two elements that have the same source

Similar Questions

• Two tasks that share the same resource in sequence

  I'm the current exception "the specified resource is reserved.". I understand that you can not run two tasks that share the same resource at the same time. However, is it true that you can not run two tasks that share the same resource, one right after the other, without the eliminate completely from the first task?

  for example

  Task1 and task2 share the same resource
  
  Task1. Start();
  
  Task1. Stop();
  
  Task2. Start(); This will always throw an exception until you have a task1 (i.e. task1.) Dispose()?
  
  Task2. Stop();

  Hi cafarm,

  You're right, the fact to just stop at one task does not allow that run you another task with the same resources immediately after. That being said, you can start the next task without having to erase or dispose of the first by making use of the task status template DAQmx. As you can see on the next page, stop a task simply going to the State of "Clerks", which means that the material is still used by the task even if she does not actively.

  http://zone.NI.com/reference/en-XX/help/370466AC-01/mxcncpts/taskstatemodel/

  Instead, you can use the control functionality of task for the first task of transition to the status "verified" after stopping it, then start the second task. This keeps the parameters of the task without having to have the task, but it should free up resources to be used by other tasks.
  http://zone.NI.com/reference/en-XX/help/370473H-01/mstudiowebhelp/HTML/verifycommitstartstopdaqtask/

• query to view separate lines that have the same values

  Hello

  I have the columns id, addr1, addr2, city and State of coding. I need to know all the coding IDS that have the same addresses and at the same time filter out the same side of the id of the request.

  For example:

  acctid addr1, addr2 City State

  WEF 1 101 101 sd

  1                         101          wef     sd

  WEF 2 101 101 sd

  WEF 3 101 101 sd

  DC 4 102 102 homeless

  From the above data, I want to get filter tier 4 and get lines 1, 2 and 3. How to achieve with a sql query

  Thank you

  Arch

  This has nothing to do with hard... coding is how you test a query without creating the table, so you'd:

  with match_finder AS (SELECT acctid, addr1, addr2, city, State, count (*) OVER (PARTITION BY addr1, addr2, city, province) AS identical_rows)

  OF )

  Select acctid, addr1, addr2, city, province

  of match_finder

  where identical_rows > 1

  /

  and the rest, I'm sure, you can find on your own...

  HTH

• How to identify columns that have the same data in a SQL query or function?

  Deal all,

  How to identify columns that have the same data in a SQL query or function? I have the sample data as below

  DEPT_ID

  EMP_ID

  Come on

  !CITYSTATECOUNTRY111 June 1983DELHIHUMAN RESOURCESIndia1218 January 1987DELHIHUMAN RESOURCESIndia1328 November 1985DELHIHUMAN RESOURCESIndia144 June 1985DELHIHUMAN RESOURCESIndia255 June 1983MUMBAIHDIndia266 June 1983MUMBAIHDIndia277 June 1983MUMBAIHDIndia288 Jun. 1983MUMBAIHDIndia399. June 1983GURGAONDLIndia31010 June 1983GURGAONDLIndia

  Now, I want to Indify columns that have the same data for the same Department ID.

  Is it possible in sql unique or do I have to write the function for this? Pls Help how to write?

  Thanks in advance.

  You can try this?

  WITH T1)

  DEPT_ID, EMP_ID, DATE OF BIRTH, CITY, STATE, COUNTRY

  ), ()

  SELECT 1, 1, TO_DATE('1.) June 1983', 'JJ. LUN. (YYYY'), 'DELHI', 'HR', 'INDIA' OF THE DUAL UNION ALL

  SELECT 1, 2, TO_DATE('18.) January 1987', 'JJ. LUN. (YYYY'), 'DELHI', 'HR', 'INDIA' OF THE DUAL UNION ALL

  SELECT 1, 3, TO_DATE('28.) November 1985', 'JJ. LUN. (YYYY'), 'DELHI', 'HR', 'INDIA' OF THE DUAL UNION ALL

  SELECT 1, 4, TO_DATE('4.) June 1985', 'JJ. LUN. (YYYY'), 'DELHI', 'HR', 'INDIA' OF THE DUAL UNION ALL

  SELECT 2.5, TO_DATE('5.) June 1983', 'JJ. LUN. (YYYY'), 'BOMBAY', 'HD', 'INDIA' OF THE DUAL UNION ALL

  SELECT 2.6, TO_DATE('6.) June 1983', 'JJ. LUN. (YYYY'), 'BOMBAY', 'HD', 'INDIA' OF THE DUAL UNION ALL

  SELECT 2.7, TO_DATE('7.) June 1983', 'JJ. LUN. (YYYY'), 'BOMBAY', 'HD', 'INDIA' OF THE DUAL UNION ALL

  SELECT 2.8, TO_DATE('8.) June 1983', 'JJ. LUN. (YYYY'), 'BOMBAY', 'HD', 'INDIA' OF THE DUAL UNION ALL

  SELECT 3, 9, TO_DATE('9.) June 1983', 'JJ. LUN. (YYYY'), 'GURGAON', 'DL', 'INDIA' OF THE DUAL UNION ALL

  SELECT 3.10, TO_DATE('10.) June 1983', 'JJ. LUN. (YYYY'), 'GURGAON', 'DL', 'INDIA' OF THE DOUBLE)

  SELECT DEPT_ID,

  RTRIM (XMLAGG (XMLELEMENT(A,VALS||',')). Extract ('//Text ()'), ',') COLUMNS_WITH_DUPLICATE

  DE)

  SELECT * FROM)

  SELECT DEPT_ID,

  EMP_ID,

  Date of birth

  CITY,

  STATE,

  COUNTRY

  DE)

  SELECT DEPT_ID,

  EMP_ID,

  Date of birth

  CITY,

  STATE,

  COUNTRIES,

  COUNT (*) OVER(PARTITION BY DEPT_ID ORDER BY EMP_ID DESC,DOB DESC,CITY DESC,STATE DESC, COUNTRY DESC) RN

  DE)

  SELECT DEPT_ID,

  CASE WHEN(CEID>1) AND THEN 'YES' ELSE 'NO' END AS EMP_ID.

