TSQL Challenge 66 - Find the most relevant advertisement based on keyw
A website wants to display most relevant ads on each of its web pages based on the keywords associated with each page.
2011-10-03
563 reads
A website wants to display most relevant ads on each of its web pages based on the keywords associated with each page.
2011-10-03
563 reads
This challenge invites you to solve one of the most common calculation challenges seen in applications that deal with financial -transactions.
2011-09-19
1,643 reads
This challenge invites you to solve a payroll challenge which requires special calculation of holidays and absences that are adjacent to holidays
2011-09-05
1,358 reads
2011-08-22
1,892 reads
Read the input string and break each sentence into groups of 5 words. A row may contain more than one sentence. Each sentence in the input should start a new group.
2011-08-08
1,211 reads
This challenge is to generate an HTML calendar based on the data stored in a table.
2011-07-25
1,833 reads
Your job is to read a string containing product hierarchy information and generate a relational table (result set) representing the hierarchy of categories.
2011-07-11
983 reads
Your job is to generate a string using Forsyth-Edwards Notation which represents the final position of pieces after performing a series of moves from the original position
2011-06-27
846 reads
Your job is to read the input string and generate a result set that represents the position of pieces in a chess board.
2011-06-13
1,758 reads
Your job is to scan the trades data and identify combination of trades that match a given rollup data.
2011-05-30
897 reads
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I am currently working with Sql Server 2022 and AdventureWorks database. First of all, let's set the "Read Committed Snapshot" to ON:
use master; go alter database AdventureWorks set read_committed_snapshot on with no_wait; goThen, from Session 1, I execute the following code:
--Session 1 use AdventureWorks; go create table ##t1 (id int, f1 varchar(10)); go insert into ##t1 values (1, 'A');From another session, called Session 2, I open a transaction and execute the following update:
--Session 2 use AdventureWorks; go begin tran; update ##t1 set f1 = 'B' where id = 1;Now, going back to Session 1, what happens if I execute this statement?
--Session 1 select f1 from ##t1 where id = 1;See possible answers