Why I have two different execution plans for the same query on two different servers

Hello everyone.

I need your help to solve the problem quickly.

In a nutshell, we have two servers that have the same version of Oracle RDBMS (11.2.0.4 EE). One of them for purposes of development and another is a production one.

We have therefore two different execution plans for the same query executed on both servers. The only case of execution is OK and another is too slow.

So I have to slow down the work of query using the same scheme to explain that young.

Fence wire.

Tags: Database

Similar Questions

explain plan for the same query diff

Hi experts,
Please, help me understand explain the plan. I have tow Server (server and two server). The server are same table, even the type of database, even version Oracle (gr 11 (2), same operating system (linux Redhat 5.5) and same table and index.
but when I explain the plan for the same query on the two server. I got diff--diff to explain the plan. reason it has different, according to my understanding, it should be same. explain please, I share the explain plan and lower indices for the two server.
Server a

SQL > col COLUMN_NAME format a20

SQL > select index_name, column_name, position_colonne from user_ind_columns where table_name = 'LOAN_RUNNING_DETAILS_SOUTH"of order 1.

INDEX_NAME COLUMN_NAME POSITION_COLONNE

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

DATE_IND1_S LOANDATETIME 1

IND_MSI_LN_LNS1_S MSISDN 1

IND_MSI_LN_LNS1_S LOANDATETIME 2

IND_MSI_LN_LNS1_S LOANSTATUS 3

LAST_INDEX L_INDX_MSISDN_S 1

MSISDN L_INDX_MSISDN_S 2

SQL > select decode (status, 'N/a', 'Part Hdr', 'Global') ind_type, index_name, NULL nom_partition, status

2 from user_indexes where table_name = 'LOAN_RUNNING_DETAILS_SOUTH '.

3 union

4. Select 'Local' ind_type, index_name, nom_partition, status

5 to user_ind_partitions where index-name in (select index_name in user_indexes where table_name = 'LOAN_RUNNING_DETAILS_SOUTH')

6 order of 1,2,3;

IND_TYPE INDEX_NAME NOM_PARTITION STATUS

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

Global DATE_IND1_S VALID

Global IND_MSI_LN_LNS1_S VALID

Global L_INDX_MSISDN_S VALID

SQL > explain plan for the small circle of MSISDN, TID, of LOAN_RUNNING_DETAILS_SOUTH where LOANDATETIME < = sysdate-2 and LOANDATETIME > sysdate-15 and LOANTYPE = 1;

He explained.

SQL > SQL > set line 200

@?/rdbms/admin/utlxpls.sql

SQL >

PLAN_TABLE_OUTPUT

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

Hash value of plan: 3659874059

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

| ID | Operation | Name                       | Lines | Bytes | Cost (% CPU). Time | Pstart. Pstop |

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

|   0 | SELECT STATEMENT |                            | 1448K |    58 M | 21973 (2) | 00:04:24 |       |       |

|* 1 | FILTER |                            |       |       |            |          |       |       |

|   2.   PARTITION LIST ALL |                            | 1448K |    58 M | 21973 (2) | 00:04:24 |     1.    11.

|* 3 |    TABLE ACCESS FULL | LOAN_RUNNING_DETAILS_SOUTH | 1448K |    58 M | 21973 (2) | 00:04:24 |     1.    11.

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

PLAN_TABLE_OUTPUT

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

Information of predicates (identified by the operation identity card):

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

1 - filter(SYSDATE@!-2>SYSDATE@!-15)

3 - filter("LOANTYPE"=1 AND "LOANDATETIME">SYSDATE@!-15 AND "LOANDATETIME"<=SYSDATE@!-2)

16 selected lines.

Second server

SQL > select index_name, column_name, position_colonne from user_ind_columns where table_name = 'LOAN_RUNNING_DETAILS_SOUTH"of order 1.

INDEX_NAME COLUMN_NAME POSITION_COLONNE

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

DATE_IND1_S LOANDATETIME 1

IND_MSI_LN_LNS1_S MSISDN 1

IND_MSI_LN_LNS1_S LOANDATETIME 2

IND_MSI_LN_LNS1_S LOANSTATUS 3

LAST_INDEX L_INDX_MSISDN_S 1

MSISDN L_INDX_MSISDN_S 2

SQL > select decode (status, 'N/a', 'Part Hdr', 'Global') ind_type, index_name, NULL nom_partition, status

2 from user_indexes where table_name = 'LOAN_RUNNING_DETAILS_SOUTH '.

