Password file path

I want to create a password file in Solaris 10, is there any requirement for its path? i.e., can I create the password file in whatever path I want?

yes you can else it will create on default folder that is <ORACLE_HOME>/dbs (in unix)
<ORACLE_HOME>/database (in windows)
Rgds,
Umair

Similar Messages

How to change the default password file's name and path when the database created?

how to change the default password file's name and path when the database created?
null

Usage: orapwd file=<fname> password=<password> entries=<users>
where
file - name of password file (mand),
password - password for SYS and INTERNAL (mand),
entries - maximum number of distinct DBA and OPERs (opt),
There are no spaces around the equal-to (=) character.

HT201262 cannot go to cmd+V or S, I cannot change my password. -error message ===boot file path system library coreservices boot.efi.... Please help.

cannot go to cmd+V or S, I cannot change my password. -error message ===boot file path system library coreservices boot.efi.... Please help.

Got it thanks macjack . Command + S on restart.
In this case Nad69-Breizh did you try restarting with the option key down and re-selecting the boot drive.
If no Recovery option, try command option R for internet recovery. Takes some time to load up.

Take Source file path in sender adapter from an environment variable .

Hi,
Is it possible to take source file path(half file path) dynamically in sender file adapter.
Please suggest any alternative to this.

Hi,
One option, use Dynamic Configuration and set the values in the Mapping.
By the ways, the File Directory and the Filename are taken runtime in the file adapters after transportation.
Only if you are using FTP will you need to key in the FTP addess user id password etc.
Refer -- same way try for File adapter
Dynamic Configuration of Some Communication Channel Parameters using Message Mapping
Ps Note : I have personally never tried it to use the Dynamic Configuration for file path. But you can try.
Thanks
Swarup

