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  • Vendor : Oracle
  • Certifications : Java and Middleware
  • Exam Name : Java SE 8 Programmer I
  • Exam Code : 1z0-808
  • Total Questions : 376 Q&As
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    NEW QUESTION: 1
    エンジニアは、802.11プロトコルの修正に基づく標準を使用して、新しい無線LAN展開がアクセスポイント間のシームレスなローミングをサポートできることを確認する必要があります。エンジニアはどのプロトコルを選択する必要がありますか?
    A. 802.11i
    B. 802.11ac
    C. 802.11e
    D. 802.11r
    Answer: D
    Explanation:


    NEW QUESTION: 2

    A. Option E
    B. Option C
    C. Option F
    D. Option A
    E. Option G
    F. Option B
    G. Option D
    Answer: B,C

    NEW QUESTION: 3
    You are preparing an SAP HANA server for use with SAP Fiori analytical apps. What can you import with
    the application lifecycle management tool?
    Please choose the correct answer. Response:
    A. KPI report
    B. Generic drill down application
    C. KPI framework
    D. Custom drill down application
    Answer: A

    NEW QUESTION: 4



    A. SELECT COUNT(*)FROM tblDepositAcct DFULL JOIN tblLoanAcct L ON D.CustNo = L.CustNo
    B. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctEXCEPTSELECT CustNoFROM tblLoanAcct) R
    C. CustNo
    D. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctUNIONSELECT CustNoFROM tblLoanAcct) R
    E. SELECT COUNT (DISTINCT D.CustNo)FROM tblDepositAcct D, tblLoanAcct LWHERE D.CustNo
    F. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctUNION ALLSELECT CustNoFROM tblLoanAcct) R
    G. SELECT COUNT (DISTINCT COALESCE(D.CustNo, L.CustNo))FROM tblDepositAcct DFULL JOIN tblLoanAcct L ON D.CustNo = L.CustNoWHERE D.CustNo IS NULL OR L.CustNo IS NULL
    H. CustNoWHERE D.CustNo IS NULL
    I. SELECT COUNT(*)FROM (SELECT AcctNoFROM tblDepositAcctINTERSECTSELECT AcctNoFROM tblLoanAcct) R
    J. SELECT COUNT(DISTINCT L.CustNo)FROM tblDepositAcct DRIGHT JOIN tblLoanAcct L ON D.CustNo
    Answer: G
    Explanation:
    Explanation
    SQL Server provides the full outer join operator, FULL OUTER JOIN, which includes all rows from both tables, regardless of whether or not the other table has a matching value.
    Consider a join of the Product table and the SalesOrderDetail table on their ProductID columns. The results show only the Products that have sales orders on them. The ISO FULL OUTER JOIN operator indicates that all rows from both tables are to be included in the results, regardless of whether there is matching data in the tables.
    You can include a WHERE clause with a full outer join to return only the rows where there is no matching data between the tables. The following query returns only those products that have no matching sales orders, as well as those sales orders that are not matched to a product.
    USE AdventureWorks2008R2;
    GO
    -- The OUTER keyword following the FULL keyword is optional.
    SELECT p.Name, sod.SalesOrderID
    FROM Production.Product p
    FULL OUTER JOIN Sales.SalesOrderDetail sod
    ON p.ProductID = sod.ProductID
    WHERE p.ProductID IS NULL
    OR sod.ProductID IS NULL
    ORDER BY p.Name ;
    References: https://technet.microsoft.com/en-us/library/ms187518(v=sql.105).aspx


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