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Aleksandrów (Opole Séng)

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Aleksandrów
—  Chng-thâu  —
Aleksandrów is located in Pho-lân
Aleksandrów
Aleksandrów
Keng-hūi-tō͘: 51°6′N 18°29′E / 51.100°N 18.483°E / 51.100; 18.483
Kok-ka  Pho-lân
Séng Opole
Kūn Olesno
Kong-siā Praszka
Sî-khu CET (UTC+1)
 - Joa̍h-tang CEST (UTC+2)

Aleksandrów sī chi̍t ê tī Pho-lân Kiōng-hô-kok Opole Séng Olesno Kūn Praszka Kong-siā ê chng-thâu.[1]

Chham-oa̍t

[siu-kái | kái goân-sí-bé]

Chham-khó

[siu-kái | kái goân-sí-bé]
  1. ↑ "Central Statistical Office (GUS) - TERYT (National Register of Territorial Land Apportionment Journal)" (ēng Pho-lân-gí). 2008-06-01. 
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