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NEW QUESTION: 1
Your network contains a System Center 2012 Configuration Manager environment.
You need to create a report that lists all of the client computers that have an application named App1 installed.
Which query should you use to create the report?
A. SELECT SYS.Netbios_Name0, ARP.DisplayName0 FROM v_R_Sytem AS SYS INNER JOIN v_GS_ADD_REMOVE_PROGRAMS AS ARP ON SYS.ResourceID = ARP.ResourceID WHERE ARP.DisplayName0 NOT IN ('App1')
B. SELECT SYS.Netbios_Name0 from v_R_System SYS WHERE SYS.ResourceID NOT IN (SELECT SYS.ResouceID FROM v_R_System AS SYS INNER JOIN v_GS_ADD_REMOVE_PROGRAMS AS ARP ON SYS.ResourceID = ARP.ResourceID WHERE ARP.DisplayName0 = 'App1')
C. SELECT SYS.Netbios_Name0, ARP.DisplayName0 FROM v_R_Sytem AS SYS INNER JOIN v_GS_ADD_REMOVE_PROGRAMS AS ARP ON SYS.ResourceID = ARP.ResourceID WHERE ARP.DisplayName0 NOT LIKE 'App1'
D. SELECT SYS.Netbios_Name0 from v_R_System SYS WHERE SYS.ResourceID IN (SELECT SYS.ResouceID FROM v_R_System AS SYS INNER JOIN v_GS_ADD_REMOVE_PROGRAMS AS ARP ON SYS.ResourceID = ARP.ResourceID WHERE ARP.DisplayName0 = 'App1')
Answer: D
Explanation:
SELECT computer name FROM table v_R_System (SQL Syntax)
WHERE resource ResourceID IN
Add Remove Programs WHERE DisplayName0 = 'App1'
SQL INNER JOIN Keyword
The INNER JOIN keyword returns rows when there is at least one match in both tables.
Reference: http://www.w3schools.com/sql/sql_join_inner.asp

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