Pass File Path into Message Box

Hello,
I’m building an application that is being used to create a report for whatever date range the user chooses. After the data is exported into Excel I have a message box pop up telling the user where the file has been saved to. However, as it is now, I have
the file path hard coded. What I want to do is have the file path chosen by the user passed into the message box and display a message with the selected path. Any help with this will be greatly appreciated.
Dave
using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Windows.Forms;
using System.Data.SqlClient;
using ClosedXML.Excel;
using DocumentFormat.OpenXml;
using System.IO;
namespace LoanOrig_FDIC_Codes
public partial class Form1 : Form
SqlCommand sqlCmd;
SqlDataAdapter sqlDA;
DataSet sqlDS;
DataTable sqlDT;
SqlCommand sqlCmdCnt;
public Form1()
InitializeComponent();
//BEGIN BUTTON LOAD CLICK EVENT
private void btnLoad_Click(object sender, EventArgs e)
string sqlCon = "Data Source=FS-03246; Initial Catalog=ExtractGenerator; User ID=myID; Password=myPW";
//Set the 2 dateTimePickers to today's date
DateTime @endDate = End_dateTimePicker.Value.Date;
DateTime @startDate = Start_dateTimePicker.Value.Date;
//Validate the values of the 2 dateTimePickers
if (endDate < startDate)
MessageBox.Show("End Date must be greater than or equal to the Start Date OR Start Date must be less than or equal to the End Date ", "Incorrect Date Selection",MessageBoxButtons.OK,MessageBoxIcon.Error);
//Reset both dateTimePickers to todays date
Start_dateTimePicker.Value = DateTime.Today;
End_dateTimePicker.Value = DateTime.Today;
return;
//End of date validation
string sqlData = @"SELECT AcctNbr,
CurrAcctStatCD,
Org,
MJAcctTypCD,
MIAcctTypCD,
NoteOriginalBalance,
ContractDate,
FDICCATCD,
FDICCATDESC,
PropType,
PropTypeDesc
FROM I_Loans
WHERE CAST(ContractDate AS datetime) BETWEEN @startdate AND @enddate ORDER BY ContractDate";
SqlConnection connection = new SqlConnection(sqlCon);
SqlCommand sqlCmd = new SqlCommand(sqlData, connection);
sqlCmd.Parameters.AddWithValue("@startDate", startDate);
sqlCmd.Parameters.AddWithValue("@endDate", endDate);
sqlDS = new DataSet();
sqlDA = new SqlDataAdapter(sqlCmd); //SqlAdapter acts as a bridge between the DataSet and SQL Server for retrieving the data
connection.Open();
sqlDA.SelectCommand = sqlCmd; //SqlAdapter uses the SelectCommand property to get the SQL statement used to retrieve the records from the table
sqlDA.Fill(sqlDS, "I_Loans"); //SqlAdapter uses the "Fill" method so that the DataSet will match the data in the SQL table
sqlDT = sqlDS.Tables["I_Loans"];
//Code section to get record count
sqlCmdCnt = connection.CreateCommand();
sqlCmdCnt.CommandText = "SELECT COUNT(AcctNbr) AS myCnt FROM I_Loans WHERE ContractDate BETWEEN @startDate AND @endDate";
sqlCmdCnt.Parameters.AddWithValue("@startDate", startDate);
sqlCmdCnt.Parameters.AddWithValue("@endDate", endDate);
int recCnt = (int)sqlCmdCnt.ExecuteScalar();
txtRecCnt.Text = recCnt.ToString();
btnExport.Enabled = true;
//End of code section for record count
connection.Close();
dataGridView1.DataSource = sqlDS.Tables["I_Loans"];
dataGridView1.ReadOnly = true;
//Reset both dateTimePickers to todays date
Start_dateTimePicker.Value = DateTime.Today;
End_dateTimePicker.Value = DateTime.Today;
//END BUTTON LOAD CLICK EVENT
//BEGIN BUTTON EXPORT CLICK EVENT
private void btnExport_Click(object sender, EventArgs e)
{ //ClosedXML code to export datagrid result set to Excel
string dirInfo = Path.GetPathRoot(@"\\FS-03250\users\dyoung\LoanOrig_FDIC_Codes");
if (Directory.Exists(dirInfo))
var wb = new XLWorkbook();
var ws = wb.Worksheets.Add(sqlDT);
ws.Tables.First().ShowAutoFilter = false;
SaveFileDialog saveFD = new SaveFileDialog();
saveFD.Title = "Save As";
saveFD.Filter = "Excel File (*.xlsx)| *.xlsx";
saveFD.FileName = "LoanOrig_FDIC_Codes_" + DateTime.Now.ToString("yyyy-MM-dd");
if (saveFD.ShowDialog() == System.Windows.Forms.DialogResult.OK)
Stream stream = saveFD.OpenFile();
wb.SaveAs(stream);
stream.Close();
//End of ClosedXML code
MessageBox.Show("File has been exported to U:\\LoanOrig_FDIC_Codes", "File Exported", MessageBoxButtons.OK, MessageBoxIcon.Information);
else
MessageBox.Show("Drive " + "U:\\Visual Studio Projects\\LoanOrig_FDIC_Codes" + " " + "not found, not accessible, or you may have invalid permissions");
return;
//END BUTTON EXPORT CLICK EVENT
private void Form1_Load(object sender, EventArgs e)
//Set dates to be today's date when the form is openend
Start_dateTimePicker.Value = DateTime.Today;
End_dateTimePicker.Value = DateTime.Today;
private void Form1_Load_1(object sender, EventArgs e)
// TODO: This line of code loads data into the 'dataSet1.I_Loans' table. You can move, or remove it, as needed.
this.i_LoansTableAdapter.Fill(this.dataSet1.I_Loans);
private void iLoansBindingSource_CurrentChanged(object sender, EventArgs e)
private void btnExit_Click(object sender, EventArgs e)
this.Close();
//END THE SAVE AS PROCESS
David Young

Assuming I have located the part of your code you are talking about, I think you just need to replace the hard code path in the message with the path from the SaveFileDialog. You should only display the message if the use clicked OK.
if (saveFD.ShowDialog() == System.Windows.Forms.DialogResult.OK)
Stream stream = saveFD.OpenFile();
wb.SaveAs(stream);
stream.Close();
MessageBox.Show("File has been exported to " + saveFD.FileName, "File Exported", MessageBoxButtons.OK, MessageBoxIcon.Information);

Problem setting up user ids to use the Oracle password file.

