email from iCloud without using the iCloud

I just created a new iCloud email address. However, I prefer not to use the iCloud to download the messages. When I disable "Mail" in the preferences system - iCloud, my new mail mailbox goes. I saw this page: iCloud: server for e-mail clients - Apple support settings and suspect that the answer may lie in configure server settings of mail by hand (email - preferences - accounts). It is the right approach, or is there another way? I am running Yosemite on an iMac. Thank you!

You can explain further. What makes you think that it is possible to use iCloud email address without using iCloud?

Tags: iCloud

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SDO_NN cannot be assessed without using the index when put inside subquery

Hi all

I met a problem when you use the function sdo_nn to find the nearest neighbor. Here is my scenario:

_ I have 2 customer and store tables.

Customer table _ a client_ID and a 2D sdo_geom point

_ Store table has store_ID and a 2D polygon sdo_geom.

In the beginning, I have this query to find the nearest store to each customer as below:

Select s.STORE_ID, c.CLIENT_ID

store customer, s c

where sdo_nn (s.MYPOLYGON, c.MYPOINT, 'sdo_num_res = 1', 1) = "TRUE";

_It works as expected when it returns a table showing the nearest store each customer.

_Now I want to count the number of customers who have the same nearest store:

Select / * + INDEX (store store_spatial_idx, client_spatial_idx client) * / count (nearest_store. CLIENT_ID)

from (select s.STORE_ID, c.CLIENT_ID

store customer, s c

where sdo_nn (s.MYPOLYGON, c.MYPOINT, 'sdo_num_res = 1', 1) = "TRUE") nearest_store

Group of nearest_store. STORE_ID;

This query generates the following error:

Error report-

SQL error: ORA-13249: SDO_NN cannot be assessed without using the index

ORA-06512: at the 'MDSYS. MD", line 1723

ORA-06512: at the 'MDSYS. MDERR", line 17

ORA-06512: at the 'MDSYS. PRVT_IDX', line 9

13249 00000 - '%s '.

I'm pretty new to spatial databases and hope get help to go further. Thank you in advance!

Hello Pinball,

Oracle space tends to be a quite complex with many variables and moving parts. We chatted about the group to a sort of FAQ or guidelines to help people like you submit questions that actually answers. First of all, you really have to tell us the version of Oracle you are using. Particularly the problems involving the optimizer, version down to the exact defined patch number is a good idea. Secondly, you took the time to submit the question so I guess you want a response. If you really want to see the answer and then providing an example is one of the most important things that you can do. I'm going to do here for you, but in general people on this forum come and go and are often pushed into lurkitude, so if you want the coax to provide you with an example of work is the key.

DROP TABLE store1 PURGE;
CREATE TABLE store1(
    store_id INTEGER NOT NULL
   ,shape    MDSYS.SDO_GEOMETRY
   ,PRIMARY KEY(store_id)
);

DROP TABLE client2 PURGE;
CREATE TABLE client2(
    client_id INTEGER NOT NULL
   ,shape    MDSYS.SDO_GEOMETRY
   ,PRIMARY KEY(client_id)
);

CREATE OR REPLACE PROCEDURE seeder(
    p_client_count IN NUMBER
   ,p_store_count IN NUMBER
)
AS
  sdo_foo MDSYS.SDO_GEOMETRY;
  int_counter NUMBER;
  FUNCTION random_point
  RETURN MDSYS.SDO_GEOMETRY
  AS
      num_x1 NUMBER;
      num_y1 NUMBER;

  BEGIN
      num_x1 := dbms_random.value(-179,179);
      num_y1 := dbms_random.value(-89,89);

      RETURN MDSYS.SDO_GEOMETRY(
          2001
         ,8265
         ,MDSYS.SDO_POINT_TYPE(
              num_x1
             ,num_y1
             ,NULL
          )
         ,NULL
         ,NULL
      );

  END random_point;

BEGIN
  int_counter := 1;
  FOR i IN 1 .. p_client_count
  LOOP
      -- Create a client point
      sdo_foo := random_point();
      INSERT INTO client2
      VALUES (
          int_counter
         ,sdo_foo
      );
      int_counter := int_counter + 1;

  END LOOP;

  int_counter := 1;
  FOR i IN 1 .. p_store_count
  LOOP
      -- Create a store polygon of some kind
      sdo_foo := MDSYS.SDO_GEOM.SDO_ARC_DENSIFY(
          MDSYS.SDO_GEOM.SDO_BUFFER(
              random_point()
             ,5000
             ,0.05
          )
         ,0.05
         ,'arc_tolerance=0.05'
      );
      INSERT INTO store1
      VALUES (
          int_counter
         ,sdo_foo
      );
      int_counter := int_counter + 1;

