Ïÿòíèöà, 30 ßíâàðÿ 2015 ã. 11:51
+ â öèòàòíèê
Çäðàâñòâóéòå. Åñòü òàêàÿ ïðîáëåìà.
Íà÷àë ðàçáèðàòüñÿ ñ äàííûìè îïåðàòîðàìè. Íî â èòîãîâûõ äàííûõ ïî [Âñåãî êëèåíòîâ] ãäå êëèåíòû äîëæíû áûòü - ïóñòûå ïîëÿ. Òàê æå Èìÿ äàííîãî ïîëÿ îòîáðàæàåòñÿ êàê "<>"
Öåëü äàííîãî çàïðîñà çàêëþ÷àåòñÿ â òîì, ÷òî áû óçíàòü ó êàêîãî êëèåíòà ñêîëüêî ñ÷åòîâ â áàíêå. Ñòîëáöû:. áàíê, áèê, îáùåå êîëè÷åñòâî êëèåíòîâ, è ñïèñîê êëèåíòîâ ñ óêàçàíèåì êîëè÷åñòâà ñ÷åòîâ.
TRANSFORM Count(INN.INN) AS [Âñåãî êëèåíòîâ]
SELECT BIK.NAME AS Áàíê, BIK.BIK AS [ÁÈÊ Áàíêà]
FROM BIK LEFT JOIN INN ON BIK.BIK = INN.BIK
GROUP BY BIK.BIK, BIK.NAME
ORDER BY BIK.NAME
PIVOT INN.NAME;
http://www.sql.ru/forum/1139987/transform-i-pivot
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