Union

3 4 Select 'Local' ind_type, index_name, nom_partition, status

5 to user_ind_partitions where index-name in (select index_name in user_indexes where table_name = 'LOAN_RUNNING_DETAILS_SOUTH')

6 order of 1,2,3;

IND_TYPE INDEX_NAME NOM_PARTITION STATUS

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

Global DATE_IND1_S VALID

Global IND_MSI_LN_LNS1_S VALID

Global L_INDX_MSISDN_S VALID

SQL > explain plan for the small circle of MSISDN, TID, of LOAN_RUNNING_DETAILS_SOUTH where LOANDATETIME < = sysdate-2 and LOANDATETIME > sysdate-15 and LOANTYPE = 1;

SQL > set line 200

@?/rdbms/admin/utlxpls.sql

SQL >

PLAN_TABLE_OUTPUT

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

Hash value of plan: 1161680601

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

| ID | Operation | Name                       | Lines | Bytes | Cost (% CPU). Time | Pstart. Pstop |

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

|   0 | SELECT STATEMENT |                            |     2.    84.     5 (0) | 00:00:01 |       |       |

|* 1 | FILTER                             |                            |       |       |            |          |       |       |

|* 2 |   TABLE ACCESS BY INDEX ROWID | LOAN_RUNNING_DETAILS_SOUTH |     2.    84.     5 (0) | 00:00:01 | ROWID | ROWID |

|* 3 |    INDEX RANGE SCAN | DATE_IND1_S |     2.       |     3 (0) | 00:00:01 |       |       |

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

PLAN_TABLE_OUTPUT

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

Information of predicates (identified by the operation identity card):

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

1 - filter(SYSDATE@!-2>SYSDATE@!-15)

2 - filter ("LOANTYPE" = 1)

3 - access("LOANDATETIME">SYSDATE@!-15 AND "LOANDATETIME"<=SYSDATE@!-2)

17 selected lines.

Reg,
Hard

Hi , HemantKChitale,

I also update statistics manual as you say, but not see 'TABLE ACCESS FULL' good result

What should I do? my need of production tuning, but I cannot able tune this...

SQL > exec dbms_stats.gather_table_stats (-online 'ttt' ownname, tabname => 'LOAN_RUNNING_DETAILS_SOUTH', cascade => TRUE, estimate_percent => NULL, method_opt => 'for all columns size 254', => of degree 4);

PL/SQL procedure successfully completed.

SQL > explain plan for the small circle of MSISDN, TID, of LOAN_RUNNING_DETAILS_SOUTH where LOANDATETIME<=sysdate-2 and="" loandatetime="">sysdate-15 and LOANTYPE = 1;

He explained.

SQL > set line 200

@?/rdbms/admin/utlxpls.sql

SQL >

PLAN_TABLE_OUTPUT

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

Hash value of plan: 3659874059

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

| ID | Operation | Name                       | Lines | Bytes | Cost (% CPU). Time | Pstart. Pstop |

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

|   0 | SELECT STATEMENT |                            | 1874K |    75 M | 19626 (2) | 00:03:56 |       |       |

|* 1 | FILTER |                            |       |       |            |          |       |       |

|   2.   PARTITION LIST ALL |                            | 1874K |    75 M | 19626 (2) | 00:03:56 |     1.    11.

|* 3 |    TABLE ACCESS FULL | LOAN_RUNNING_DETAILS_SOUTH | 1874K |    75 M | 19626 (2) | 00:03:56 |     1.    11.

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

PLAN_TABLE_OUTPUT

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

Information of predicates (identified by the operation identity card):

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

1 - filter(SYSDATE@!-2>SYSDATE@!-15)

3 - filter("LOANDATETIME">SYSDATE@!-15 AND "LOANTYPE"=1 AND "LOANDATETIME")<>

16 selected lines.

Why I have Photo and Photos open at the same time?

