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DRM Web Client 500.19 issue on windows 2008(x64) ERROR

825689Aug 31 2011 — edited Sep 4 2013
Hi,

I installed DRM version 11.1.2 on windows server 2008(x64), according to the documents, I pre-installed .Net 3.5.1 and modified the applicationHost.conf file.

I found this in epm_install_troubleshooting

Web Client Access Failure
Issue: After installing Data Relationship Management on Windows 2008 64-bit platform, you
get this error message when attempting to access the Web client:
HTTP Error 500.19 - Internal Server Error The requested page cannot be
accessed because the related configuration data for the page is invalid.

Solution: In the IIS configuration file (C:/Windows/System32/inetsrv/config/
applicationHost.config), replace the two occurrences of Deny in the following section with
Allow:
<configuration>
<configSections>
<sectionGroup name="system.webServer">
<section name="handlers" overrideModeDefault="Deny" />
<name="modules" allowDefinition="MachineToApplication"
overrideModeDefault="Deny" />

However when I access the web client I got the 500.19 error, can't find C:\inetpub\wwwroot\drm-web-client\web.config, the web.config file do not exist.

Config File \\?\C:\inetpub\wwwroot\drm-web-client\web.config

Comments

BluShadow
Not sure about a clean regular expression for it. I'm sure CD will be along soon to show us how... in the meantime, I managed to get it to deal with that specific format of string...
SQL> ed
Wrote file afiedt.buf

  1  with t as (select 'a, b, c (x, y, z)' as txt from dual)
  2  --
  3  select txt, replace(regexp_replace(txt, '(^,)*\(.*\)','\1'),',')||regexp_replace(txt,'.*(\(.*\))','\1') as new_txt
  4* from t
SQL> /

TXT
-----------------
NEW_TXT
--------------------------------------------------------------------------------
a, b, c (x, y, z)
a b c (x, y, z)
But if your format of string differs then it's gonna be more complex as far as I can tell.
cd_2
lurchmodeon You rang? lurchmodeoff

How about this?
WITH t AS (SELECT 'a, b, c (x, y, z)' col1 
             FROM dual)
SELECT t.col1
     , REGEXP_REPLACE(t.col1, '(\(.*\))|,', '\1') new_col
  FROM t
;               

COL1              NEW_COL
----------------- --------------------
a, b, c (x, y, z) a b c (x, y, z)
C.
564484

Hi cd and BluShadow,

Thanks very much for your valuable help!

One other question - what if I want to replace the non-() enclosed commas with another string - such as "#". I tried your suggestion and got this:

WITH t AS (SELECT 'a, b, c (x, y, z)' col1 
             FROM dual)
SELECT t.col1
     , REGEXP_REPLACE(t.col1, '(\(.*\))|,', '\1') cds_awesome_solution
     , REGEXP_REPLACE(t.col1, '(\(.*\))|,', '#\1') pdaddys_futile_attempt1
     , REGEXP_REPLACE(t.col1, '(\(.*\))|,', '\1#') pdaddys_futile_attempt2
  FROM t
;

Result:

"COL1"	             "CDS_AWESOME_SOLUTION"   "PDADDYS_FUTILE_ATTEMPT1"   "PDADDYS_FUTILE_ATTEMPT2"
"a, b, c (x, y, z)"    "a b c (x, y, z)"        "a# b# c #(x, y, z)"        "a# b# c (x, y, z)#"

Notice how the pound is place before the left paren in attempt1 and placed after the right paren in attempt2? I know I could nest another regexp_replace - but I was wondering if a single pattern could suffice...

Thanks - and sorry I didn't make my question clear at all in my original post...

You guys ARE geniuses

cd_2
Not sure about a single pattern, but at least without a 2nd regex_replace:
WITH t AS (SELECT 'a, b, c (x, y, z)' col1
             FROM DUAL)
SELECT t.col1,
       REPLACE(REGEXP_REPLACE(t.col1, '(\(.*\))|(,)', '\1\2\2'), ',,', '#') new_col
  FROM t;

COL1              NEW_COL
----------------- --------------------
a, b, c (x, y, z) a# b# c (x, y, z)
C.
564484
Thanks!

That works perfectly!

(though I don't know HOW it works :( - I don't understand how the \1 backreference gets the part of the string before the parentheses... )

oh well - sometimes you have to just have faith :) - seeing is believing.

Message was edited by:
PDaddy
cd_2
(though I don't know HOW it works :( - I don't
understand how the \1 backreference gets the part of
the string before the parentheses... )
You have to think of regular expressions as a parser. Each position of the searched string is compared to each available pattern. The first "()" search pattern sort of protects that string against the second pattern.
oh well - sometimes you have to just have faith :) -
seeing is believing.
Or you start thinking in regular expressions. ;-)

C.
nemecj
Excelent solution! You may consider to add a ? (non greedy, restrictive mode) to cover the case there are more parentheses in the string.

<pre>
WITH t AS (SELECT 'a, b, c (x, y, z), a, (xx, yy, zz), x,' col1
FROM dual)
SELECT t.col1
, REGEXP_REPLACE(t.col1, '(\(.*?\))|,', '\1') new_col
FROM t
;

COL1 NEW_COL
--------------------------------------
a, b, c (x, y, z), a, (xx, yy, zz), x, a b c (x, y, z) a (xx, yy, zz) x
</pre>
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Post Details

Locked on Oct 2 2013
Added on Aug 31 2011
#business-intelligence-applications, #hyperion-drm-analytical-master-data, #system-management-and-integration
2 comments
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