Out to condition

Similar Questions

• How to find out whose Conditions are brought together in a book?

  I'm new to FM and I inherited a bookfile with 50 component files. Each has 20 available conditions, but it * seems * that only about 5 are never used.

  Is there a way to search or report on which Conditions are in use in the book (so I can check exactly what conditions are never used)?

  There are two lists you can build that will show you all the tags used in a book condition (and where they are in the game).

  1. from your book file, add > list of > references... and select Condition Tags. This will give you a list in order of occurrence in the entire book.

  2. from your address book to add > Index of > references... and select Condition Tags. This will provide a more concise list of individual tags Condition in alphabetical order, i.e. an Index. [This is probably what you want].

• After using the reset (out-of-condition box) computer laptop Toshiba Windows 8 should 79 updates. What I do before moving on to Windows 8.1?

  My Toshiba Satellite C855D with preinstalled Windows 8 has been reset to a State factory (eveything to remove and reinstall Windows). Everything is good so far. Now updates Windows says to download/install 79 important updates and many of those that are recommended. Is it better to install all updates of recommened front the upgrade to Windows 8.1? Thank you for the answers.

  Yes. Some updates are needed for the 8.1 update, so I would just leave your computer install what it needs.

• Outer join condition

  Hello
  
  Scenario:
  If tableA is related to the tableB, and then get all rows from tableA, excluding the lines in table B with col2 = 'P '.
  If to tableB, tableA is not linked, then get all rows from tableA.

  create table tableA as (
  select '1' col1 from dual
  union all
  select '2' from dual
  union all
  select '3' from dual
  union all
  select '4' from dual
  )
  
  create table tableB as (
  select 'A' Pkey, '1' col1, 'S' col2 from dual
  union all
  select 'B' Pkey, '2' col1, 'S' col2 from dual
  union all
  select 'E' Pkey, '2' col1, null col2 from dual
  union all
  select 'C' Pkey, '3' col1, null col2 from dual
  union all
  select 'D' Pkey, null col1, 'S' col2 from dual
  union all
  select 'F' Pkey, '3' col1, 'P' col2 from dual
  )

  select A.*, B.*
  from tableA A, tableB B
  where A.col1 = B.col1 (+)
  --This is the output I need to get
  1     A     1     S
  2     B     2     S
  2     E     2     (null)
  3     C     3     (null)
  3     F     3     P      --     <---- excluding this row
  4     (null)     (null)     (null)

  and to exclude this line, I added this to my request

  and B.col2 (+) = 'S'

  but after adding only at my request, the result became

  1     A     1     S
  2     B     2     S
  3     (null)     (null)     (null)
  4     (null)     (null)     (null)

  Some of the fields that involve having a value becomes zero.
  What I need to change to get the result I wanted above.
  
  
  Thank you.
  
  Published by: Spongebob 28 June 2011 19:08

  with tableA as
  (
  select '1' col1 from dual
  union all
  select '2' from dual
  union all
  select '3' from dual
  union all
  select '4' from dual
  ),
  tableB as (
  select 'A' Pkey, '1' col1, 'S' col2 from dual
  union all
  select 'B' Pkey, '2' col1, 'S' col2 from dual
  union all
  select 'E' Pkey, '2' col1, null col2 from dual
  union all
  select 'C' Pkey, '3' col1, null col2 from dual
  union all
  select 'D' Pkey, null col1, 'S' col2 from dual
  union all
  select 'F' Pkey, '3' col1, 'P' col2 from dual
  )
  select A.col1 a_col1, B.pkey b_pkey, b.col1 b_col1, b.col2 b_col2
  from tableA A, tableB B
  where A.col1 = B.col1 (+)
    and nvl(b.col2,'S')='S';
  

• Join with conditions of outer join statement

  Hello
  The following was extracted from the 11 GR 2 document guide performance, which I am not quite understand.
  Can anyone provide some examples to clearly explain this. Thank you
  
  For join instructions examples with outer join conditions, the table with the outer join operator must come after the other table in the condition in the join order. The optimizer does not consider join orders that violate this rule. Similarly, when a subquery was transformed into an antijoin or a semi-join, arrays of the subquery must come after these tables in the outer query to which they were connected or block in correlation. However, semijoins and antijoins of hash are override this order condition in certain circumstances
  

• A left outer join

  create table test1 (col1 varchar2 (10));

  create table test2 (col1 varchar2 (10));

  Insert into test1 values('1');

  Insert into test1 values('2');

  Insert into test1 values('3');

  Insert into test2 values('1');

  Select * from test1, test2 where test1.col1 = test2.col1 (+) and test1.col1 = 1;

  Select * from test1 test2 outer join left on test1.col1 = test2.col1 and test1.col1 = 1;

  Why are these two different results?

