What is the setVariable syntax for AS3

Hi,
I am having issues with setting a variable in AS3 SWF playing inside a Dir12
spriteObjRef.setVariable(variableName, newValue)
In the context of ActionScript 3, this method is only supported on a Flash object, not on a
spriteObjectRef.
What would the Flash Object syntax be??
Thanks,
Jim

Jusclark-Oracle wrote:
What is the correct syntax for referencing a page item (P23_ID, for example) in a PL/SQL page process, after submit?
See "About Referencing Session State" in the documentation.
Within an anonymous block in an APEX page process, use the standard bind variable syntax:
:P23_ID
In general, use bind variable syntax when referencing session state values in SQL or DML:
select
    ename
into
    :p23_name
from
    emp
where
    empno = to_number(:p23_id);
and obviously when assigning values to the item:
:P23_ID := foo.nextval;
PL/SQL function references v('p23_id') and nv('p23_id') [for number values] must be used to access values in stored program units called from APEX, although it's usually better to pass the values as parameters: foo(:p23_id).

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    Hi, John,
    Not that you asked, but as you proabably know, analytic functions are much better at doing this kind of thing.
    You may be amazed (as I continually am) by how simple and efficient these queries can be.
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         SELECT     proj_id, proj_start, proj_end
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         FROM     v
    SELECT       ROW_NUMBER () OVER (ORDER BY grp)     AS proj_grp
    ,       MIN (proj_start)                         AS proj_start
    ,       MAX (proj_end)               AS proj_end
    FROM       got_grp
    GROUP BY  grp
    ORDER BY  proj_start
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      PROJ_GRP PROJ_START  PROJ_END
             1 01-Jan-2005 05-Jan-2005
             2 06-Jan-2005 07-Jan-2005
             3 16-Jan-2005 20-Jan-2005
             4 21-Jan-2005 22-Jan-2005
             5 26-Jan-2005 30-Jan-2005This is problem is an example of Neighbor-Defined Groups . You want to GROUP BY something that has 5 distinct values, to get the 5 rows above, but there's nothing in the table itself that tells you to which group each row belongs. The groups are not defined by any column in hte table, but by relationships between rows. In this case, a row is in the same group as its neighbor (the row immediatly before or after it when sorted by proj_start or proj_end) if proj_end of the earlier row is the same as proj_start of the later row. That is, there is nothing about 03-Jan-2005 that says the row with proj_id=2 is in the first group, or even that it is in the same group with its neighbor, the row with proj_id=3. Only the relation between those rows, the fact that the earlier row has end_date=03-Jan-2005 and the later row has start_date=03-Jan-2003, that says these neighbors belong to the same group.
    You're figuring out when a new group starts, and then counting how many groups have already started to see to which group each row belongs. That's a prefectly natural procedural way of approaching the problem. But SQL is not a procedural language, and sometimes another approach is much more efficient. In this case, as in many others, a Constant Difference defines the groups. The difference between proj_end (or proj_start, it doesn't matter in this case) and the total duratiojn of the rows up to that date determines a group. The actual value of that difference means nothing to you or anybody else, so I used ROW_NUMBER in the query above to map those distinct values into consecutive integers 1, 2, 3, ... which are a much simpler way to identify the groups.
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    If you used analytic functions (LEAD or LAG) to compute flag, and then to compute proj_grp (COUNT or SUM), you would need two sub-queries, one for each analytic function, but you would still only need one pass through the table. Also, those sub-queries could be in-line views; yiou would not need to use a WITH clause.

