Expression regular/replace - Oracle 7.3

Similar Questions

• Expression regular replacement

  Hi gurus,
  
  I have as a result string and replace the string with tags as follows.
  

   |3 String1 |0 some text  |4 String2 |0 
  
  to
  
   < i >  String1 < /i > some text < b > String2 < /b >
  
  i.e. Want to replace |3 to < i >
                               |0  to  < /i >      -- First occurrence of |0
                       
                               |4 to < b > 
                               |0 to < /b >      -- Second occurrence of |0

  Thanks in advance,

  with t as (
             select '|3 String1 |0 some text  |4 String2 |0' str from dual
            )
  select regexp_replace(
                        regexp_replace(
                                       str,
                                       '\|3(.*?)\|0',
                                       '\1'
                                      ),
                        '\|4(.*?)\|0',
                        '\1'
                       )
    from  t
  /
  
  REGEXP_REPLACE(REGEXP_REPLACE(STR,'\|3(.*?)\
  --------------------------------------------
   String1  some text   String2 
  
  Elapsed: 00:00:00.02
  SQL> 
  

  SY.

• Query Sub behavior strange when using Expressions regular Oracle

  I met a strange "inconsistent" when you use an Expression regular Oracle with subqueries in SQL. Here are some details on how to reproduce what I have observed. We managed to find a solution to this "problem" using a database index; I'm writing this case hoping to better understand why Regular Expressions Oracle do not seem to work in the same way that the older, standard functions integrated as INSTR, SUBSTR, AS, etc..
  
  Environment settings:
  This test has been done using Oracle XE (GR 11, 2) on 32-bit Windows operating system. For my test, I used the HR schema (which is delivered pre-completed with the installation of this product) with some modifications of my own.
  
  Make the Test objects:
  

  create table hr.emp_test as
  select to_char(employee_id) as emp_id_char, employee_id as emp_id,
  first_name, last_name, hire_date
  from hr.employees;

  To illustrate my test, I inserted mixed alphanumeric values and a null value in the column of my emp_id_char for good measure:
  

  insert into hr.emp_test (emp_id_char, first_name, last_name, hire_date)
  values ('XKCD123','TEST','LASTNAME',sysdate);
  insert into hr.emp_test (emp_id_char, first_name, last_name, hire_date)
  values (null,'TEST1','LASTNAME2',sysdate);
  commit;

  * (1) this request fails once a nonnumeric value is inserted into the emp_test table.*
  

  with
     sub1 as ( select to_number(emp_id_char) as emp_id, first_name, last_name, hire_date 
        from hr.emp_test  )
  select * from sub1
  where emp_id between 100 and 110

  * (2) this query works OK.*
  

  with
     sub1 as ( select to_number(emp_id_char) as emp_id, first_name, last_name, hire_date 
        from hr.emp_test where emp_id_char not like 'X%'  )
  select * from sub1
  where emp_id between 100 and 110

  * (3) this query works OK.*
  

  with
     sub1 as ( select to_number(emp_id_char) as emp_id, first_name, last_name, hire_date 
        from hr.emp_test where instr(emp_id_char,'X',1) = 0 )
  select * from sub1
  where emp_id between 100 and 110

  * (4) this query Fails.*
  

  with
     sub1 as ( select emp_id_char, first_name, last_name, hire_date
                 from hr.emp_test
                where regexp_instr(emp_id_char, '[^[:digit:]]') = 0 ),
     sub2 as ( select to_number(emp_id_char) as emp_id, first_name, last_name, hire_date
                from sub1 )
  select *
  from sub2
  where emp_id between 100 and 110
  
  ERROR:
  ORA-01722: invalid number

  * (5) even down the results of the rational expression of 3rd under the query in sequential processing order also FAILS *.
  

  with
     sub1 as ( select emp_id_char, first_name, last_name, hire_date
                 from hr.emp_test
                where regexp_instr(emp_id_char, '[^[:digit:]]') = 0 ),
     sub2 as ( select to_number(emp_id_char) as emp_id, first_name, last_name, hire_date
                from sub1 ),
     sub3 as ( select emp_id, first_name, last_name, hire_date from sub2 )
  select *
  from sub3
  where emp_id between 100 and 110
  