I want to set up my database users so that a password file is used for connecting to the database. I have completed these steps successfully.
BTW - DB is 11.2.0.3 on AIX power 64
1. Created the password file for the database using the orapwd command. Allowing 20 entries. Confirmed the file was created in the $ORACLE_HOME/dbs directory
2. Created a database user, sbrower
CREATE USER SBROWER IDENTIFIED BY <password> DEFAULT TABLESPACE USERS TEMPORARY TABLESPACE TEMP PROFILE DEFAULT ACCOUNT UNLOCK ;
GRANT DBA TO SBROWER;
ALTER USER SBROWER DEFAULT ROLE ALL;
GRANT UNLIMITED TABLESPACE TO SBROWER;
ALTER USER SBROWER QUOTA UNLIMITED ON USERS;
3. Connected to the database as SYS and granted sysoper to SBROWER
4. Using putty, ssh'ed into the server where the database resides.
5. Set the oracle variables (ORACLE_HOME, ORACLE_BASE, etc.) and PATH
6. Was able to connect to the database using sqlplus / as syoper
THE PROBLEM
For another user, EA_RDX_ORACLE1, I follow the same steps (2-6) bu when I execute step 6, it does not allow the connection
ERROR:
ORA-01031: insufficient privileges
   but, if I use sqlplus ea_rdx_oracle1/thepassword as sysoper it works
Looking at v$pwfile_users on the database:
USERNAME SYSDBA   SYSOPER      SYSASM
SYS      TRUE      TRUE      FALSE
SBROWER FALSE     TRUE      FALSE
EA_RDX_ORACLE1 FALSE TRUE FALSE
3 rows selected.
There is one thing that is different for the ea_rdx_oracle1 id's:
- The users who use this id, use a took called CyberVault to check out the id. The password for the id changes each time the id is checked out; however, the way the id is set up on the DB servers, us user does not have to enter the password when they log in (ssh).
I have sent an email to our unix admin asking his how the id was set up so that it can ssh into the server. It is not included in the list of users in any group in the /etc/ogroup file and it is not included in the /etc/opassword file.

The OS authentication ( sqlplus / as sysdba ) does not require the password file.
The problem may be related to the OS user you are connecting to that server - it is not a member of OSDBA group ( usually DBA ).

Deploying war file on Tomcat: file path problem

Hello all,
i am using my eclipse ide for automatically deploy my webapplication to Tomcat. For Database connectivity i configured hibernate. To configure hibernate i am using the following code:
package de.wfm.hibernate.hibernateUtil;
import org.hibernate.HibernateException;
import org.hibernate.Session;
import org.hibernate.SessionFactory;
import org.hibernate.cfg.Configuration;
public class ESDBFactory {
     private static SessionFactory sf;
     private static Session session = null;
     private static String pathToCfgFile = "de\\"
                         + "wfm\\hibernate\\hibernateUtil\\esdb.cfg.xml";
     public static synchronized Session getSession() throws HibernateException {
          if (session==null) {
               if (sf==null) {
                    sf = new Configuration()
                         .configure(pathToCfgFile).
                         buildSessionFactory();
          session = sf.openSession();
          return session;
}When i use this in the ide all is fine. But when i export my application to a war file and deploy it on Tomcat i got the exception:
2006-07-28 10:06:10 StandardWrapperValve[Urlaubsplanung Servlet]: Servlet.service() for servlet Urlaubsplanung Servlet threw exception
javax.faces.FacesException: #{User.doLogin}: javax.faces.el.EvaluationException: org.hibernate.HibernateException: de\wfm\hibernate\hibernateUtil\esdb.cfg.xml not foundHow to avoid this? What do i have to do to set the file path correctly?
Regards,
ak

hmm.. ic
The .war file when drop in the webapps dir and u restart tomcat, it auto create a dir with your .war filename and the content of .war file is copied to there.
Am i rite to say so?
And wat about the web.xml?
If i need to add users and password how can I accomplished it in tomcat web.xml?
Last but not least, thanks for taking the time to reply. :D