  END LOOP;

  COMMIT;

END seeder;
/

BEGIN
  seeder(10000,200);
END;
/

BEGIN
  INSERT INTO user_sdo_geom_metadata(
      table_name
     ,column_name
     ,diminfo
     ,srid
  ) VALUES (
      'STORE1'
     ,'SHAPE'
     ,MDSYS.SDO_DIM_ARRAY(MDSYS.SDO_DIM_ELEMENT('X',-180,180,.05),MDSYS.SDO_DIM_ELEMENT('Y',-90,90,.05))
     ,8265
  );

  COMMIT;

EXCEPTION
  WHEN OTHERS
  THEN
      NULL;

END;
/

BEGIN
  INSERT INTO user_sdo_geom_metadata(
      table_name
     ,column_name
     ,diminfo
     ,srid
  ) VALUES (
      'CLIENT2'
     ,'SHAPE'
     ,MDSYS.SDO_DIM_ARRAY(MDSYS.SDO_DIM_ELEMENT('X',-180,180,.05),MDSYS.SDO_DIM_ELEMENT('Y',-90,90,.05))
     ,8265
  );

  COMMIT;

EXCEPTION
  WHEN OTHERS
  THEN
     NULL;

END;
/

CREATE INDEX store1_spx ON store1
(shape)
INDEXTYPE IS MDSYS.SPATIAL_INDEX
NOPARALLEL;

CREATE INDEX client2_spx ON client2
(shape)
INDEXTYPE IS MDSYS.SPATIAL_INDEX
NOPARALLEL;

/* Works as expected */
SELECT
s.store_id
,c.client_id
,MDSYS.SDO_NN_DISTANCE(1)
FROM
store1 s
,client2 c
WHERE
MDSYS.SDO_NN(
    s.shape
   ,c.shape
   ,'sdo_num_res=1'
   ,1
) = 'TRUE';

/* No worky? Works for me */
SELECT
ns.store_id
,COUNT(ns.client_id)
FROM (
   SELECT
    s.store_id
   ,c.client_id
   FROM
    store1 s
   ,client2 c
   WHERE
   MDSYS.SDO_NN(
       s.shape
      ,c.shape
      ,'sdo_num_res=1'
      ,1
   ) = 'TRUE'
) ns
GROUP BY
ns.store_id
ORDER BY
ns.store_id;

So I wrote this about 12 c (12.1.0.2.0) and everything works fine for me. Then I moved back from 11 GR 2 (11.2.0.4.0) and of course, there are questions. So I guess that you don't use flavor of 11g. So at this point we can look at the docs and see for 11g, have you often need to specify which table is the head and that is the one that has the spatial index to use.
http://docs.Oracle.com/CD/E11882_01/AppDev.112/e11830/sdo_operat.htm#SPATL1032

Its rather interesting that the optimizer of 12 c knows what you want, when I had to squint myself at your request and to play a little with the refining. Note that SDO_NN is sensitive, because the geometry of the main table should come second in the operator. I did not know that on the top of my head.


SELECT
/*+ LEADING(c) INDEX(s store1_spx)  */
 s.store_id
,c.client_id
,MDSYS.SDO_NN_DISTANCE(1)
FROM
 store1 s
,client2 c
WHERE
MDSYS.SDO_NN(
    s.shape
   ,c.shape
   ,'sdo_num_res=1'
   ,1
) = 'TRUE';

SELECT
 ns.store_id
,COUNT(ns.client_id)
FROM (
   SELECT
   /*+ LEADING(c) INDEX(s store1_spx)  */
    s.store_id
   ,c.client_id
   ,MDSYS.SDO_NN_DISTANCE(1)
   FROM
    store1 s
   ,client2 c
   WHERE
   MDSYS.SDO_NN(
       s.shape
      ,c.shape
      ,'sdo_num_res=1'
      ,1
   ) = 'TRUE'
) ns
GROUP BY
ns.store_id
ORDER BY
ns.store_id;

So I think that is your answer. Give it a shot and see if this fits the Bill. Of course, moving to 12 c would be useful for such things. It would be interesting to collect more examples of this kind of space thing where 12 c is the answer. Also, would be nice if we could mark somehow this discussion as applying only to 11g and earlier versions.

See you soon,.