  ERROR:
  ORA-01722: invalid number

  * (6) that it does not like previous query as well *.
  

  with
     sub1 as ( select to_number(emp_id_char) as emp_id, first_name, last_name, hire_date,
         regexp_instr(emp_id_char, '[^[:digit:]]') as reg_x 
         from hr.emp_test 
         where reg_x = 0),
     sub2 as ( select emp_id, first_name, last_name, hire_date, reg_x
        from sub1 where reg_x = 0 )
  select * from sub2
  where emp_id between 100 and 110
  
  ERROR:
  ORA-00904: "REG_X": invalid identifier

  Our Solution...
  Add a hint to the query of sup that 'hiding' the result of sub1 in memory. That did the trick. This suspicion resembles only viable workaround for this behavior. Other old built-in functions (INSTR, AS, etc.) an they were automatically following the execution plan (results of cache memory) that he had to use a 'hint' to force with the function of the regular expression.
  
  The conclusion, which is what I would like to help to understand or explain is that:
  If you create a series of queries/sup queries or inline views, values depend on "regular expression" type built-in sql functions do not seem to stick or maintain when implemented in complex query logic.
  
  Any idea is appreciated!
  Thank you!
  
  Published by: 870810 on July 6, 2011 15:47

  870810 wrote:
  I met a strange "inconsistent" when you use an Expression regular Oracle with subqueries in SQL.

  This is the expected behavior and has nothing to do with regular expressions - much less directly (I'll explain later). Main rule: there is no WHERE clause predicate order. Even if you use views, views online, subquery factoring, optimizer etc. can extend your display, display online, a subquery factoring. And it's Optimizer who decides the order of execution predicate unless you use the ORDERED_PREDICATES key. Now, I can explain. Regular expressions are powerful enough but also everywhere in life to pay for it with the higher cost of execution. That's why optimizer decides to apply emp_id between 100 and 110 first and regexp_instr (emp_id_char, "[^ [: digit:]]'") = 0 later.

  explain plan for
  with
  sub1 as ( select emp_id_char, first_name, last_name, hire_date
  from emp_test
  where regexp_instr(emp_id_char, '[^[:digit:]]') = 0 ),
  sub2 as ( select to_number(emp_id_char) as emp_id, first_name, last_name, hire_date
  from sub1 )
  select *
  from sub2
  where emp_id between 100 and 110;
  
  SQL> @?\rdbms\admin\utlxpls
  
  PLAN_TABLE_OUTPUT
  -------------------------------------------------------------------------------------
  Plan hash value: 3124080142
  
  ------------------------------------------------------------------------------
  | Id  | Operation         | Name     | Rows  | Bytes | Cost (%CPU)| Time     |
  ------------------------------------------------------------------------------
  |   0 | SELECT STATEMENT  |          |     1 |    57 |     3   (0)| 00:00:01 |
  |*  1 |  TABLE ACCESS FULL| EMP_TEST |     1 |    57 |     3   (0)| 00:00:01 |
  ------------------------------------------------------------------------------
  
  Predicate Information (identified by operation id):
  ---------------------------------------------------
  
  PLAN_TABLE_OUTPUT
  -------------------------------------------------------------------------------------
  
     1 - filter(TO_NUMBER("EMP_ID_CHAR")>=100 AND
                TO_NUMBER("EMP_ID_CHAR")<=110 AND  REGEXP_INSTR
                ("EMP_ID_CHAR",'[^[:digit:]]')=0)
  
  15 rows selected.
  
  SQL> 
  

  As you can see, optimizer uses a FULL SCAN to read data from the table and apply emp_id between 100 and 110 which translates TO_NUMBER ("EMP_ID_CHAR") > = 100 AND TO_NUMBER ("EMP_ID_CHAR")<=110. and="" obviously="" it="" fails="" trying="" to="" convert="" xkcd123="" to="" number.="" now="" cost="" of="" instr(emp_id_char,'x',1)="0" is="" lower="" and="" optimizer="" decides="" to="" apply="" instr="" first.="" therefore="" xkcd123="" is="" filtered="" out="" before="" to_number="" is="">

  SQL> explain plan for
    2  with
    3  sub1 as ( select emp_id_char, first_name, last_name, hire_date
    4  from emp_test
    5  where instr(emp_id_char, 'X') = 0 ),
    6  sub2 as ( select to_number(emp_id_char) as emp_id, first_name, last_name, hire_date
    7  from sub1 )
    8  select *
    9  from sub2
   10  where emp_id between 100 and 110;
  
  Explained.
  
