dante
- 81'Mica PintoHarroui Abdou
- Eiting CarelGravenberch Ryan76'
- 74'Piroe Joel(2-1)
- Traore LassinaBabel Ryan70'
- 69'Piroe JoelJoosten Patrick
- Gravenberch Ryan(2-0)60'
- Ekkelenkamp JurgenZiyech Hakim59'
- van de Beek Donny(1-0)15'
Face-à-face
13 dernières confrontations
10
Victoires Ajax
3
nuls
0
Victoires Sparta Rotterdam
- 2026-03-14Ajax vs Sparta RotterdamEredivisieV
- 2025-10-04Sparta Rotterdam vs AjaxEredivisieN
- 2025-04-27Ajax vs Sparta RotterdamEredivisieN
- 2024-12-22Sparta Rotterdam vs AjaxEredivisieV
- 2024-03-17Sparta Rotterdam vs AjaxEredivisieN
Ajax — forme récente
VVNNV
Sparta Rotterdam — forme récente
NDDNV
Où regarder Ajax - Sparta Rotterdam
- ESPN SelectUS · OTTOfficiel
- OneSoccerCA · TVOfficiel
- ESPN NetherlandsNL · TVOfficiel
- Triller TV+AU · OTTOfficiel
- Australie:Triller TV+
- Canada:OneSoccer
- Pays-Bas:ESPN Netherlands, ESPN 2 Netherlands
- Royaume-Uni:Triller TV+
- États-Unis:ESPN Select
Questions fréquentes
＋ What time does Ajax vs Sparta Rotterdam kick off?
Ajax host Sparta Rotterdam at 13:30 UTC on Sunday 19 January 2020.
＋ Where can I watch Ajax vs Sparta Rotterdam on TV or online?
Broadcast coverage for this Eredivisie fixture includes ESPN Select, OneSoccer, ESPN Netherlands and Triller TV+.
＋ What is the head-to-head record between Ajax and Sparta Rotterdam?
Ajax have won 10, Sparta Rotterdam have won 0, and 3 matches have ended level between the two sides.
＋ What form are Ajax and Sparta Rotterdam in coming into this match?
Ajax arrive in WDDWW form, while Sparta Rotterdam come into this Eredivisie fixture in WDLLD form.