2007 ASB Classic – Doubles

Doubles
2007 ASB Classic
Final
Champion	Janette Husárová Paola Suárez
Runner-up	Hsieh Su-wei Shikha Uberoi
Score	6–0, 6–
Details
Draw	16
Seeds	4
Events
Singles Doubles
← 2006 · WTA Auckland Open · 2008 →

Elena Likhovtseva and Vera Zvonareva were the defending champions, but neither chose to participate this year.

Janette Husárová and Paola Suárez won the title, defeating Hsieh Su-wei and Shikha Uberoi 6–0, 6–2 in the final.^[1]

Seeds

Draw

Key

Draw

	First round							Quarterfinals							Semifinals							Final
	1	J Husárová P Suárez	6	6
	WC	M Erakovic S Errani	1	2				1	J Husárová P Suárez	6	6
		É Loit Á Szávay	6	6					É Loit Á Szávay	2	1
		C Ji T Li	4	4											1	J Husárová P Suárez	6	7
	3	E Daniilidou J Wöhr	77	2	6										3	E Daniilidou J Wöhr	4	5
		C-w Chan V Uhlířová	65	6	3			3	E Daniilidou J Wöhr	6	6
		E Birnerová J Craybas	5	3					J Janković T Križan	2	3
		J Janković T Križan	7	6																		1	J Husárová P Suárez	6	6
		S Sun T Sun	7	6																			S-w Hsieh S Uberoi	0	2
		R Oprandi J Schruff	5	0					S Sun T Sun	7	6
		L Baker N Kriz	3	6	6				L Baker N Kriz	5	2
	4	M Domachowska J Kostanić	6	2	3											S Sun T Sun
		L Granville C Gullickson	6	6												S-w Hsieh S Uberoi	w/o
		E Bychkova M Müller	2	3					L Granville C Gullickson	5	2
		S-w Hsieh S Uberoi	1						S-w Hsieh S Uberoi	7	6
	2	G Dulko M Tu	2	r

References

External links

• WTA tournament draws