Paul

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How can I download Adobe Premiere elements 11 without using the installer from the DVD

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alternatives without using the regexp_replace function

SELECT
REGEXP_REPLACE (phone_number,
"([[: digit:]] {3})------." ([[: digit:]] {3}). ([[: digit:]] {4})',
"\2-\3 (\1)") "REGEXP_REPLACE.
Employees;
is there another without using the regexp_replace function...

Another way

(eliminating instr... function as your problem focuses on one length fixed)

WITH t AS (SELECT '112.345.6789' FROM DUAL str).

TT AS (SELECT REPLACE (str, '.') t str)

SELECT ' (')

|| SUBSTR (str, 1, 3)

|| ') '

|| SUBSTR (str, 4, 3)

|| '-'

|| SUBSTR (str, 7)

TT;

See you soon,.

Manik.

SDO_NN giving ORA-13249: SDO_NN cannot be assessed without using the index
Hi people,

I do not understand why the SDO_NN gives ORA-13249 in circumstances.

SQL > SELECT SlavaTest WHERE SDO_NN s s.title (s.geometry, SDO_GEOMETRY (2001, 4326, SDO_POINT (14.0, 49.0, NULL), null, null)) = 'TRUE' and title like '%' and rownum < 10;

TITLE
--------------------------------------------------------------------------------
MultiPoint_305199
LineString_691779
MultiPolygon_180478
MultiPolygon_358113
MultiPolygon_53008
MultiPolygon_249905
MultiPolygon_204076
MultiPolygon_636994
MultiPoint_464514

9 selected lines.

SQL > SELECT SlavaTest WHERE SDO_NN s s.title (s.geometry, SDO_GEOMETRY (2001, 4326, SDO_POINT (14.0, 49.0, NULL), null, null)) = 'TRUE' and timestamp > = to_timestamp (January 6, 2011 ', ' dd/mm/yyyy') and rownum < 10;
SELECT SlavaTest WHERE SDO_NN s s.title (s.geometry, SDO_GEOMETRY (2001, 4326, SDO_POINT (14.0, 49.0, NULL), null, null)) = 'TRUE' and timestamp > = to_timestamp (January 6, 2011 ', ' dd/mm/yyyy') and rownum < 10
*
ERROR on line 1:
ORA-13249: SDO_NN cannot be assessed without using the index
ORA-06512: at the 'MDSYS. MD", line 1723
ORA-06512: at the 'MDSYS. MDERR", line 17
ORA-06512: at the 'MDSYS. PRVT_IDX', line 49

The spatial index is created with:
CREATE the INDEX SlavaTest_geometry_idx_spatial ON SlavaTest (geometry) INDEXTYPE IS mdsys.spatial_index;

'Title' and 'timestamp' columns have an index.

Note the query comes from Hibernate and I can't change it's arbitrary.
Slava2 wrote:
What this means - there is a bug in Oracle?

Well, it could probably be considered a, but [url http://docs.oracle.com/cd/E11882_01/appdev.112/e11830/sdo_operat.htm#i78067] documentation on SDO_NN warns you:

Documentation says:
However, if the column in the WHERE clause predicate specifies a non-space column in the table for geometry1 with an associated index, make sure that this index is not used by specifying the NO_INDEX indicator for this index.

See you soon,.
Stefan

help in the application without using the pivot function

Hi gurus,

Can you please help write the query without using the PIVOT function.

 WITH indata AS
         (SELECT 1 sn, '123:456,789,323:456,213,344,345:5454' x
            from dual
          UNION ALL
          SELECT 2, 'abcd:fgrfr,rfrf,rfred,tg:tg,tg:ophhh,op,vdfgbh:poijn' x from dual),
        instr AS
         (SELECT a.SN,
                 rownum RN,
                 B.column_value || CASE
                   WHEN B.column_value NOT LIKE '%:%' THEN
                    ':'
                 END column_value 
            FROM indata a,
                 TABLE(CAST(multiset
                            (SELECT trim(SUBSTR(x,
                                                (CASE LEVEL
                                                  WHEN 1 THEN
                                                   1
                                                  ELSE
                                                   instr(x, ',', 1, LEVEL - 1) + 1
                                                END),
                                                (CASE instr(x, ',', 1, LEVEL)
                                                  WHEN 0 then
                                                   4000
                                                  ELSE
                                                   instr(x, ',', 1, LEVEL) - 1 -
                                                   (CASE LEVEL
                                                      WHEN 1 THEN
                                                       0
                                                      ELSE
                                                       instr(x,
                                                             ',',
                                                             1,
                                                             level - 1)
                                                    END)
                                                END)))
                               FROM dual
                             CONNECT BY level <=
                                        LENGTH(x) -
                                        LENGTH(REPLACE(x, ',', '')) + 1) AS
                            sys.odcivarchar2list)) b)
        SELECT  col1_val2,
                col1_val1,
                col2_val2,
                col2_val1,              
               systimestamp
          FROM (select SN,
                      ROW_NUMBER() over(partition by SN order by RN) RN,
                       SUBSTR(column_value, INSTR(column_value, ':') + 1) VAL1,
                       substr(column_value, 1, instr(column_value, ':') - 1) val2
                  FROM instr ) PIVOT(MAX(VAL1) VAL1, MAX(VAL2) VAL2 FOR SN IN(1 as col1,2 as col2));

