Ask help for OID on Linux

I installed OID which come from the download version of Oracle 8i 8.1.7 on redhat linux 6.2. I use universal installer installed the software code and run postcfg script in $ORACLE_HOME/ldap/postcfg to generate the database and LDAP schema. After correct some
bugs manually in the postcfg(it is better to add one sentence of oidmon -start) and bugs in oidadmin(script in $ORACLE_HOME/bin, one is it point to a wrong place of jre, the other is it use a switch that only worked on sparc), I finally started the oidadmin.
BUT, I still can not login the OID from oidadmin. Can any guy kind give me some info on what is the initial password for oidadmin of OID? I have tried all combos that listed on the on-line help, like: cn=orcladmin, oracladmin, etc. it doesn't work. I also use sqlplus as system to search it in the tables
of the OID database, but find nothing.
Thanks for any info or hint!
null

Hello:
Sorry it took so long to respond to your question. I was exhibiting OID at AppsWorld in Paris and New Orleans for the past 2 weeks.
Assuming you have followed the installation procedures from the "OID Installation Guide" there is one small bug patch that needs to be applied to the Linux versions. You can obtain a copy of this patch from the Oracle Metalinks web site:
http://www.oracle.com/support/metalink/index.html
Assuming that OID is up and running you should be able to start a gui JAVA tool from the command line prompt by typing in "oidamdin" or from your Linux Window console. The super user account name is "orcladmin" and the default password is "welcome". The default port number is 389. Also make sure you put in the host name of your machine.
Let me know if any of this helps. If not Ill try to help you reinstall OID to make sure it is installed properly. I am unable to fully understand where you are stuck from your description of the problem.
thanks,
Jay
null

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Oidldapd service does not start for OID on LINUX

Hi,
I have installed OCS 9.0.3.0.0 on LINUX, all components i.e. Infrastructure, Storage and MiddleTier are installed properly (without giving any error). While trying to start OID with oidmon and oidctl (through command line), oidmon is started but OID is not available.
Surprisingly oidldapd01.log file is not found (as log registers), where to locate this file ?
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2003/05/01:17:37:19ODISRV Instance 1 deleted from registry
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2003/05/01:17:41:41Updating Process Table...
2003/05/01:17:41:51OIDLDAPD server terminated normally
2003/05/01:17:41:51Check log file oidldapd01.log
2003/05/01:17:41:51Deleting OIDLDAPD instance 1 from Process table
2003/05/01:17:41:52OIDLDAPD Instance 1 deleted from registry
2003/05/01:18:10:15Updating Process Table...
2003/05/01:18:10:15Starting OIDLDAPD Server, PID=1620
2003/05/01:18:10:15Exec of OIDLDAPD Server failed with error=2
2003/05/01:18:10:15OIDLDAPD Instance 1 added into registry
2003/05/01:18:10:15Updating Process Table...
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2003/05/01:18:10:25Check log file oidldapd01.log
2003/05/01:18:10:25Deleting OIDLDAPD instance 1 from Process table
2003/05/01:18:10:27OIDLDAPD Instance 1 deleted from registry
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Regards
Ashwani

Hi Manuel,
I tested on my client, and I also keep getting the error 1053, which indicates timeout and the new created service can not be started during the limit time, so you can try to change the service time limit, please note this needs to change registry value,
Using Registry Editor incorrectly can cause serious problems:
In Registry Editor, locate, and then right-click the following registry subkey:
HKEY_LOCAL_MACHINE\SYSTEM\CurrentControlSet\Control
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Refer to:
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Please Note: Since the web site is not hosted by Microsoft, the link may change without notice. Microsoft does not guarantee the accuracy of this information.
If there is anything else regarding this issue, please feel free to post back.
Best Regards,
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Hi,
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Ask help for a question about Caesar cipher