Why I have Photo and Photos open at the same time?

When what's going on? When you connect a camera or a card reader?

Or after the Mac is restarted?

If this occurs after startup, check system preferences > users and groups.

Select your account and open the connection. Deselect the applications you want to open at startup.

If this happens when you connect a camera, open the iPhoto preferences > general.

Set the preference "Connecting camera opens" to "no application".

The execution plan changes for the same query.

Hi all

This issue was raised before also, but still not able to find the real cause of this.

Thread1:
Re: Research of fragmentation of the table in Oracle 8.1.6.3.0

Thread2:
CBC latch and buffer busy await you on the same table.

It comes, sometimes hammers server 100% CPU utilization with free latch and buffer busy wait events.

We found a single query consumes high CPU usage that is run by different sessions.

This query have two types of execution plans, where one is accurate and is not (its primary key hit index index no appropriate means present on the table)

Because its primary key index hit repeatedly at various sessions, some sessions are powerful db file sequential read and a few sessions waiting buffer busy waits for event. Also during this time a few sessions waiting for latch free event.

My doubt is how to sql even with different literal values execution plan changes and causes a prob.

select count(*),event from v$session_wait group by event;
  COUNT(*) EVENT
---------- ----------------------------------------------------------------
       165 SQL*Net message from client
         1 SQL*Net message to client
         3 buffer busy waits
         2 db file parallel read
        18 db file sequential read
        10 latch free
         5 log file sync
         1 pmon timer
         6 rdbms ipc message
         1 smon timer

SQL> select sid from v$session_wait where event='db file sequential read';
       SID
----------
        26
        58
        82
       107
       116
       223
       212
       203
       192
       173
       161
       157
       150
       147
       254
       238
       229
       112
       101
        81
        68

SQL> select spid, sid, s.serial#, p.program from v$session s, v$process p where paddr=addr and sid=&SID;
Enter value for sid: 161
old   1: select spid, sid, s.serial#, p.program from v$session s, v$process p where paddr=addr and sid=&SID
new   1: select spid, sid, s.serial#, p.program from v$session s, v$process p where paddr=addr and sid=161

SPID             SID    SERIAL# PROGRAM
--------- ---------- ---------- ------------------------------------------------
4231             161      49569 oracle@tfrdb3 (TNS V1-V3)


SQL> select sql_text
from v$process a,
     v$session b,  2    3
     v$sql c
where a.addr = b.paddr and
     b.sql_hash_value = c.hash_value and
    a.spid = &PID;  4    5    6    7
Enter value for pid: 4231
old   7:     a.spid = &PID
new   7:     a.spid = 4231

SQL_TEXT
--------------------------------------------------------------------------------
SELECT ERROR,TIME_STAMP,O_RESOURCE,QUEUE,NEW_QUEUE FROM LOG WHERE ID = '09292AMR
10B41FE' AND TYPE IN (11, 28, 25, 18, 60, 13) AND (LOG_SEQ>'234225222' OR TYPE =
 18 AND LOG_SEQ='234225222') ORDER BY TIME_STAMP ASC

SQL> set autotrace traceonly exp
SQL> SELECT ERROR,TIME_STAMP,O_RESOURCE,QUEUE,NEW_QUEUE FROM amrwf1.LOG WHERE ID = '09292AMR10B41FE' AND TYPE IN (11, 28, 25, 18, 60, 13) AND (LOG_SEQ>'234225222' OR TYPE =18 AND LOG_SEQ='234225222') ORDER BY TIME_STAMP ASC;

Execution Plan
----------------------------------------------------------
   0      SELECT STATEMENT Optimizer=CHOOSE (Cost=11 Card=2 Bytes=126)
   1    0   SORT (ORDER BY) (Cost=11 Card=2 Bytes=126)
   2    1     CONCATENATION
   3    2       TABLE ACCESS (BY INDEX ROWID) OF 'LOG' (Cost=4 Card=1
          Bytes=63)

   4    3         INDEX (UNIQUE SCAN) OF 'PK_LOG_LOG_SEQ' (UNIQUE) (Co
          st=3 Card=1)

   5    2       TABLE ACCESS (BY INDEX ROWID) OF 'LOG' (Cost=4 Card=1
          Bytes=63)