Thanks in advance!...

Any help would be appreciated.

user590978 wrote:

Can you please help write the query without using the PIVOT function.

with indata as (
                select 1 sn,'123:456,789,323:456,213,344,345:5454' x from dual union all
                select 2, 'abcd:fgrfr,rfrf,rfred,tg:tg,tg:ophhh,op,vdfgbh:poijn' x from dual
               ),
         t1 as (
                select  x
                  from  indata
                  where sn = 1
               ),
         t2 as (
                select  x
                  from  indata
                  where sn = 2
               ),
         t3 as (
                select  level lvl,
                        substr(
                               x,
                               instr(',' || x,',',1,level),
                               instr(x || ',',',',1,level) - instr(',' || x,',',1,level)
                              ) col1
                  from  t1
                  connect by level <= length(x) - length(translate(x,'A,','A')) + 1
               ),
         t4 as (
                select  level lvl,
                        substr(
                               x,
                               instr(',' || x,',',1,level),
                               instr(x || ',',',',1,level) - instr(',' || x,',',1,level)
                              ) col2
                  from  t2
                  connect by level <= length(x) - length(translate(x,'A,','A')) + 1
               ),
         t5 as (
                select  lvl,
                        substr(col1,1,instr(col1 || ':',':') - 1) col1_val1,
                        substr(col1,instr(col1 || ':',':') + 1) col1_val2
                  from  t3
               ),
         t6 as (
                select  lvl,
                        substr(col2,1,instr(col2 || ':',':') - 1) col2_val1,
                        substr(col2,instr(col2 || ':',':') + 1) col2_val2
                  from  t4
               )
select  col1_val1,
        col1_val2,
        col2_val1,
        col2_val2,
        systimestamp
  from      t5
        full join
            t6
        on (
            t6.lvl = t5.lvl
           )
  order by coalesce(t5.lvl,t6.lvl)
/

COL1_VAL1  COL1_VAL2  COL2_VAL1  COL2_VAL2  SYSTIMESTAMP
---------- ---------- ---------- ---------- -------------------------------------
123        456        abcd       fgrfr      23-OCT-12 11.50.12.015000 AM -04:00
789                   rfrf                  23-OCT-12 11.50.12.015000 AM -04:00
323        456        rfred                 23-OCT-12 11.50.12.015000 AM -04:00
213                   tg         tg         23-OCT-12 11.50.12.015000 AM -04:00
344                   tg         ophhh      23-OCT-12 11.50.12.015000 AM -04:00
345        5454       op                    23-OCT-12 11.50.12.015000 AM -04:00
                      vdfgbh     poijn      23-OCT-12 11.50.12.015000 AM -04:00

7 rows selected.

SQL>

SY.

How to view the line of columns without using the pivot keyword

Hello
could someone help me how to display lines in columns without using the keyword pivot and actuall is my scenario, iam having two tables with names and examples of data is shown below

MIDDLE MINAME TASKID TASKNAME IDENTIFICATION PROJECT

1 PROJ1 1 AA 100 PR1_TASK1
1 PROJ1 3 CC PR1_TASK3 102
1 PROJ1 DD 4 103 PR1_TASK4
1 PROJ1 EE 5 104 PR1_TASK5
1 PROJ1 6 105 FF PR1_TASK6
2 PROJ2 EE 5 114 PR2_TASK1
2 PROJ2 6 115 FF PR2_TASK2
2 PROJ2 GG 7 116 PR2_TASK3
2 PROJ2 HH 8 117 PR2_TASK4
2 PROJ2 9 118 PR2_TASK5 JJ
2 PROJ2 KK 10 119 PR2_TASK6
2 PROJ2 1 AA PR2_TASK7 120

