<!-- /* Style Definitions */ p.MsoNormal, li.MsoNormal, div.MsoNormal {mso-style-parent:""; margin:0cm; margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:12.0pt; font-family:"Times New Roman"; mso-fareast-font-family:"Times New Roman";} code {font-family:"Courier New"; mso-ascii-font-family:"Courier New"; mso-fareast-font-family:"Times New Roman"; mso-hansi-font-family:"Courier New"; mso-bidi-font-family:"Courier New";} pre {margin:0cm; margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:10.0pt; font-family:"Courier New"; mso-fareast-font-family:"Times New Roman";} @page Section1 {size:612.0pt 792.0pt; margin:2.0cm 42.5pt 2.0cm 3.0cm; mso-header-margin:36.0pt; mso-footer-margin:36.0pt; mso-paper-source:0;} div.Section1 {page:Section1;} -->
Какие данные вернет запрос: SELECT DISTINCT COUNT(T2.F_VAL) AS F, COUNT(DISTINCT T1.P_VAL) AS PFROM T1, T2 WHERE T1.P_KEY=T2.F_KEY
Для следующих таблиц: T1( P_KEY INTEGER, P_VAL VARCHAR(20))
P_KEY
P_VAL
1
UK
2
Russia
NULL
Ukraine
T2( P_KEY INTEGER, F_KEY INTEGER, F_VAL VARCHAR(20))
P_KEY
F_KEY
F_VAL
1
1
London
2
2
Moscow
NULL
2
St. Petersburg
NULL
NULL
Kiev