  Hello

  user13648517 wrote:

  create table test1 (col1 varchar2 (10));

  create table test2 (col1 varchar2 (10));

  Insert into test1 values('1');

  Insert into test1 values('2');

  Insert into test1 values('3');

  Insert into test2 values('1');

  Select * from test1, test2 where test1.col1 = test2.col1 (+) and test1.col1 = 1;

  Select * from test1 test2 outer join left on test1.col1 = test2.col1 and test1.col1 = 1;

  Why are these two different results?

  In the first query, the condition

  test1.Col1 = 1

  is not part of the join condition.  The outer join is done first and then the lines where the test1.col1 <> 1 are deleted.

  In the second query, this condition is part of the join condition.  Test1 lines are not removed results just don't not outer join conditions.

  This shows one of the many benefits of the ANSI join syntax, especially when well formatted:

  Select *.

  of test1

  outer join test2 left test1.col1 = test2.col1

  and test1.col1 = 1;

  This clearly shows that the condition "test1.col1 = 1" is part of the join condition.

• Doubt with OUTER JOIN behavior

  I thought I knew all about OUTER JOINS to this scenario. Can pls someone explain how Oracle behaves in the following SQL code:

  WITH t1 AS
  (SELECT 1 as empid from dual
  union all
  select 2 from dual),
  t2 as
  (select 1 as empid from dual)
  select *
  from t1 
  left outer join
  t2
  on (
  t1.empid = t2.empid AND t1.empid=2
  );
  

  Expected O/P

  EMPID	EMPID_1
  2	NULL VALUE

  Real O/P

  EMPID	EMPID_1
  1	NULL VALUE
  2	NULL VALUE

  In this way...

  SQL > ed
  A written file afiedt.buf

  1 t1 WITH (SELECT empid from all double union 1
  2. Select 2 double),
  3 t2 (select 1 as double empid)
  4  --
  5 Select
  6 T1
  7 * a left outer join t2 (t1.empid = t2.empid) and t1.empid = 2
  SQL > /.

  EMPID EMPID
  ---------- ----------
  2
  1

  the "t1.empid = 2' is part of the OUTER JOIN condition.

  In this way...

  SQL > ed
  A written file afiedt.buf

  1 t1 WITH (SELECT empid from all double union 1
  2. Select 2 double),
  3 t2 (select 1 as double empid)
  4  --
  5 Select
  6 T1
  left outer join 7 t2 on (t1.empid = t2.empid)
  8 * where t1.empid = 2
  SQL > /.

  EMPID EMPID
  ---------- ----------
  2

  the "t1.empid = 2" belongs to the filtering conditions

  When it is part of the OUTER JOIN condition, he says that he should join t2 records where all conditions are met, otherwise the join should not be made, it does not say it to filter the overall results.

  Maybe this helps clarify...

  SQL > ed
  A written file afiedt.buf

  1 t1 WITH (SELECT empid from all double union 1
  2. Select 2 double),
  3 t2 (select empid Union 1 double all the)
  4. Select 2 of the double)
  5  --
  6 select *.
  7 T1
  8 * a left outer join t2 (t1.empid = t2.empid) and t1.empid = 2
  SQL > /.

  EMPID EMPID
  ---------- ----------
  2          2
  1

  Here we have 1 and 2 records in both tables, but the records in t2 are joined only where the record of t1 has an empid 2, so, although t2 has an empid 1, which is not reached because the join condition is not met.

• Different Outer Join query result

  Hi all
  
  Under query gives the expected result
  
  
  Select DECODE (FPOVR. PROFILE_OPTION_VALUE,
  NULL, FPOVS. PROFILE_OPTION_VALUE, FPOVR. ZZ PROFILE_OPTION_VALUE)
  OPS FND_PROFILE_OPTIONS, FND_PROFILE_OPTION_VALUES FPOVR,.
  FND_PROFILE_OPTION_VALUES FPOVS where 1 = 1 and FPOVS. LEVEL_ID (+) = 10001
  and FPOVS. LEVEL_VALUE (+) = 0
  and FPOVS. PROFILE_OPTION_ID (+) IS FRENCH POLYNESIA. PROFILE_OPTION_ID
  and FPOVR. (+) LEVEL_ID = 10003 and FPOVR. LEVEL_VALUE (+) = 21533
  and FPOVR. PROFILE_OPTION_ID (+) IS FRENCH POLYNESIA. PROFILE_OPTION_ID
  and fpo.profile_option_name = 'GL_SET_OF_BKS_ID. '
  
  
  But the following query does not give the expected result.
  
