List vm name, IP address, and the folder path

Hello

DOE anyone at - it a script that will list the following?

VM name, IP address, path to folder.

Thank you

Try the attached script

Tags: VMware

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How to read the files in a particular directory with PL/SQL, without knowing the exact name? My application interface with another system that puts the files in a specific directory on the server. UTL_FILE method reads a file when you know the name of the file, but I don't know the name in advance. Is it possible to output the name of the file? I also used the method of Tom "http://asktom.oracle.com/pls/asktom/f?p=100:11:0:P11_QUESTION_ID:439619916584" (related to the Java stored procedure) but am getting error of procedure "ORA-00900: invalid SQL statement. Here's the code I used Tom - forums

CREATE or REPLACE AND compile java source named 'DirList '.
AS
import java. IO;
import java.sql. *;

public class DirList
{
Public Shared Sub getList (string directory)
throws SQLException
{
Path = new file (directory);
List of strings [] = path.list ();
Element string;

for (int i = 0; i < list.length; i ++)
{
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#sql {INSERT INTO DIR_LIST (FILENAME)
{VALUES (: element)};
}
}

}
/
**********************************************************************
CREATE OR REPLACE PROCEDURE get_dir_list (p_directory in VARCHAR2) AS java language
name ' DirList.getList (java.lang.String) htp.p (p_directory) ";
**********************************************************************
get_dir_list exec (' / CERT_XML');
**********************************************************************

I would like to know is there any alternative approach to retrieve all the names of the files in the folder of the server and move these files to a different folder?

Thank you
Amit Bhandari
Hello

It is a limitation of UTL_FILE in Oracle that we can do with Java stored procedures in Oracle.

We can read the files in Server directory using the following steps:

Please follow the steps below:

1. creating a Type of Type Varchar2.
```
  CREATE OR REPLACE TYPE file_list AS TABLE OF VARCHAR2(255);
```
2. next, we need create a library of Java file
```
CREATE OR REPLACE AND COMPILE JAVA SOURCE NAMED "ListVirtualDirectory" AS
  import java.io.*;
  import java.security.AccessControlException;
  import java.sql.*;
  import oracle.sql.driver.*;
  import oracle.sql.ArrayDescriptor;
  import oracle.sql.ARRAY;

  public class ListVirtualDirectory {
    public static ARRAY getList(String path) throws SQLException, AccessControlException {

    Connection conn = DriverManager.getConnection("jdbc:default:connection:");

    File directory = new File(path);

    ArrayDescriptor arrayDescriptor = new ArrayDescriptor("FILE_LIST",conn);

    ARRAY listed = new ARRAY(arrayDescriptor,conn,((Object[])directory.list()));
  return listed; }}
```
The most advanced method overrides by removing information about java.properties of exception management settings. You can do this by catching the exception thrown in native mode and rethrow it or ignore it. The example it up again.

3. then, you must create a Wrapper function:
```
CREATE OR REPLACE FUNCTION list_files(path VARCHAR2) RETURN FILE_LIST IS
LANGUAGE JAVA
NAME 'ListVirtualDirectory.getList(java.lang.String) return oracle.sql.ARRAY';
```
4 grant permissions to the Driectory:
```
BEGIN
  DBMS_JAVA.GRANT_PERMISSION('USER_NAME'
                             ,'SYS:java.io.FilePermission'
                             ,'C:\JavaDev\images'
                             ,'read');
  END;
```
5. then you can read the contents of the directory
```
SELECT column_value FROM TABLE(list_files('C:\JavaDev\images'));
```
That displays all the files in the mentioned directory.

Thank you
Shankar

List vm name, IP address, and the folder path

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