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Aleksandrów
sī
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Opolskie
séng
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kūn ê chi̍t ê chng-thâu.
Aleksandrów
—
Chng-thâu
—
Aleksandrów
Aleksandrów tī Pho-lân ê tāi-lio̍k ūi-tì
Keng-hūi-tō͘
:
51°5′46″N
18°28′20″E
/
51.09611°N 18.47222°E
/
51.09611; 18.47222
Kok-ka
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Kūn
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Che sī chi̍t phiⁿ koan-î
Pho-lân
tē-lí ê
phí-á-kiáⁿ
. Chhiáⁿ tàu
khok-chhiong
.