  
  Select DECODE (FPOVR. PROFILE_OPTION_VALUE,
  NULL, FPOVS. PROFILE_OPTION_VALUE, FPOVR. ZZ PROFILE_OPTION_VALUE)
  OPS FND_PROFILE_OPTIONS, FND_PROFILE_OPTION_VALUES FPOVR,.
  FND_PROFILE_OPTION_VALUES FPOVS where 1 = 1 and FPOVS. LEVEL_ID (+) = 10001
  and FPOVS. LEVEL_VALUE (+) = 0
  and FPOVS. PROFILE_OPTION_ID (+) IS FRENCH POLYNESIA. PROFILE_OPTION_ID
  and FPOVR. (+) LEVEL_ID = 10003 and FPOVR. LEVEL_VALUE = 21533
  and FPOVR. PROFILE_OPTION_ID (+) IS FRENCH POLYNESIA. PROFILE_OPTION_ID
  and fpo.profile_option_name = 'GL_SET_OF_BKS_ID. '
  
  Difference between them is a join with external FPOVR. LEVEL_VALUE and hard coded value. Why these query above return different results even if the outer join is made with hard coded in value?
  
  Thank you
  SK

  SK says:
  Hi all

  Under query gives the expected result

  Select DECODE (FPOVR. PROFILE_OPTION_VALUE,
  NULL, FPOVS. PROFILE_OPTION_VALUE, FPOVR. ZZ PROFILE_OPTION_VALUE)
  OPS FND_PROFILE_OPTIONS, FND_PROFILE_OPTION_VALUES FPOVR,.
  FND_PROFILE_OPTION_VALUES FPOVS where 1 = 1 and FPOVS. LEVEL_ID (+) = 10001
  and FPOVS. LEVEL_VALUE (+) = 0
  and FPOVS. PROFILE_OPTION_ID (+) IS FRENCH POLYNESIA. PROFILE_OPTION_ID
  and FPOVR. (+) LEVEL_ID = 10003 and FPOVR. LEVEL_VALUE (+) = 21533
  and FPOVR. PROFILE_OPTION_ID (+) IS FRENCH POLYNESIA. PROFILE_OPTION_ID
  and fpo.profile_option_name = 'GL_SET_OF_BKS_ID. '

  But the following query does not give the expected result.

  Select DECODE (FPOVR. PROFILE_OPTION_VALUE,
  NULL, FPOVS. PROFILE_OPTION_VALUE, FPOVR. ZZ PROFILE_OPTION_VALUE)
  OPS FND_PROFILE_OPTIONS, FND_PROFILE_OPTION_VALUES FPOVR,.
  FND_PROFILE_OPTION_VALUES FPOVS where 1 = 1 and FPOVS. LEVEL_ID (+) = 10001
  and FPOVS. LEVEL_VALUE (+) = 0
  and FPOVS. PROFILE_OPTION_ID (+) IS FRENCH POLYNESIA. PROFILE_OPTION_ID
  and FPOVR. (+) LEVEL_ID = 10003 and FPOVR. LEVEL_VALUE = 21533
  and FPOVR. PROFILE_OPTION_ID (+) IS FRENCH POLYNESIA. PROFILE_OPTION_ID
  and fpo.profile_option_name = 'GL_SET_OF_BKS_ID. '

  Difference between them is a join with external FPOVR. LEVEL_VALUE and hard coded value.

  In fact, the difference is that the condition relating to FPOVR. LEVEL_VALUE to 21533 is part of the outer join condition in the first query, but it is not in the second query.

  Why these query above return different results even if the outer join is made with hard coded in value?