  SQL> @?\rdbms\admin\utlxpls
  
  PLAN_TABLE_OUTPUT
  -----------------------------------------------------------------------------------------------
  Plan hash value: 3124080142
  
  ------------------------------------------------------------------------------
  | Id  | Operation         | Name     | Rows  | Bytes | Cost (%CPU)| Time     |
  ------------------------------------------------------------------------------
  |   0 | SELECT STATEMENT  |          |     1 |    57 |     3   (0)| 00:00:01 |
  |*  1 |  TABLE ACCESS FULL| EMP_TEST |     1 |    57 |     3   (0)| 00:00:01 |
  ------------------------------------------------------------------------------
  
  Predicate Information (identified by operation id):
  ---------------------------------------------------
  
  PLAN_TABLE_OUTPUT
  -----------------------------------------------------------------------------------------------
  
     1 - filter(INSTR("EMP_ID_CHAR",'X')=0 AND
                TO_NUMBER("EMP_ID_CHAR")>=100 AND TO_NUMBER("EMP_ID_CHAR")<=110)
  
  14 rows selected.
  
  SQL> 
  

  The bottom LINE: With the help of strings to store non-chaine (numeric in your case) is never a good idea and shows design problems.

  SY.

• Expression regular slash before?

  I'm trying to use a regular expression to replace all the slashes in a selection.

  I tried the following, my performance desired is 0 m [div] s + s 16.67 m [div].

  var str = "0 m/s + 16.67 m/s";
  var output = str.contents.replace(/[/]/gi,"[div]"); //output "0 m/s + 16.67 m/s"
  

  Thanks in advance for any help is to solve it.

  Hi KuddRoww,

  simply to escape the slash should help (and chain is not necessary: content)

  var str = "0 m/s + 16.67 m/s";
  var output = str.replace(/\//gi,"[div]"); // "0 m[div]s + 16.67 m[div]s"
  alert(output);
  

  Have fun

• Since version 18.0.0, Firefox regularly replace some parts of my screen when it is enlarged but not active (Outlook or Word); is there any solution for this?

  Any time I can get Firefox enlarged but not active (so it is behind another active window / application) on Windows XP Pro SP3, it regularly replace significant portions of my screen. If I move the mouse, type, scroll up/down or anything to change the screen, the active application is slowly starting to appear. For example, in Office 2007, when this happens, I can continue typing a sentence in Word or Outlook and this line is displayed, but the rest of the message area displays remnants of Firefox. If I move the mouse along the Ribbon of Office, it will be displayed in pieces are the reached mouse. This problem has been held regularly since the version 18.0.0 and persists in 18.0.1. I prefer not to minimize Firefox so I can move conveniently via Alt - Tab from Firefox and the other application, in that I work, but this problem that is almost impossible. FYI, when I'm in Firefox, no other pop application on top, not other applications show otherwise, so I know that the problem is caused by Firefox and/or interaction weird with the new version and installed plugins, drivers or other software.

  Could you try to disable graphics hardware acceleration? (I have trouble determining your "more system information" if it is enabled or disabled). As this feature has been added to Firefox, it has gradually improved, but there are still some problems.

  You need to restart Firefox for it to take effect, so save any work first (e.g. you compose mail, documents online that you are editing, etc.).

  Orange Firefox button or classic menu Tools > Options > advanced

  In the mini ' General' tab, uncheck the box for "use hardware acceleration when available.

  If you restart Firefox, the problem is solved?

• How can we replace Oracle form when you used by another user

  Hi Experts,

  Is it possible to replace Oracle form while in use by another user?

  Some time need deliver urgent fixes on the production environment, but it must wait for downtime, and the form does not use.

  I searched the Forum, but able to get the answer to this topic, I will be grateful if you could provide no alternative to do this.

  Env:

  Windows Server 2008 R2

  WebLogic Server 10.3.5

  Oracle form & report 11.1.2

  Hello

  For Production environment, it will be handled with tactics that led to minor maintenance work as well.