   6    5         INDEX (RANGE SCAN) OF 'PK_LOG_LOG_SEQ' (UNIQUE) (Cos
          t=3 Card=1)


SQL> select spid, sid, s.serial#, p.program from v$session s, v$process p where paddr=addr and sid=&SID;
Enter value for sid: 147
old   1: select spid, sid, s.serial#, p.program from v$session s, v$process p where paddr=addr and sid=&SID
new   1: select spid, sid, s.serial#, p.program from v$session s, v$process p where paddr=addr and sid=147

SPID             SID    SERIAL# PROGRAM
--------- ---------- ---------- ------------------------------------------------
6255             147      38306 oracle@tfrdb3 (TNS V1-V3)

SQL> select sql_text
from v$process a,
     v$session b,
     v$sql c  2    3
where a.addr = b.paddr and
     b.sql_hash_value = c.hash_value and
    a.spid = &PID;  4    5    6    7
Enter value for pid: 6255
old   7:     a.spid = &PID
new   7:     a.spid = 6255

SQL_TEXT
--------------------------------------------------------------------------------
SELECT ERROR,TIME_STAMP,O_RESOURCE,QUEUE,NEW_QUEUE FROM LOG WHERE ID = '09273AMR
62B4894' AND TYPE IN (11, 28, 25, 18, 60, 13) AND (LOG_SEQ>'223324996' OR TYPE =
 18 AND LOG_SEQ='223324996') ORDER BY TIME_STAMP ASC


SQL> set autotrace traceonly exp
SQL> SELECT ERROR,TIME_STAMP,O_RESOURCE,QUEUE,NEW_QUEUE FROM amrwf1.LOG WHERE ID = '09273AMR62B4894' AND TYPE IN (11, 28, 25, 18, 60, 13) AND (LOG_SEQ>'223324996' OR TYPE =18 AND LOG_SEQ='223324996') ORDER BY TIME_STAMP ASC;

Execution Plan
----------------------------------------------------------
   0      SELECT STATEMENT Optimizer=CHOOSE (Cost=1538 Card=736 Bytes=
          46368)

   1    0   SORT (ORDER BY) (Cost=1538 Card=736 Bytes=46368)
   2    1     TABLE ACCESS (BY INDEX ROWID) OF 'LOG' (Cost=1527 Card=7
          36 Bytes=46368)

   3    2       INDEX (RANGE SCAN) OF 'LOG_ID' (NON-UNIQUE) (Cost=32 C
          ard=736)



SQL> select spid, sid, s.serial#, p.program from v$session s, v$process p where paddr=addr and sid=&SID;
Enter value for sid: 82
old   1: select spid, sid, s.serial#, p.program from v$session s, v$process p where paddr=addr and sid=&SID
new   1: select spid, sid, s.serial#, p.program from v$session s, v$process p where paddr=addr and sid=82

SPID             SID    SERIAL# PROGRAM
--------- ---------- ---------- ------------------------------------------------
6172              82      45378 oracle@tfrdb3 (TNS V1-V3)


SQL> select sql_text
from v$process a,
     v$session b,
     v$sql c
where a.addr = b.paddr and
     b.sql_hash_value = c.hash_value and  2
  3      a.spid = &PID;  4    5    6    7
Enter value for pid: 6172
old   7:     a.spid = &PID
new   7:     a.spid = 6172

SQL_TEXT
--------------------------------------------------------------------------------
INSERT INTO LOG (ID,TIME_STAMP,TYPE,ERROR,INSTANCE,RULE_NUM,RULE_TYPE,PRIORITY,F
LAGS,NAME,BATCH,O_RESOURCE,QUEUE,NEW_QUEUE,SERVER,FORM,WORKSET) VALUES (:V001,:V
002,11,0,0,3,1,0,1,:V003,:V004,:V005,:V006,:V007,:V008,:V009,:V010)

INSERT INTO LOG (ID,TIME_STAMP,TYPE,ERROR,INSTANCE,RULE_NUM,RULE_TYPE,PRIORITY,F
LAGS,NAME,BATCH,O_RESOURCE,QUEUE,NEW_QUEUE,SERVER,FORM,WORKSET) VALUES (:V001,:V
002,11,0,0,3,1,0,1,:V003,:V004,:V005,:V006,:V007,:V008,:V009,:V010)