  When you say

  FPOVR.LEVEL_VALUE(+) = 21533
  

  (with a sign +) only the lines of fpovr with level_value = 21533 will join the other table. If there is no such line in fpovr, then a row of output will be generatedd for each row in the other table anyway.
  When you say

  FPOVR.LEVEL_VALUE = 21533
  

  (without a sign +) then the outer join is made using the other conditions, which may result in some lines of the other table who appear in the result set with NULL in places where the values of fpovr would normally appear. Then when the join is complete, the condition is applied, and all these lines with the NULL value instead of fpovr.level_value are deleted because they do not meet the condition "FPOVR. LEVEL_VALUE = 21533 ". The effect is identical to an inner join.

  When you use the old rating of outer join (i.e. with signs +), if any condition involving the fpovr table has a + sign, and then all of the conditions involving table fpovr must have a + sign, otherwise the result will be the same if none of the conditions were a sign +.

• outer join: difference between two queries

  Below two queries which should give the same results in my opinion. I want that all records of u_protocol and the pval.u_protocol_variable_value value if necessary.
  
  Why the outer join in query2 does not work as in query1?
  
  
  Query1:

  select p.u_protocol_id, i.u_protocol_variable_value
  from lims_sys.u_protocol p,
  (
     select pval.u_protocol_id, pval.u_protocol_variable_value
     from lims_sys.u_protocol_variable pvar, lims_sys.u_protocol_value_user pval
     where pvar.u_protocol_variable_id = pval.u_protocol_variable_id
     and pvar.name = 'VALUE_Protocol_Group'
  ) i   
  where p.u_protocol_id  = i.u_protocol_id (+)

  Query2:

  select prt.u_protocol_id, pval.u_protocol_variable_value
  from lims_sys.u_protocol prt, lims_sys.u_protocol_variable pvar, lims_sys.u_protocol_value_user pval
  where pvar.u_protocol_variable_id = pval.u_protocol_variable_id
  and prt.u_protocol_id = pval.u_protocol_id (+)
  and pvar.name = 'VALUE_Protocol_Group'

  Hello

  When you perform an outer join, all conditions on the tables in option must be outer join conditions. Otherwise, the effect is an inner join.
  There is one exception: cascade of outer joins, external-junction b of a and c is outside attached to b. In this case, in the b - c join only columns of c are marked with the sign +.

  Re - write to obtain the same results as query1, query2 like this:

  select  prt.u_protocol_id
  ,      pval.u_protocol_variable_value
  from     lims_sys.u_protocol          prt
  ,      lims_sys.u_protocol_variable     pvar
  ,      lims_sys.u_protocol_value_user pval
  where     pvar.u_protocol_variable_id (+)     = pval.u_protocol_variable_id     -- cascading outer join
  and     prt.u_protocol_id           = pval.u_protocol_id (+)
  and     pvar.name (+)               = 'VALUE_Protocol_Group'     -- (+) added
  ;
  

  I find the syntax ANSI easier to use for all the joins and much easier for outer joins.

• Satellite 1800-314-replacement of the CD-ROM disc

  We have an old Toshiba Satellite 1800-314 the CD-ROM drive appears to be failing - it often will not recognize a CD, or run the applications of one.

  First of all, the CD-ROM drive is not removable, so the laptop must be open. I'm familiar enough with opening, laptops, but in a hardware store or repair is not currently available, I thought I would ask if there is something to know, such as coins out of condition precedent, when I open the case?

  I would also like to take this opportunity to replace the old CD with a DVD - ROM drive. But I read that there is a problem with a possible conflict of IDE bus:
  [REPLACE CD-ROM DRIVE WITH DVD OF TOSHIBA SATELLITE 1800-314 | http://www.tomshardware.co.uk/forum/page-28372_20_0.html] (Their caps, not mine ;))

  Apart from "shorting pin 45 and 47" on the connector, there are other suggestions?

  The laptop belongs to my little girl, who will give me + hell + if it is out of use for too long, or worse, break... X (so I would like to correct it at once.

  Thanks in advance.

  Hello

  The fact is that each optical disc drive cannot be used on every laptop. you are right and there is nothing other than the compatibility issue. The assignment of pins is the same for all readers and ratings and because of this, you must use compatible discs only.

  The good thing is that Toshiba has listed 6-disc compatible optical disc that can be used with this old Satellite notebook.
  All of them are listed in the paper manuals user s. check it out.