  You must have two physical directories (real and temporary) including the same fmx/plx/pll/mmx, files etc.

  and a file reference of in the FORMS_PATH in default.env, who have the latest version of the source.

  Copy the last source file into a temporary directory, when the user take the new session, it will access the new version of the source of the temporary path and who obviously last source version.

  Meanwhile, once the actual source published by the user version, you can copy the latest version of the source to the real directory and reset the FORMS_PATH with the actual path.

  In this way, you can solve your problem.

• Expression regular-not clear for the result

  I have written 3 applications using regular expressions, but I'm not very clear on the outcome. Looking for help to understand the logic.

  Here are the queries:

  1 > SELECT REGEXP_SUBSTR ('APT-101, BLDG 34, community xyz, XYZ ROAD, BANGALORE - 560066', '([[:alpha:]]+)') ADDR FROM DUAL;

  2 > SELECT REGEXP_SUBSTR ('APT-101, BLDG 34, community xyz, XYZ ROAD, BANGALORE - 560066', ' [[: alpha:]] +') ADDR FROM DUAL;

  3 > SELECT REGEXP_SUBSTR ('APT-101, BLDG 34, community xyz, XYZ ROAD, BANGALORE - 560066',(^'[[:alpha:]]+) ') ADDR FROM DUAL;

  All requests above returns the same result, i.e. 'APT '.

  Question: My understanding is:

  a > the regular expression [[: alpha:]] + represents one or more continuous occurrences of alphabets.

  b > parenthesis (i.e.) - represents the exact mactch

  c > carrot i.e. ^ represents the negation, that is to say. NOT (xyz)

  I can somehow convince myself that regular expressions - ' ([[: alpha:]] +)' and ' [[: alpha:]] +' in the given scenario may return the same string that is 'APT', but in the last query the regex ([[: alpha:]] + is preceded by ^, which according to my understanding should return something that isn't ONE or MORE CONTINUOUS OCCURRENCE OF ALPHABETS, but even this query is retune "APT".) Ask for help to understand this.

  Thank you

  Amrit Pandey

  
  

  Hi, Amrit,

  ba1fbc36-dd1f-46af-80EE-a9cedf91e344 wrote:

  I have written 3 applications using regular expressions, but I'm not very clear on the outcome. Looking for help to understand the logic.

  Here are the queries:

  1 > SELECT REGEXP_SUBSTR ('APT-101, BLDG 34, community xyz, XYZ ROAD, BANGALORE - 560066', '([[:alpha:]]+)') ADDR FROM DUAL;

  2 > SELECT REGEXP_SUBSTR ('APT-101, BLDG 34, community xyz, XYZ ROAD, BANGALORE - 560066', ' [[: alpha:]] +') ADDR FROM DUAL;

  3 > SELECT REGEXP_SUBSTR ('APT-101, BLDG 34, community xyz, XYZ ROAD, BANGALORE - 560066',(^'[[:alpha:]]+) ') ADDR FROM DUAL;

  All requests above returns the same result, i.e. 'APT '.

  Question: My understanding is:

  a > the regular expression [[: alpha:]] + represents one or more continuous occurrences of alphabets.

  b > parenthesis (i.e.) - represents the exact mactch

  c > carrot i.e. ^ represents the negation, that is to say. NOT (xyz)

  I can somehow convince myself that regular expressions - ' ([[: alpha:]] +)' and ' [[: alpha:]] +' in the given scenario may return the same string that is 'APT', but in the last query the regex ([[: alpha:]] + is preceded by ^, which according to my understanding should return something that isn't ONE or MORE CONTINUOUS OCCURRENCE OF ALPHABETS, but even this query is retune "APT".) Ask for help to understand this.

  Thank you

  Amrit Pandey

  Be careful.  Cut and paste the exact code you run.

  I do not get the same results for alI these queries.  I get an error message ' ORA-00936: missing expression. "for the 3rd.  Maybe you wanted to have the circumflex accent (^) inside the single quotes, like this:

  SELECT REGEXP_SUBSTR (' APT-101, 34 BLDG, community xyz, XYZ ROAD, BANGALORE - 560066' ")

  , ('^ [[: alpha:]] +')

  ) AS addr

  DOUBLE;

  a > you're right;

  [[: alpha:]] +.

  refers to a group of characters 1 or more adjacent, which are all letters of the alphabet.

  b > an expression can almost always be enclosed in parentheses free.  In other words, almost anywhere, you can use an expression

  x you can also use

  (x)               or

  ((x))                  or

  (((x))) and so on.  Each of them the same meaning.