  • Hi guru's what is the diff between for all entries & joins

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    A
    B
    C
    D
    D
    E
    F
    G
    H
    a1
    b1
    c1
    1
    1
    e1
    f1
    g1
    h1
    a2
    b2
    c2
    1
    3
    e2
    f2
    g2
    h2
    a3
    b3
    c3
    2
    4
    e3
    f3
    g3
    h3
    a4
    b4
    c4
    3
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    A
    B
    C
    D
    D
    E
    F
    G
    H
    a1
    b1
    c1
    1
    1
    e1
    f1
    g1
    h1
    a2
    b2
    c2
    1
    1
    e1
    f1
    g1
    h1
    a4
    b4
    c4
    3
    3
    e2
    f2
    g2
    h2
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    CONNID LIKE SFLIGHT-CONNID.
    SELECT FCARRID FCONNID F~FLDATE
    INTO (CARRID, CONNID, DATE)
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    ON FCARRID = PCARRID AND
    FCONNID = PCONNID
    WHERE P~CITYFROM = 'FRANKFURT'
    AND P~CITYTO = 'NEW YORK'
    AND F~FLDATE BETWEEN '20010910' AND '20010920'
    AND FSEATSOCC < FSEATSMAX.
    WRITE: / DATE, CARRID, CONNID.
    ENDSELECT.
    If there are columns with the same name in both tables, you must distinguish between them by prefixing the field descriptor with the table name or a table alias.
    Note
    In order to determine the result of a SELECT command where the FROM clause contains a join, the database system first creates a temporary table containing the lines that meet the ON condition. The WHERE condition is then applied to the temporary table. It does not matter in an inner join whether the condition is in the ON or WHEREclause. The following example returns the same solution as the previous one.
    Example
    Output of a list of all flights from Frankfurt to New York between September 10th and 20th, 2001 that are not sold out:
    DATA: DATE LIKE SFLIGHT-FLDATE,
    CARRID LIKE SFLIGHT-CARRID,
    CONNID LIKE SFLIGHT-CONNID.
    SELECT FCARRID FCONNID F~FLDATE
    INTO (CARRID, CONNID, DATE)
    FROM SFLIGHT AS F INNER JOIN SPFLI AS P
    ON FCARRID = PCARRID
    WHERE FCONNID = PCONNID
    AND P~CITYFROM = 'FRANKFURT'
    AND P~CITYTO = 'NEW YORK'
    AND F~FLDATE BETWEEN '20010910' AND '20010920'
    AND FSEATSOCC < FSEATSMAX.
    WRITE: / DATE, CARRID, CONNID.
    ENDSELECT.
    Note
    Since not all of the database systems supported by SAP use the standard syntax for ON conditions, the syntax has been restricted. It only allows those joins that produce the same results on all of the supported database systems:
    Only a table or view may appear to the right of the JOIN operator, not another join expression.
    Only AND is possible in the ON condition as a logical operator.
    Each comparison in the ON condition must contain a field from the right-hand table.
    If an outer join occurs in the FROM clause, all the ON conditions must contain at least one "real" JOIN condition (a condition that contains a field from tabref1 amd a field from tabref2.
    Note
    In some cases, '*' may be specified in the SELECT clause, and an internal table or work area is entered into the INTO clause (instead of a list of fields). If so, the fields are written to the target area from left to right in the order in which the tables appear in the FROM clause, according to the structure of each table work area. There can then be gaps between table work areas if you use an Alignment Request. For this reason, you should define the target work area with reference to the types of the database tables, not simply by counting the total number of fields. For an example, see below:
    Variant 3
    ... FROM tabref1 LEFT [OUTER] JOIN tabref2 ON cond
    Effect
    Selects the data from the transparent database tables and/or views specified in tabref1 and tabref2. tabref1 und tabref2 both have either the same form as in variant 1 or are themselves join expressions. The keyword OUTER can be omitted. The database tables or views specified in tabref1 and tabref2 must be recognized by the ABAP-Dictionary.
    In order to determine the result of a SELECT command where the FROM clause contains a left outer join, the database system creates a temporary table containing the lines that meet the ON condition. The remaining fields from the left-hand table (tabref1) are then added to this table, and their corresponding fields from the right-hand table are filled with ZERO values. The system then applies the WHERE condition to the table.
    Left outer join between table 1 and table 2 where column D in both tables set the join condition:
    Table 1 Table 2
    A
    B
    C
    D
    D
    E
    F
    G
    H
    a1
    b1
    c1
    1
    1
    e1
    f1
    g1
    h1
    a2
    b2
    c2
    1
    3
    e2
    f2
    g2
    h2
    a3
    b3
    c3
    2
    4
    e3
    f3
    g3
    h3
    a4
    b4
    c4
    3
    |--|||--|
    Left Outer Join
    A
    B
    C
    D
    D
    E
    F
    G
    H
    a1
    b1
    c1
    1
    1
    e1
    f1
    g1
    h1
    a2
    b2
    c2
    1
    1
    e1
    f1
    g1
    h1
    a3
    b3
    c3
    2
    NULL
    NULL
    NULL
    NULL
    NULL
    a4
    b4
    c4
    3
    3
    e2
    f2
    g2
    h2
    |--||||||||--|
    Example
    Output a list of all custimers with their bookings for October 15th, 2001:
    DATA: CUSTOMER TYPE SCUSTOM,
    BOOKING TYPE SBOOK.
    SELECT SCUSTOMNAME SCUSTOMPOSTCODE SCUSTOM~CITY
    SBOOKFLDATE SBOOKCARRID SBOOKCONNID SBOOKBOOKID
    INTO (CUSTOMER-NAME, CUSTOMER-POSTCODE, CUSTOMER-CITY,
    BOOKING-FLDATE, BOOKING-CARRID, BOOKING-CONNID,
    BOOKING-BOOKID)
    FROM SCUSTOM LEFT OUTER JOIN SBOOK
    ON SCUSTOMID = SBOOKCUSTOMID AND
    SBOOK~FLDATE = '20011015'
    ORDER BY SCUSTOMNAME SBOOKFLDATE.
    WRITE: / CUSTOMER-NAME, CUSTOMER-POSTCODE, CUSTOMER-CITY,
    BOOKING-FLDATE, BOOKING-CARRID, BOOKING-CONNID,
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