"Caesar ciphers are among the simplest devised, and rely solely on remapping characters to others in the alphabet using a constant shift modulo the size of the alphabet. The amount shifted is the key used to encipher, or decipher, the message. This remapping is usually restricted to letters, so that with a key of 2, `A' is replaced by `C', `B' by `D', ..., `X' by `Z', `Y' by `A', and `Z' by `B'. Lower case letters are mapped in an identical way to give their lower case replacements. Thus here the key is an integer in the range 1 to 25 (not 0 to 25, as there is no point replacing every `A' by an `A' etc.).
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L M N O P
Q R S T U
V W X Y Z
A B C D E
F G H I K
and the uppercase alphabet upon encryption becomes: LQVAFMRWBBGNSXCHOTYDIPUZEK Note that here the `B' is doubled up because it represents both `I' and `J'. Thus, with this key of 10, the following line is enciphered to the one after it.
Caesar ciphers are child's play.
Vlfylt vbhwfty ltf vwbna'y hnle.
An acceptable drawback of the treatment of `I' and `J' in the grid is that a decoded message will have all `J's mapped to `I's, and all `j's to `i's"
this is a part of requirement in the whole program,I almost finished it, but I do not know thata how to make" a decoded message will have all `J's mapped to `I's, and all `j's to `i's". "Treating `I' and `J' as one letter from the start"?what does that mean?
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if(ch>='a'&&ch<='z')
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return ch;
public static char decode(char ch,int n) {
while (!Character.isLetter(ch))
return ch;
if(ch>='A'&&ch<='Z')
return (char)('A'+(ch-'A'-n)%26);
if(ch>='a'&&ch<='z')
return (char)('a'+(ch-'a'-n)%26);
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}but
how to make" a decoded message will have all `J's mapped to `I's, and all `j's to `i's". "Treating `I' and `J' as one letter from the start"?

Your teacher/instructor/professor would obviously be the final authority on this, but I think all it's saying is that when you encode, treat each 'j' you come across as an 'i', and therefore, when you decode, because you've lost information that it was a 'j' to start with, it will be mapped back to an 'i'.
One easy way to accomplish this (that would save special cases later in your for-loop) would simply be a:
String toEncode = "Djibouti"
toEncode = toEncode.replaceAll("j","i"); //obviously handle both cases instead
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System.out.println(decoded); //this would print "Diibouti"The other way is to simply add a special case to your encode/decode (char) methods.
Edited by: endasil on 28-Sep-2009 9:35 AM

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I am using anydataset as a column type. after referring to the demo anydset.sql:
can run then following correctly:
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as1 Sys.AnyDataSet;
at1 Sys.AnyType ;
cl1 clob ;
begin
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for i in 1..2 loop
     as1.AddInstance();
     cl1 := to_clob(rpad('Clob',i*10,'Clob')) ;
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end loop ;
as1.EndCreate();
--anydataset_display_value(as1);
end ;
while if just added one sql to update the cell value to as1 as following:
declare
as1 Sys.AnyDataSet;
at1 Sys.AnyType ;
cl1 clob ;
begin
Sys.AnyDataSet.BeginCreate(DBMS_TYPES.TYPECODE_CLOB,at1,as1);
for i in 1..2 loop
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end ;
I got :
ERROR at line 1:
ORA-22370: incorrect usage of method AnyData Insert
ORA-06512: at line 14
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Also, you should always close your resources in finally blocks. So not this:
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Since you are a beginner, I'd advice you to take a
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%

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HT204088 Dear sirs, Could you please help me on my problem with my apple ID I create new account with security questions, and when I try to purchase paid application,it ask me for answers the questions but it's not working, My account have 50$ and stil

Dear sirs,
Could you please help me on my problem with my apple ID
I create new account with security questions, and when I try to purchase paid application,it ask me for answers the questions but it's not working,
My account have 50$ and still didn't buy anything

You need to ask Apple to reset your security questions; ways of contacting them include clicking here and picking a method for your country, phoning AppleCare and asking for the Account Security team, and filling out and submitting this form.
(100546)

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