INSERT INTO LOG (ID,TIME_STAMP,TYPE,ERROR,INSTANCE,RULE_NUM,RULE_TYPE,PRIORITY,F
LAGS,NAME,BATCH,O_RESOURCE,QUEUE,NEW_QUEUE,SERVER,FORM,WORKSET) VALUES (:V001,:V
002,11,0,0,3,1,0,1,:V003,:V004,:V005,:V006,:V007,:V008,:V009,:V010)

SQL_TEXT
--------------------------------------------------------------------------------

INSERT INTO LOG (ID,TIME_STAMP,TYPE,ERROR,INSTANCE,RULE_NUM,RULE_TYPE,PRIORITY,F
LAGS,NAME,BATCH,O_RESOURCE,QUEUE,NEW_QUEUE,SERVER,FORM,WORKSET) VALUES (:V001,:V
002,11,0,0,3,1,0,1,:V003,:V004,:V005,:V006,:V007,:V008,:V009,:V010)

INSERT INTO LOG (ID,TIME_STAMP,TYPE,ERROR,INSTANCE,RULE_NUM,RULE_TYPE,PRIORITY,F
LAGS,NAME,BATCH,O_RESOURCE,QUEUE,NEW_QUEUE,SERVER,FORM,WORKSET) VALUES (:V001,:V
002,11,0,0,3,1,0,1,:V003,:V004,:V005,:V006,:V007,:V008,:V009,:V010)

INSERT INTO LOG (ID,TIME_STAMP,TYPE,ERROR,INSTANCE,RULE_NUM,RULE_TYPE,PRIORITY,F
LAGS,NAME,BATCH,O_RESOURCE,QUEUE,NEW_QUEUE,SERVER,FORM,WORKSET) VALUES (:V001,:V

SQL_TEXT
--------------------------------------------------------------------------------
002,11,0,0,3,1,0,1,:V003,:V004,:V005,:V006,:V007,:V008,:V009,:V010)

INSERT INTO LOG (ID,TIME_STAMP,TYPE,ERROR,INSTANCE,RULE_NUM,RULE_TYPE,PRIORITY,F
LAGS,NAME,BATCH,O_RESOURCE,QUEUE,NEW_QUEUE,SERVER,FORM,WORKSET) VALUES (:V001,:V
002,11,0,0,3,1,0,1,:V003,:V004,:V005,:V006,:V007,:V008,:V009,:V010)

How to avoid this... why its different execution plan using (I mean bad index PK)

Is it possible to avoid this?

If any details please check out some of my previous post on this specific URL (above)

-Yasser

My doubt is how to sql even with different literal values execution plan changes and causes a prob.

Different literal values cause analysis difficult.
Hard analysis includes the re-evaluation of the best path.
Literal value is included in the assessment of the selectivity for the scan interval (log_seq >...)

See
http://www.centrexcc.com/A%20Look%20under%20The%20Hood%20Of%20CBO%20-%20The%2010053%20Event.ppt.PDF
http://www.centrexcc.com/fallacies%20Of%20The%20Cost%20Based%20Optimizer.PDF
more the book of Jonathan Lewis which other threads, I believe that you already have.

You must lower your CPU.
Previous discussions, if the situation is still the same, it sounded like hard analysis particularly with this SELECTION against the NEWSPAPER plays an important role in that.

How to avoid this... why its different execution plan using (I mean bad index PK)

The points raised in the previous discussion remain valid.
-Do you have access to this SQL to change?
for example using bind variable or trick it if necessary due to problems caused by data as discussed in the previous thread.
- Or you could it repoint the view to a view and a hint?
-If a particular user makes this sql, could affect you cursor_sharing just for this user. If not, you should consider implementing pan-Canadian database.

Oracle 8.1.6 still?

Different plan for the same sql id

Hello
Our team of application reported recently that a query ran longer than usual.
Search in tables AWR (dba_hist_active_sess_history & dba_hist_sqlstat), it was a change of plan_hash_value. The next day, without intervention from anyone (collect statistics did not run, no change in the tables involved) the sql used the former plan and completed quickly.
The differences in the plan was the index that was used to access the table.
What could be the reason for the optimizer to choose a bad plan and return once more the good thing. ?
The code sql that has been run twice & three in one day and then it years off cursor cache.

