Alexander Clark (sinh ngày 26 tháng 6 n\u0103m 1992) là m\u1ed9t c\u1ea7u th\u1ee7 bóng \u0111á chuyên nghi\u1ec7p ng\u01b0\u1eddi Scotland, ch\u01a1i \u1edf v\u1ecb trí th\u1ee7 môn cho câu l\u1ea1c b\u1ed9 Heart of Midlothian và \u0111\u1ed9i tuy\u1ec3n qu\u1ed1c gia Scotland.<\/P>
Sau khi thi \u0111\u1ea5u bóng \u0111á tr\u1ebb cùng Hamilton Academical và St Johnstone, Clark b\u1eaft \u0111\u1ea7u s\u1ef1 nghi\u1ec7p chuyên nghi\u1ec7p c\u1ee7a mình trong mùa gi\u1ea3i 2011–12 \u0111\u01b0\u1ee3c cho m\u01b0\u1ee3n t\u1ea1i Elgin City. Anh c\u0169ng \u0111ã có hai k\u1ef3 ngh\u1ec9 cho m\u01b0\u1ee3n t\u1ea1i Queen of the South tr\u01b0\u1edbc khi tr\u1edf thành m\u1ed9t c\u1ea7u th\u1ee7 chính th\u1ee9c c\u1ee7a St Johnstone sau khi ra m\u1eaft cho h\u1ecd vào n\u0103m 2015. Anh là m\u1ed9t ph\u1ea7n c\u1ee7a \u0111\u1ed9i St Johnstone giành c\u1ea3 hai cúp qu\u1ed1c n\u1ed9i trong mùa gi\u1ea3i 2020–21. Clark r\u1eddi St Johnstone sau mùa gi\u1ea3i 2021–22 và ký h\u1ee3p \u0111\u1ed3ng v\u1edbi Heart of Midlothian vào tháng 9 n\u0103m 2022.<\/P>
Clark ra m\u1eaft \u0111\u1ed9i tuy\u1ec3n qu\u1ed1c gia Scotland vào tháng 10 n\u0103m 2023, \u1edf 31 tu\u1ed5i.<\/P>","0C852134AA40B2DD":"1","8D04D562CDE8A422":"Zander Clark","A16D5E6475634BB1C2248B183B7E41495B218039EB2EF0A3":"1","18E9F76C307DDD99":"359","C6E00DEB351621C4":"98032","A16D5E6475634BB126EF698FB3550B8C":"700000.00","2B5572269355781B0007104B7181FA88":"1992-6-26","DA5E7C1A0F1CB8295E941203AD41B403":"Scotland","30A86FC0880FF75E":"191cm","F27CBDD6332F20A59321B0F4E9CF2AB8":"28","28017CE41B75854B":"Hearts FC","8DC362CCE065919BEFF67A555F929850":"
Câu l\u1ea1c b\u1ed9<\/U><\/STRONG><\/P> St Johnstone<\/STRONG>
Cúp Qu\u1ed1c gia Scotland: 2020–21
Cúp Liên \u0111oàn Scotland: 2020–21<\/P>","09702F2B27619389F07D77F05427B135":"2022-9-2","82CBDCB828DBDF89E0DAB2151FFA57BC":"","FA149E2002A43ED77C24C62CDFF094A5":"3","0A9B0394F31973599466C3FC5C90F503":"St. Johnstone,Queen of the South FC[Thu\u00ea],Elgin City FC[Thu\u00ea],St. Johnstone(U20),Hamilton Academical FC","5E8F53D0A504BF86":"Alexander Clark","e_index":1};