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NEW QUESTION: 1
You want to select data from two tables and store the result in as structure.
Table PARTNER contains the fields PART_ID and KIND.
Table CONTRACT contains the fields CONT_ID, CONT_TYPE and DIVISION.
The structure is defined as follows
DATA: BEGIN OF wa_result,
Part_id type partner-part_id, cont_id type contract-cont_id,
Cont_type TYPE contract-cont_type,
END of wa_result,
Lt_result type table of wa_result.
How can you replace the following SELECT statement with an outer join?
SELECT part_id from partner INTO wa_result WHERE kind = 'Residential'.
SELECT cont_id from CONTRACT into wa_result-cont_id WHERE part EQ
wa_partner-part_id And DIVISION eq 'Water'.
Append wa_result to lt_result.
ENDSELECT.
If sy-subrc<>0. CLEAR wa_result-cont_id
APPEND wa_result TO lt_result. ENDIF.
ENDSELECT.
Please choose the correct answer.
Response:
A. SELECT part_idcont_id from partner LEFT JOIN contract on partner-part_id = contract-part_id AND contract-division EQ 'Water' INTO TABLE lt_result WHERE kind EQ 'Residential'.
B. SELECT part_idcont_id from partner AS A LEFT JOIN contract AS b ON a~part_id = b~part_id INTO CORRESPONDING FIELDS OF TABLE lt_result WHERE kind = 'Residential' and AND division EQ 'Water'.
C. SELECT part_idcont_id from partner LEFT JOIN contract on partner-part_id = contract-part_id AND partner-kind EQ 'Residential' INTO CORRESPONDING FIELDS OF TABLE lt_result WHERE division eq
'Water'.
D. SELECT part_idcont_id from partner AS A LEFT JOIN contract AS b ON a~part_id = b~part_id AND b~division EQ 'Water' INTO TABLElt_result WHERE kind = 'Residential'
Answer: B

NEW QUESTION: 2

To allow or prevent load balancing to network 172.16.3.0/24, which of the following commands could be used in R2? (Choose two.)
A. R2(config-if)#ip ospf priority
B. R2(config-if)#ip ospf cost
C. R2(config-if)#clock rate
D. R2(config-router)#distance ospf
E. R2(config-if)#bandwidth
Answer: B,E
Explanation:
Explanation/Reference:
Explanation:
http://www.cisco.com/en/US/tech/tk365/technologies_white_paper09186a0080094e9e.sht ml#t6 The cost (also called metric) of an interface in OSPF is an indication of the overhead required to send packets across a certain interface. The cost of an interface is inversely proportional to the bandwidth of that interface. A higher bandwidth indicates a lower cost. There is more overhead (higher cost) and time delays involved in crossing a 56k serial line than crossing a 10M Ethernet line. The formula used to calculate the cost is:
Cost = 10000 0000/bandwidth in bps
For example, it will cost 10 EXP8/10 EXP7 = 10 to cross a 10M Ethernet line and will cost 10 EXP8/1544000 =64 to cross a T1 line.
By default, the cost of an interface is calculated based on the bandwidth; you can force the cost of an interface with the ip ospf cost <value> interface subconfiguration mode command.

NEW QUESTION: 3

A. Option B
B. Option D
C. Option A
D. Option E
E. Option C
Answer: A

NEW QUESTION: 4
Refer to the exhibit.

multiple hosts on the 10.2.2.0/24 network are sending traffic to the web server,
Which configuration can you apply to R2 so that traffic from host 1 uses the path R2-R1-R3 to reach the
web server, without affecting other hosts?
A. access-list 1 permit 10.2.2.3 0.0.0.0
B. interface FastEthernet2/0 ip policy route-map POLICY-ROUTE
C. access-list 1 permit 10.2.2.0 0.0.0.255
D. ip local policy route-map POLICY-ROUTE
E. access-list 1 permit 10.2.2.4 0.0.0.0
F. access-list 1 permit 10.2.2.3 255.255.255.255
Answer: A