錯誤軟刪除

Question

我有以下的JPA實體

@SQLDelete(sql="UPDATE service SET date_deletion = CURRENT_DATE() WHERE id = ?") 
@Where(clause="date_deletion IS NULL ") 
public class Service { 
... 
} 

選擇工作確定所有與date_deletion告知沒有示明，但元素，當我嘗試刪除....

16:38:26,836 DEBUG SQL:111 - UPDATE service SET date_deletion = CURRENT_DATE() WHERE id = ? 
16:38:26,836 DEBUG AbstractBatcher:418 - about to close PreparedStatement (open PreparedStatements: 1, globally: 1) 
16:38:26,836 DEBUG JDBCExceptionReporter:225 - could not delete: [com.foo.domain.Service#1] [UPDATE service SET date_deletion = CURRENT_DATE() WHERE id = ?] 
java.sql.SQLException: Parameter index out of range (2 > number of parameters, which is 1). 

SQL中有什麼問題？看起來像嘗試處理CURRENT_DATE（）作爲參數，並期望2參數，而不是1 ...

Answer 1

固定。 我正在使用Spring Roo在內部處理作爲參數發送的「版本」字段，正確的註釋是：

@SQLDelete(sql="UPDATE service SET date_deletion=CURRENT_DATE WHERE id=? and version=? ") 
@Where(clause="date_deletion IS NULL ") 
public class Service { 
...