  This has nothing to do with regular expressions.

  In expressions regular, curved brackets can be used for the Backreferences, for example \1 can be used to return to exactly what corresponded to the subexpression inside the 1st pair of parentheses.  You are not using anything like \1 here, so this backreferences do not apply to this question.

  c > sign means negation only when it comes immediately after [.]  For example

  SELECT REGEXP_SUBSTR (' APT-101, 34 BLDG, community xyz, XYZ ROAD, BANGALORE - 560066' ")

  ", ' [^ [: alpha:]] +"

  ) AS addr

  DOUBLE;

  product

  ADDR

  ------

  -101,

  in other words, the characters 1 or more consecutive which are NOT letters of the alphabet.

  Outside square brackets, the circumflex accent means the beginning of the string.

• Weblogic Server will replace Oracle Application Server in the future!

  Weblogic Server will replace Oracle Application Server in the future!

  Roadmap for Oracle application server

• expression regular-insert to create a 'new' string expression in the correct position

  I have sample data in table T, and I sample how I want more than data output of the query output.
  

  with T as
  (
  select 'CREATE OR REPLACE PACKAGE BODY "YYY"."PACKAGEONE" IS' s from dual union all
  select 'Create or REPLACE PACKAGE BODY "ZZZ"."PACKAGETWO" IS' s from dual
  )   
  select REGEXP_REPLACE(T.s, '^.PACKAGE BODY$','(\1)_new',1,1,'i') as s from T;
  /*
  Expected output:
  CREATE OR REPLACE PACKAGE BODY "YYY"."PACKAGEONE_new" IS
  Create OR REPLACE PACKAGE BODY "ZZZ"."PACKAGETWO_new" IS
  */

  All data as a result of model:

  CREATE OR REPLACE PACKAGE BODY "[owner]"."[name]" IS

  Where [owner] can be any string. In the examples of data we have for example XXX and YYY values here.
  And [name] can be any string. In the sample data we have for example values PACKAGEONE and PACKAGETWO here.
  Other parts of the chain is fixed and rest as you see.
  In accordance with the request of expected results should replace substring '[owner] '. ' ' [name] ' '[owner] '. "" [name] _new.
  
  How can I write this query?
  I think that I have in some way should in regular expression counts the positions of double quotes to achieve the expected result, but I don't know how.

  SQL> ed
  Wrote file afiedt.buf
  
    1  with T as
    2  (
    3  select 'CREATE OR REPLACE PACKAGE BODY "YYY"."PACKAGEONE" IS' s from dual union all
    4  select 'Create or REPLACE PACKAGE BODY "ZZZ"."PACKAGETWO" IS' s from dual
    5  )
    6* select REGEXP_REPLACE(T.s, '"([^.]+\.[^.]+)"','"\1_new"') as s from T
  SQL> /
  
  S
  ----------------------------------------------------------------------------------------
  CREATE OR REPLACE PACKAGE BODY "YYY"."PACKAGEONE_new" IS
  Create or REPLACE PACKAGE BODY "ZZZ"."PACKAGETWO_new" IS
  
  SQL>
  

• Outlook Express 6: replacement of the Archives after problem

  How can I replace/reinstate my archive after that my OE6 was corrupted and newly rebooted.

  Archive is saved completely (in .dbx) but I can't find the solution to make the message readable again in my OE!  Help!

  Outlook Express problems are dealt with elsewhere: http://social.answers.microsoft.com/Forums/en-US/xpnetwork/threads Russ Valentine

• Expression to replace the image according to the text of the layer.

  Hello, I wanted to ask you is it possible to replace the image in the function composition text text layer. That is the text layer contains code # 34 and expression indicates what kind of image should be placed according to conditions in expression of image.

  Sincerely, Deimantas

  Hello

  Expressions can not add/replace images.

  All images must be here somewhere, in the form of layers in the model or in a precomp, but not only as items in the project Panel.

