firebird language reference 30 rus
запрос вызовет ошибку "Column does not belong to referenced table" SELECT * FROM TA, TB JOIN TC ON TA.COL1 = TC.COL1 WHERE TA.COL2 = TB.COL2 Это происходит потому запрос будет выполнен без ошибок, поскольку изоляция не нарушена. SELECT * FROM TA, TB JOIN TC ON TB.COL1 = TC.COL1 WHERE TA.COL2 = TB.COL2 Перекрёстное соединение0 码力 | 719 页 | 3.98 MB | 1 年前3firebird 40 language reference
will raise the error “Column does not belong to referenced table” SELECT * FROM TA, TB JOIN TC ON TA.COL1 = TC.COL1 WHERE TA.COL2 = TB.COL2 That is because the explicit join cannot see the TA table. However complete without error, since the restriction is not violated. SELECT * FROM TA, TB JOIN TC ON TB.COL1 = TC.COL1 WHERE TA.COL2 = TB.COL2 A Note on Equality This note about equality and inequality0 码力 | 778 页 | 3.43 MB | 1 年前3firebird 30 sprachreferenz
den Fehler “Spalte gehört nicht zur referenzierten Tabelle” aus. SELECT * FROM TA, TB JOIN TC ON TA.COL1 = TC.COL1 WHERE TA.COL2 = TB.COL2 Das liegt daran, dass der explizite Join die Tabelle TA nicht Fehler abgeschlossen, da die Einschränkung nicht verletzt wird. SELECT * FROM TA, TB JOIN TC ON TB.COL1 = TC.COL1 WHERE TA.COL2 = TB.COL2 Ein Hinweis zu Gleichheit Dieser Hinweis zu Gleichheits-0 码力 | 701 页 | 3.12 MB | 1 年前3
共 3 条
- 1