• Drift of signal on SCC-SG24

  We have equipment OR measurements of charge 3 cell, it consists of a NOR-6220, SC-2345, SCC-SG24 (2) and (2) SCC-LP02. During a recent calibration system, I noticed that all channels showed considerable drift when the load cells were out load condition (and probably under load as well but I didn't not test to check). For example, a cell of weighing capacity of 25,000 pounds was a no load measure 142 pounds and an hour and forty minutes later, it measures 56 pounds. The degree of drift seems to be linked to the section how long the computer containing the material was turned on, but continues well beyond the long warm-up period specified in the material master records. in the previous example, the load cell was measured after an hour and a half additional 25 pounds. I ran a test using a simulator and has seen the same drift, it must ensure the material and not the load cells causing the drift. I also have one of the load cells connected to different materials (cDAQ-9178 and NOR-9237) and have not seen the same degree of drift.

  Basically, after warming of the system for 4 hours stabilized enough steps to make all without consequence drift continues for the duration of the test. This amount of warming seems excessive in relation to the other material OR that we have. Should I be worried? Others know about this?

  Thank you for contributing to knowledge or information on the question, as I am experiencing.
  Tessa

  I was finally able to remove the drift by removing the low pass filter of the installation. Appearently the SG24 and LP02 don't play well together.

• How do I either set channel name of a graph of cluster or get a 1 d waveform chart to start different time (t0) for each line of the waveform?

  I have a chart single 2D out a conditional of a loop indexing output such that over the rows of the table are not the same size. -Basically, they have different starting and ending points and sizes. -Not so concerned by the endpoints, because as soon as I get correctly starting points, everything shows fall in place.

  The challenge is that when I try to have the variable t0 (start time) for each line, the wave of construction vi form would always keep each line at the same starting point.

  I used the approach bundled with success (the cluster approach ensures that each waveform starts at different times according to the guidelines of my t0 defined for each line) but then I'm not able to get in the channel names I could make using the graphical approach (wave generation) waveform.

  In essence what I get here, I'm losing here.

  Because I don't want one of my mentors, Bob and Altenbach have fed up with me I have attached a vi this time

  Attached VI shows a combination of the two attempts (first with the waveform graph) and then with the graph of cluster

  1. with the first (graphical waveform), I get my channel names as you wish, but the alignment of the wave is not correct.

  2. with the second (graphical cluster), the alignment is good, but I can't do the names of channel in the chart, even if they are present in the cluster.

  I read some reviews that mentions that attributes can be displayed with waveform and data Dynamics (not clsuters) so I guess that's why.

  I saw another report indicating the start time for a waveform 1 d will always remain the same for the rest lines defined for the first line even if changed for the following lines in a loop.

  So I guess my question is: what is the way around questions like that?

  First of all, let me be the first to the congratulate and thank you for finally posting a code!  I'm not 100% certain I understand your question or your code, but I have an idea, perhaps, of what you want to do, so I wrote a little VI who made something simple that could be relevant.

  You mention waveforms of different lengths and beginning at different times.  You also want everyone to have a unique attribute (although I'm not sure what you want to do with the attribute).  So, I did the following:

  1. Generated an array of 100-sample random to represent one second of a waveform.
  2. Created 4 waveforms on this 100-sample basis.  The first waveform (channel 0) is just these 100 points.  The second, 1 channel, is the concatenation of string 0 with the base of 100 samples, or a waveform "double".  Channel 2 is 1 string concatenated with the base, and channel 3 is 2 string concatenated with the base.
  3. In order to trace the four channels that they rest 'on' the other, the waveform has the number of the channel added to it.  Channel 3 is 3 + (4 copies of the basis of 100 points), a waveform 400-point random centered around a shift of 3.
  4. All channels have dt value 0.01 (but I guess I could have varied, as well).
  5. To make the channels start at different times, I started channel N N seconds before channel 0 (by subtracting the index of the loop, I, T0).
  6. For each channel, I created an attribute called "Chan" equal to "Channel N" (where N = 0, 1, 2 or 3, as the case may be).

  This is the plot that results. Scale X is the absolute time value (no Date) using the 24-hour HH: mmS format.  You can see that the plots are 1, 2, 3 and 4 seconds of time, and are offset from each other by a second.  I used the trace attributes to change the name to the respective attribute.

  

  The code to do this is very simple - I almost don't need to show it, because I think it is completely described by the text above, but this is here:

  

  Now, it was not that much faster that some of your previous posts, when you refused to your postcode, "guess us" what you wanted (but not to not correctly guess), you tried to "push" us in the right direction (still refuses to post code), and no one seemed very happy?

