# Mach3 R3042036 [TOP] Full Versiontorrent

## Mach3 R3042036 [TOP] Full Versiontorrent

Mach3 R3042036 Full Versiontorrent

A: Ctrl+U or Edit->Undo should let you get back to the previous version. Q: SQL Server: Updating the Max ID for a table in the database Im trying to find the id for the largest record in the database and then update it to one higher than that record. However I can’t find the MAX function to do this. I’ve tried CAST, but that doesn’t seem to work either. Is there any functions that I’m overlooking? I’ve already tried… UPDATE TABLE SET MAX(ID) = (SELECT MAX(ID) FROM TABLE) A: There’s no max function or any other aggregate function – the only functions that work in this context are: SELECT TOP (1) with ties –TOP(1) is the “standard” syntax, but there are others SELECT TOP (1) WITH TIES * SELECT TOP (1 WITH TIES) * There’s no reason to update it, just do SELECT MAX(ID) FROM TABLE To update the existing record to make it the current record: UPDATE TABLE SET ID = MAX(ID) + 1 If you do this, you will delete the data rows you were using before. If you just want to get the data in a form that can be sorted and retrieved, use: SELECT * FROM ( SELECT ID, COL1, COL2, COL3 FROM TABLE ) T ORDER BY ID DESC The part in parentheses returns the rows, and the other part orders the rows by ID and returns the columns as the next record to be selected. Q: How to determine $P(X\le a|Y\ge b)$ when $Y=max\{X,b\}$? In the following formula: $$P(X\le a|Y\ge b)=\dfrac{P(X\le a,Y\ge b)}{P(Y\ge b)}$$ What is the correct way to determine $P(X\le a|Y\ge b)$? I tried it several ways but to no avail. Thank you very much in advance. A: P(X \leq a|Y \geq b)=\frac{P(X \ 37a470d65a