File path problem in jsp

Hi techies,
I am having a jsp ,which is having the following fields,
url
driver
userName
Password
Location of the file
Here is my jsp code
<table rows=6 cols=2>
               <tr>
                    <td>Enter url</td>
                    <td> <input type="text" name="url" value ="<%= url%>"></input></td>
               </tr>
               <tr>
                    <td>Enter DriverName </td>
                    <td><input type="text" name="driver" value="<%= driver%>"></td>
               </tr>
               <tr>
                    <td>Enter UserName </td>
                    <td><input type="text" name="userName"></td>
               </tr>
               <tr>
                    <td>Enter Password </td>
                    <td><input type="password" name="passWord"></td>
               </tr>
               <tr>
                    <td>choose the backupXml file</td>
                    <td><input type ="file" name ="xmlFileName"></td>
                    <input type="hidden" name ="xmlFile" >
               </tr>
               <input type="hidden" name ="resubmit" >
          </table>
               <br>
               <input type="button" value="submit" onClick ="doSubmit()" >
                 For the first time I am sending request to the servlet and after checking the conditions we will get one confirmation box.
If we click ok (i.e true) ,we have to again send the same data to the servlet.
I am keeping all the variables in a session in servlet and retrieving in to jsp, but the problem is ,when again sending request to servlet , I am missing file name,but I am getting file path(i.e c:/sun/appserver/bin/) only.
I need to get c://sun/appserver/bin How can i get entire path??
Here is my servlet code
File f = new File(xmlFileName);
String     xmlFilePath = f.getAbsolutePath();
System.out.println("xmlFilePath"+xmlFilePath);
session.setAttribute("xmlFile",xmlFilePath);Here is my java script code
function init() {
     <%
          if(message !=null && messageType != null && messageType.equals("ALERT")){ %>
          var mss = confirm("<%=message%>");
        // alert(mss);
           if(mss == true)
      var user = '<%=session.getAttribute("user")!=null?session.getAttribute("user"):"0"%>';
        var pass = '<%=session.getAttribute("pass")!=null?session.getAttribute("pass"):"0"%>';
     var xmlF = '<%=session.getAttribute("xmlFile")!=null?session.getAttribute("xmlFile"):"0"%>';
                alert("inside if"+xmlF);
                document.forms[0].resubmit.value = "true"
                document.forms[0].userName.value =user;
                alert(document.forms[0].userName.value);
                document.forms[0].passWord.value=pass;
                alert(document.forms[0].passWord.value);
                document.forms[0].xmlFile.value=xmlF;
                alert(document.forms[0].xmlFile.value);
                doSubmit();
          else
               alert("else");
     <% } %>
function doSubmit() {
                   alert(document.forms[0].userName.value);
                   alert(document.forms[0].passWord.value);
                    alert(document.forms[0].xmlFileName.value);
            document.forms[0].submit();
                     }

Thanks for u r reply Mr.Prasad.
Sorry for the delay for reply.
My scenario is in 2 phases
Phase 1. I am taking input from the front end using jsp and writing the contents in to a file and then downloading that file on to my desktop.(Now let us say this one as file downloading)
Phase 2. Now I had designed a page,which locates the downloaded file by using the following tag.
                    <td><input type ="file" name ="xmlFileName"></td>
    In the downloaded xml file I am having applicationid. I am checking whether the application id already xists in the database (or)not.
If already exists in the database , the end user will get one confirmation msg,saying application already exists in the database.
If the user clicks ok again the request has to go to the servlet.
Now the actual problems are
1. When I locate the file on the desktop let us say c:\desktop\krish and click on submit button, it is locating to the file that is present in the server(this is the case for the first time means before checking the application id in the database).
2. when the end user clicks "ok" on the confirmation box , I am not getting the file name,that has to be submitted to the servlet.
Can u plz help me how to resolve this issue.
regards,
Krish

ORA-01991: invalid password file

I used producation backup file to restore in development side... but I get the following errors:
RMAN> restore controlfile from "E:\RMAN\C-922101420-20090628-01.BCK";
Starting restore at 29-JUN-09
using target database controlfile instead of recovery catalog
allocated channel: ORA_DISK_1
channel ORA_DISK_1: sid=14 devtype=DISK
channel ORA_DISK_1: restoring controlfile
channel ORA_DISK_1: restore complete
replicating controlfile
input filename=C:\ORACLE\ORADATA\HUBDB\CONTROL01.CTL
output filename=C:\ORACLE\ORADATA\HUBDB\CONTROL02.CTL
output filename=C:\ORACLE\ORADATA\HUBDB\CONTROL03.CTL
Finished restore at 29-JUN-09
RMAN> startup mount;
database is already started
RMAN-00571: ===========================================================
RMAN-00569: =============== ERROR MESSAGE STACK FOLLOWS ===============
RMAN-00571: ===========================================================
RMAN-03002: failure of startup command at 06/29/2009 15:57:11
ORA-01991: invalid password file 'C:\oracle\ora92\DATABASE\PWDhubdb.ORA'
I don't have any ideas?
Amy

Dear Amy!
Symptom:
When trying to open (or mount) a database the following error is generated:
ORA-01991: invalid password file '...PWDOrcl.ora'Where ‘Orcl’ is the SID of the database instance.
Cause:
Typically some corruption of the password file, or possibly the password file is for another database instance?
Possible Remedies:
If encountered when creating the database then it is because Oracle expects to create the password file but it already exists. In this instance database creation will have failed. Delete the password file and try again.
If it is suspected that the password file is corrupt then try deleting it and recreating it:
1. Delete the existing password file.
2.Create a new/replacement password file:
orapwd file=<path>\PWD<sid>.ora password=oracleUse the same path as given in the Oracle error. The password specified will be the new internal password.
Yours sincerely
Florian W.