The output should show the project and County of tasks at a given stage as shown below

project AA BB CC DD EE FF GG HH JJ KK
1 2 0 1 5 3 2 0 2 1 0
2 1 2 0 2 1 0 2 4 3 1

Thanks in advance,
VVR

CREATE TABLE pivot_test (
ID           NUMBER,
PROJECT      VARCHAR2(10),
MID          NUMBER,
MINAME       VARCHAR2(5),
TASKID       NUMBER,
TASKNAME     VARCHAR2(10)
);

INSERT INTO pivot_test VALUES (1, 'PROJ1', 1,  'AA', 100, 'PR1_TASK1');
INSERT INTO pivot_test VALUES (1, 'PROJ1', 3,  'CC', 102, 'PR1_TASK3');
INSERT INTO pivot_test VALUES (1, 'PROJ1', 4,  'DD', 103, 'PR1_TASK4');
INSERT INTO pivot_test VALUES (1, 'PROJ1', 5,  'EE', 104, 'PR1_TASK5');
INSERT INTO pivot_test VALUES (1, 'PROJ1', 6,  'FF', 105, 'PR1_TASK6');
INSERT INTO pivot_test VALUES (2, 'PROJ2', 5,  'EE', 114, 'PR2_TASK1');
INSERT INTO pivot_test VALUES (2, 'PROJ2', 6,  'FF', 115, 'PR2_TASK2');
INSERT INTO pivot_test VALUES (2, 'PROJ2', 7,  'GG', 116, 'PR2_TASK3');
INSERT INTO pivot_test VALUES (2, 'PROJ2', 8,  'HH', 117, 'PR2_TASK4');
INSERT INTO pivot_test VALUES (2, 'PROJ2', 9,  'JJ', 118, 'PR2_TASK5');
INSERT INTO pivot_test VALUES (2, 'PROJ2', 10, 'KK', 119, 'PR2_TASK6');
INSERT INTO pivot_test VALUES (2, 'PROJ2', 1,  'AA', 120, 'PR2_TASK7');

SELECT ID as PROJECT,
       SUM(DECODE(miname, 'AA', 1, 0)) AS AA,
       SUM(DECODE(miname, 'BB', 1, 0)) AS BB,
       SUM(DECODE(miname, 'CC', 1, 0)) AS CC,
       SUM(DECODE(miname, 'DD', 1, 0)) AS DD,
       SUM(DECODE(miname, 'EE', 1, 0)) AS EE,
       SUM(DECODE(miname, 'FF', 1, 0)) AS FF,
       SUM(DECODE(miname, 'GG', 1, 0)) AS GG,
       SUM(DECODE(miname, 'HH', 1, 0)) AS HH,
       SUM(DECODE(miname, 'JJ', 1, 0)) AS JJ,
       SUM(DECODE(miname, 'KK', 1, 0)) AS KK
FROM   pivot_test
GROUP BY ID;

PROJECT AA BB CC DD EE FF GG HH JJ KK
------- -- -- -- -- -- -- -- -- -- --
      1  1  0  1  1  1  1  0  0  0  0
      2  1  0  0  0  1  1  1  1  1  1

count the number of lines without using the count function
Hello
How count the number of lines without using the count function?

Thank you
Assuming that your statistics table are updated:
```
select num_rows
from   all_tables
where  table_name = 'YOUR_TABLE_NAME_IN_UPPERCASE';  
```
http://download.Oracle.com/docs/CD/B19306_01/server.102/b14237/statviews_2105.htm#REFRN20286

email from iCloud without using the iCloud

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Select s.STORE_ID, c.CLIENT_ID

store customer, s c

where sdo_nn (s.MYPOLYGON, c.MYPOINT, 'sdo_num_res = 1', 1) = "TRUE";

Select / * + INDEX (store store_spatial_idx, client_spatial_idx client) * / count (nearest_store. CLIENT_ID)

from (select s.STORE_ID, c.CLIENT_ID

store customer, s c

where sdo_nn (s.MYPOLYGON, c.MYPOINT, 'sdo_num_res = 1', 1) = "TRUE") nearest_store

Group of nearest_store. STORE_ID;

Error report-

SQL error: ORA-13249: SDO_NN cannot be assessed without using the index

ORA-06512: at the 'MDSYS. MD", line 1723

ORA-06512: at the 'MDSYS. MDERR", line 17

ORA-06512: at the 'MDSYS. PRVT_IDX', line 9

13249 00000 - '%s '.

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