Regime shifts are normal.

Flipfloppping plans are quite normal.

I'd be willing to bet that a large percentage of your SQL shows variations in implementation plans and you never notice.

And there are several reasons why you might get a different plan.

1 bind variable peeking - it is normal SQL for the age from the cache and then get analyzed once again with different bind variable leading to different cardinality estimates and plans

2. the statistics change - it is normal that the statistics of change over time causing different cardinality estimates and the various plans, especially if the histograms are involved and changing buckets

3. data change - especially if you use dynamic sampling it is normal to get a different sample of data leading to different cardinality estimates and plans

4 SYSDATE - it's normal for queries with SYSDATE in them to recognize that time is never on leave and which, in conjunction with statistics or data, can lead to different cardinality estimates and plans

5. integrated optimization features - it's normal for features like ACS to kick in and notice that they got forecasts wrong in an execution plan and force the recalculation to a different plan with an adjusted cardinality.

Using DBMS_XPLAN. DISPLAY_CURSOR or DISPLAY_AWR you can get different plans (provided that they are in memory or AWR). The notes section should indicate cardinality comments have been a factor. You can also get links peeked with format mask of "+ PEEKED_BINDS".

Different output for the same query in SQL Server and Oracle
I have two tables table1 and table2

-table1 has two columns c1 int and varchar c2. There are no constraints added in it. It contains data as shown below

C1 c2
-------------------
d 6
5 d
102 g
6%
f 103
5.
501 j
1 g
601 n
2 m

-table2 has only a single column c1 int. There is no added in constraints. It contains data as shown below

C1
----
6
1
4
3
2

now when I run below, given the query in sql server and oracle it gives me different results

Select *.
FROM table1
table2 on table2.c1 = table1.c1 inner join (SELECT ROW_NUMBER() (any ORDER by ASC c1) AS c1 from table2)

output of SQL server
------------------------
C1 c2 c1
--------------------------------
1 g 1
2 m 2
3 h 3
4 g 4
5 d 5

release of Oracle
----------------------
C1 C2 C1
---------------------------------
5 d 5
4 g 4
3 h 3
1 g 1
2 m 2

If you notice the first column in the two outputs. It is sorted in sql server and no oracle.

Why he behaves differently in oracle? Is there a way I can fix this problem in oracle?

Thank you
Jigs
It is NOT a behavior "differently" in Oracle; you did not specify just an order that you expect of your results, so you'll get output in what order the fantasies of the database showing (ie. no guaranteed order). It is an artifact of the way the database chooses to collect the data and databases (or same sets of data within the same database) can and will most likely behave differently.

Same SQL Server will not be guaranteed to always get your data in an orderly manner if you exclude the order by clause, even if you think that there always display the data in an orderly manner.

Your solution is to add an order by clause, in the TWO databases, to force the order of the data output.

Hello! I have to cancel my membership for the creative cloud plan but I find no different contact of live chat. I need TO cancel before March 22. Please help me!

Hello! I have to cancel my membership for the creative cloud plan but I find no different contact of live chat. I need TO cancel before March 22. Please help me!

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Set a style different cases for the same placeholder (data fusion) used two times in the same text

Hello.
I have a document with some parties updated with the fusion of data (via placeholders). Is it possible to set a style of different cases for the same placeholder in the same text. See the example:
Mike runs its < < vehicule_placeholder > > in the city. But its < < VEHICULE_PLACEHOLDER > > is meant to be shared.
If the value of this placeholder in the file merger is 'bike', the sentence should be:
Mike runs his bike in the city. But his BIKE is meant to be shared.
Is this possible? How?
Thank you.

You can do it. A character style for this configuration placeholder text, set in all caps.

I would spend a "plan of photography" "plan of Applications." But I did have the 'photography plan' for a year yet. Can I still sign up for the same year or is this a new subscription?

I would spend a "plan of photography" "plan of Applications." But I did have the 'photography plan' for a year yet. Can I still sign up for the same year or is this a new subscription?