GRANT failed: password file missing or disabled

Hi,
I created password file using
orapwd FILE=orapwprj ENTRIES=30 (where prj is the instance name)
SQL> show parameter remote_login_passwordfile
NAME TYPE VALUE
remote_login_passwordfile string EXCLUSIVE
SQL> grant sysdba to rempar;
grant sysdba to rempar
ERROR at line 1:
ORA-01994: GRANT failed: password file missing or disabled

Command to be used as
orapwd file=password_file_name_with_complete path password=the_passwordBy default, the password file should be located in $ORACLE_HOME/dbs location for UNIX platform with password file format as "orapw<SID>"
For windows, the password file should be located in %ORACLE_HOME%\database location with password file format as "PWD<SID>"

How get password file's derectory?

I want to copy the password file of current database to another host so as to configure a satandby database.Where can I get the password file's derectory?

I am not sure I understand your question. The minimum command should be:
orapwd file=OH/dbs/orapwsidname password=sys_pwd_you_want
replace OH with actual path. On Windows, dbs should be database instead.

Common URL format for sharing file paths between Mac and Windows?

Hi -
I have to integrate several Macs into a Windows environment. We need to be able to copy and paste file paths to share between all machines. Is there a common URL format that can be used between the platforms?
On Windows, I have:
\\server\share\file
On the Mac:
smb://server/share/file
Thank you for your help,
Steve

On Windows you have what is known as a "UNC".
On the Mac, you have a "URL". It starts with the protocol to use "smb" then gives the path to the item using standard conventions.
UNC is Microsoft mainly.
The only way I know for Windows to use a URL is with an application like a web browser or ftp client.
On the Mac, you can use UNCs but they must be modified as follows.
In Terminal:
smbclient \\\\servername\\sharename\\filename -U username
And enter a password if prompted.
Notice that you must double up the slashes. This is due to how UNIX shells treat the backslash.
You will connect to the share, but it will only be in Terminal.

Minimum entries in password file

Hello,
I have created password file using following command ( on WXP ):
orapwd file=path entries=1 password=password
File was succesfully created, but server allows me to grant dba, sysdba, sysoper to 5 different users, until it displayd error message ORA-01996: ...password file ...is full.
How can I restrict minimum entries in password file to 1?
Thank you.
Tibor

Maybe this extract from Oracle documentation can explain :
ENTRIES
This parameter specifies the number of entries that you require the password file to accept. This number corresponds to the number of distinct users allowed to connect to the database as SYSDBA or SYSOPER. The actual number of allowable entries can be higher than the number of users, because the ORAPWD utility continues to assign password entries until an operating system block is filled. For example, if your operating system block size is 512 bytes, it holds four password entries. The number of password entries allocated is always a multiple of four.
Multiple of four has to be intended besides the first one. Entries=1 allows 5 entries, Entries=6 allows 9 entries.

"efiboot from device: Acpi (PNP0AP0,0)/pci(1F12)/SATA(0,0)/HD(Part2,Sig1FC891DD-9674-42C9- AA78-207D0FBE67E9) boot file path: \System\Library\CoreServices\boot.efi

Hi!
My macbookpro 15" retina crashed down, it does not recognize my password anymore.
cmd+S gives :
"efiboot from device: Acpi (PNP0AP0,0)/pci(1F12)/SATA(0,0)/HD(Part2,Sig1FC891DD-9674-42C9-
AA78-207D0FBE67E9)
boot file path: \System\Library\CoreServices\boot.efi
I tried recovery OS without success.
Help please.

Got it thanks macjack . Command + S on restart.
In this case Nad69-Breizh did you try restarting with the option key down and re-selecting the boot drive.
If no Recovery option, try command option R for internet recovery. Takes some time to load up.

Search c:\ drive and return file path for winword.exe and save as variable

Hi all, here is what I'm trying to do;
1. Search C:\ drive for winword.exe
2. take the file path and save it as a variable.
3. Then based on the path value, use the switch statement to run "some command"
Essentially I'm trying to find what the file path for winword.exe is, then run a command to modify the registry. I already have the script that will modify the registry like I want but the problem it, the path is hard coded in the script, I want to
now look for all versions of word and set the right file path so I can make the right registry changes.

This should get you started:
http://ss64.com/ps/get-childitem.html
Don't retire TechNet! -
(Don't give up yet - 13,085+ strong and growing)

Password file path

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