  Bob Schor

• driven by the structure of the events

  Hi all

  Please take a look at my VI...

  what I do in it use the same structure to control something that is that when I press the botton that I start getting the random elements of the array that I have bulid. but I can't seem to stop inside any looping in the first inning when I press the stop botton... What am I doing worng... and if a remove the while loop it only run the loop once. I want the event up to the time that I support permanently on the stop or any botton.

  concerning

  All you need is a loop: the outer loop. You can manipulate the time-out to conditionally generate your values.

  Here's a quick example. See if it makes sense.

• Impossible to put the select statement

  Oracle 10g

  Hi gurus

  I appreciate if someone can help me out, I have the sample data...

  Examples of data

  cases like
  (
  Select case_id 100 of all the double union
  Select 200 double
  )
  ,
  commands like
  (
  Select case_id 100, 111 order_id in union double all the
  Select double 200 112
  ),
  conditions that
  (
  Select 300 order_id, 'TTT' double description
  Union of all the
  Select 111, 'LTS' double
  Union of all the
  Select 111, 'TTT' from dual
  Union of all the
  Select 111, "DAN" of the double
  Union of all the
  Select 111, 'XXX' of the double
  )

  Select * from ca case, ordered oo, conditions cc
  where ca.case_id = oo.case_id
  and = oo.order_id cc.order_id (+)
  ;

  The query result

  DESCRIPTION, ORDER_ID, CASE_ID, ORDER_ID_1, CASE_ID_1

  100                100                111           111                     LTS

  100                100                111           111                     TTT

  100                100                111           111                     DAN

  100                100                111           111                     XXX

  200 200 112 NULL NULL

  My output desired is as below

  Desired output

  DESCRIPTION, ORDER_ID, CASE_ID, ORDER_ID_1, CASE_ID_1

  100                100                111           111                     LTS

  100                100                111           111                     DAN

  200 200 112 NULL NULL

  So far, I've written the following query, but I get the error message...

  Query so far

  cases like
  (
  Select case_id 100 of all the double union
  Select 200 double
  )
  ,
  commands like
  (
  Select case_id 100, 111 order_id in union double all the
  Select double 200 112
  ),
  conditions that
  (
  Select 300 order_id, 'TTT' double description
  Union of all the
  Select 111, 'LTS' double
  Union of all the
  Select 111, 'TTT' from dual
  Union of all the
  Select 111, "DAN" of the double
  Union of all the
  Select 111, 'XXX' of the double
  )

  Select * from ca case, ordered oo, conditions cc

  where ca.case_id = oo.case_id

  and (cc.order_id (+) = oo.order_id

  and description in ("LTS", "DAN")

  - or (+) description = "DAN."

  )

  ;

  Rules

  If order_id find in the condition array then only to search the records of the array of conditions that have conditions.description in ("LTS", "DAN")

  Note

  I think I can achieve this by using decode on select level, but I don't want to use decode...

  Hello

  You just do cc.description IN ("LTS", "DAN") is part of the outer join condition.  Here's one way:

  SELECT *.

  From case ca

  JOIN Orders oo ON oo.case_id = ca.case_id

  LEFT OUTER JOIN conditions cc ON cc.order_id = oo.order_id

  AND cc.description IN ('LTS', 'DAN')

  ;

  Out, as requested:

  CASE_ID CASE_ID ORDER_ID ORDER_ID DESCRIPTION

  ------- ------- -------- -------- ------------

  100 100 111 111 LTS

  DAN 100 100 111 111

  200 200 112

  What is the problem with DECODE?  As you can see, you need this problem, but if it was the easiest way and the most effective to get the results you want, why not use it?

• Change the display

  I created a view as follows:
  

  CREATE OR REPLACE VIEW plsql101_sales_per_person_v AS
  SELECT pers.first_name || ' ' || pers.last_name SALESPERSON,
         purc.product_name,
         purc.purchase_date,
         purc.quantity
  FROM   plsql101_person  pers,
         plsql101_purchase purc
  WHERE  pers.person_code = purc.salesperson (+)
  ;

  Is it possible to change the view above so that I can remove the outer join condition to be fair as below
  
  >
  WHERE pers.person_code = purc.salesperson

  No, you cannot MODIFY a view to modify the query.

  You would just take your statement CREATE or REPLACE, remove the outer join and run to replace the view.

  Justin