  Then you can play with the opacity of these layers, with an expression of opacity to each of them:

  textLayer = (pickwhip the text layer that transports information);

  If (textLayer.sourceText == thisLayer.name) 100

  0 otherwise;

  The condition can be something different, for example "if (textLayer.sourceText is thisLayer.index"), etc, etc.

  Not clear what should be this condition in your case.

  Xavier.

• Validation of object Expression regular email address

  Hello
  
  Does anyone know of a regular expression, that I can use to validate an e-mail address field? I use Apex 3.2.
  
  Your help would be much appreciated.
  
  Thank you
  ca84

  ^((\s*[a-zA-Z0-9\._%-]+@[a-zA-Z0-9\.-]+\.[a-zA-Z]{2,4}\s*[,;:]){1,100}?)?(\s*[a-zA-Z0-9\._%-]+@[a-zA-Z0-9\.-]+\.[a-zA-Z]{2,4})*$
  

• Database 10 g Express Edition and Oracle 10 g install

  I have Oracle 10 g client on my desktop and it connects to a server Oracle 10g on a database server by using a tnsnames.ora file. I feel that to run Oracle directs that a client server and database is required. Is this correct? What is the database 10g Express Edition? It is the client, the database server, or both? How should I use the tnsnames.ora file? If it is a database server where the tables are stored? Please help me I'm confused. Thank you.
  
  Este

  >
  I have Oracle 10 g client on my desktop and it connects to a server Oracle 10g on a database server by using a tnsnames.ora file. I feel that to run Oracle directs that a client server and database is required. Is this correct?
  >

  Not necessarily; in fact, you can run the Oracle database commands using pass (local) connections. You would of course be connected to the operating system where it is installed your Oracle database to establish local connections without using tnsnames.

  >
  What is the database 10g Express Edition? It is the client, the database server, or both? How should I use the tnsnames.ora file?
  >

  Oracle Database XE is a database server. Client software must be installed on client systems for remote access to Oracle Database 10g Express Edition. You need not install the client on the same computer as Oracle Database 10 g Express Edition.

  >
  If it is a database server where the tables are stored? Please help me I'm confused. Thank you.
  >

  The tables are logical structures that are in your data files; the data files are physically stored on the operating system on which the Oracle database server is installed.

  Kind regards
  Phiri

• expression regular {}

  Can you please explain the below two statement

  SELECT COUNT (*)
  OF THE DOUBLE
  WHERE REGEXP_LIKE ('670013432423615 ', '^(6700[0-9]{11,15}$)')

  SELECT COUNT (*)
  OF THE DOUBLE
  WHERE REGEXP_LIKE ('670013432423615 ', '^(6700[0-9]{12,15}$)')

  We do match and another does not.

  What is the diff?

  Hello

  2989211 wrote:

  Can you please explain the below two statement

  SELECT COUNT (*)
  OF THE DOUBLE
  WHERE REGEXP_LIKE ('670013432423615 ', '^(6700[0-9]{11,15}$)')

  SELECT COUNT (*)
  OF THE DOUBLE
  WHERE REGEXP_LIKE ('670013432423615 ', '^(6700[0-9]{12,15}$)')

  We do match and another does not.

  What is the diff?

  The difference is that the former has 11 where the second has 12:

  ' ^ (6700 [0-9] {11, 15} $) '

  ' ^ (6700 [0-9] {12, 15} $) '

  Both are looking for '6700' at the beginning of the string, followed immediately of N digits, and then the end of the string.  The only difference is N.

  In the first expression, N is a number between 11 and 15.  In this case, there are exactly 11 digits after '6700' in the string, so that it corresponds.

  In the second expression, N is a s number 12 and 15.  11 is not between 12 and 15, so it does not.

• Oracle Weblogic Server will replace Oracle Application Server in the future!

  I would like to know if the Oracle Weblogic Server will become the technology stack into future releases on the EBS instead of the Oracle ACE...

  Yes - that's the plan. WebLogic's application server included in the merger applications.

  http://technology.AMIS.nl/Blog/5645/Oracle-Fusion-applications-it-is-for-real-and-impressive-too
  http://www.theregister.co.UK/2009/10/13/oracle_weblogic_roadmap/

  As far as I KNOW, R12.x will continue to use Oracle AS - this is a personal opinion.

  HTH
  Srini