In addition to the other suggestions here, you can also consult the plan switch option (described here):
Update or change your plan of Adobe Creative Cloud

How to capture the execution plan for a query
Hi all

Can someone help me please to find the command to capture the execution plan for a query.

Execution plan for select * from EMP where < condition >

Is getting executed successfully, but I need to get the appropriate for the same execution plan.

Thank you
Read the Doc

Plan for the upgrade to complete our special $ 29.99

I have bought a lot of perpetual licenses during the last 15 years and I am now in terms of cloud photography. Two questions:
1. Why do I get only 2 GB of storage on creative cloud?
2. how to switch my plan for the special Plan for $29.99 per month?
Thank you.

Hello

29.99 is for customers who have bought a creative suite CS3, CS4 or CS5 before. If you have one of its products, you must register it on your account, the system will detect and make the discount automatically prior to payment.

I have Windows xp and Vista on the same computer, my fligth Simulator is on xp. How do I play for vista or I need to install Island all on vista

I have Windows xp and Vista on the same computer, my fligth Simulator is on xp. How do I play for vista or I need to install Island all on vista

Hi, CPC,.

Welcome to the Microsoft Answers Community!

You must install Flight simulator on vista as well as Windows Xp and vista are two different operating systems.

You log in to vista, you may not use any of the XP application and vice versa.

Thank you, and in what concerns:
Azeez Nadeem
Visit our http://social.answers.microsoft.com/Forums/en-US/answersfeedback/threads/ Microsoft answers feedback Forum and let us know what you

Both inputs of printer with different names for the same printer

I try to add a new printer in Windows 7 64 bit laptop. I chose to add a printer in printers and peripherals and selected add a network, wireless or Bluetooth printer. I have two tickets for the same printer

Why are there 2 inputs and one which should I use?

These entries are actually at the same network address (192.168.1.10). The second address Specifies a particular port (3910). Software print server HP Jetadmin 'web-based' uses the 3910 port to receive the print requests. See http://h20564.www2.hp.com/hpsc/doc/public/display?docId=emr_na-c03975926

The second is probably 'discovery of Web Services' (see http://en.wikipedia.org/wiki/WS-Discovery). You probably don't want to use it from the "Add printer" Wizard

I suggest to select the first entry. This is the normal way to network printers appear in the add a Printer Wizard. For example:

If you have any problems with printing, post back and I'll see if I can diagnose the things for you.

I'm a customer of CS5.5, why say that I am not eligible for the discount?

I'm a customer of CS5.5, why say that I am not eligible for the discount?

There are two price for Lightroom: full or upgrade.

The upgrade price is the same for all owners of a previous version of Lightroom.

So, if you have purchased a version upgrade then you got your discount.

Payment plan for the photographer and the creative shift and cannot use photoshopPayment for the photographer and the creative shift plan and may not use photoshop

Payment plan for the photographer and the creative cloud and can't use photoshop

Your subscription to cloud shows correctly on your account page?

If you have more than one email, you will be sure that you use the right Adobe ID?

https://www.adobe.com/account.html for subscriptions on your page from Adobe

.

If Yes

Sign out of your account of cloud... Restart your computer... Connect to your paid account of cloud

-Connect using http://helpx.adobe.com/x-productkb/policy-pricing/account-password-sign-faq.html

-http://helpx.adobe.com/creative-cloud/kb/sign-in-out-creative-cloud-desktop-app.html

-http://helpx.adobe.com/x-productkb/policy-pricing/activation-network-issues.html

-http://helpx.adobe.com/creative-suite/kb/trial--1-launch.html

-ID help https://helpx.adobe.com/contact.html?step=ZNA_id-signing_stillNeedHelp

-http://helpx.adobe.com/creative-cloud/kb/license-this-software.html

.

If no

This is an open forum, Adobe support... you need Adobe personnel to help

Adobe contact information - http://helpx.adobe.com/contact.html

Chat/phone: Mon - Fri 05:00-19:00 (US Pacific Time)<=== note="" days="" and="">

-Select your product and what you need help with

-Click on the blue box "still need help? Contact us. "

Why I have two different execution plans for the same query on two different servers

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