  CASE WHEN(CDOB>1) THEN 'YES' ELSE 'NO' END AS DATE OF BIRTH,

  CASE WHEN(CCITY>1) AND THEN 'YES' ELSE 'NO' END AS CITY.

  CASE WHEN(CSTATE>1) AND THEN 'YES' ELSE 'NO' END AS STATE.

  CASE WHEN(CCOUNTRY>1) THEN 'YES' ELSE 'NO' END AS A COUNTRY

  DE)

  SELECT DISTINCT

  DEPT_ID,

  CEID,

  CDOB,

  CITY,

  CSTATE,

  CCOUNTRY

  DE)

  SELECT DEPT_ID,

  COUNT (*) TO THE CEID (DEPT_ID PARTITION, EMP_ID),.

  COUNT (*) ON CDOB (DEPT_ID SCORE, DATE OF BIRTH),

  COUNT (*) ON THE CITY (DEPT_ID PARTITION, CITY),

  COUNT (*) ON CSTATE (DEPT_ID PARTITION, STATE).

  COUNT (*) ON CCOUNTRY (DEPT_ID, COUNTRY PARTITION)

  FROM T1)))

  WHERE RN = 1)

  UNPIVOT (CLO FOR (VALS) IN (EMP_ID, DATE OF BIRTH, CITY, STATE, COUNTRY)))

  WHERE COLS = "YES".

  DEPT_ID GROUP;

  OUTPUT:

  DEPT_ID COLUMNS_WITH_DUPLICATE
  --------- ------------------------

  1 CITY, COUNTRY, STATE
  2 CITY, COUNTRY, STATE
  3 CITY, COUNTRY, STATE

  Post edited by: Parth272025

• How to stop the connectivity between two computers that use the same IP address... ?

  Today, I managed to connect two computers with the same IP address. But now it is to be feared. for example, a person can access my computers using my IP and MAC address. So now, how can I stop the connectivity between these two computers. should what setting I use to stop.

  Tahnks.

  Hello

  1 have. what measures you taken to connect 2 computers with the same IP address?

  2. How are the 2 computers connected?

  3. how exactly you want to stop the connectivity between 2 computers?

  I want to tell you that 2 computers on the same network cannot have the same address. Each machine must have a unique address to identify it. A private network, as a home network connected to the internet, can be connected to a router to connect to the network, will have only 1 address, public IP address.

  In addition, if we know the IP address of a computer along the user ID and password, we can access the computer/shared files.

   
  If you are referring to the change of the IP address of the two computers, you can follow these steps and check.
   
  a. network connections open. Click the Start button, and then click Control Panel. In the search box, type 'adapter' and then, under center network and sharing, click view network connections.
  b. right click on the connection you want to change. Click on properties. If you are prompted for an administrator password or a confirmation, type the password or provide confirmation.
  c. click on the network tab. Under this connection uses the following items, click Internet Protocol Version 4 (TCP/IPv4) or Internet Protocol Version 6 (TCP/IPv6), and then click the properties button. To specify your IP for IPv4, click on use the following IP address, and then, at the address IP, subnet mask, and gateway boxes by default, type IP address settings.
   
  A typical IPv4 address looks like 192.168.0.2; a typical subnet mask is 255.255.255.0. The default gateway is usually the address of your router.
  For example, IP address: 192.168.0.2 by 192.168.0.12 (for computers)
  Subnet mask: 255.255.255.0
  DNS: 192.168.0.1
   

  I hope this helps. If you have any other queries/issues related to Windows, write us and we will be happy to help you further.

• Two iPhones, will have the same content

  Two iPhones, it will have the same content, like NOTES, FB, WHATSAPP AND YAHOO MAIL?

  I am worried because my friend is using my old iPhone and it can receive notifications the same as what I get.

  If you distributed or sold your phone to your friend, you should follow the steps here before releasing the device. If you did, you have no worries.

  What to do before you sell or give away your iPhone, iPad or iPod touch - Apple Support

• Only GET records that have the same values of field has the same value in field B

  Have a hard time with below, please help.

  Here's the situation:

  create table cord (identification NUMBER, CM VARCHAR2 (3), PM VARCHAR2 (2));
  insert into string values (1, '002', 'H1');
  insert into string values (2, '006', 'H1');
  insert into string values (3, '004', 'H2');
  insert into string values (4, '006', 'H2');
  insert into string values (5, '004', 'H3');
  insert into string values (6, '004', 'H3');

  I just need to select the folders which, for the SAME value of PM have the same value in CM, in example above, those are recordings with ID (5,6).

  1 and 2 fail because for them CM and PM are different for the same H1, similar on the 3 and 4.

  I don't know if this will help but records are always in 'pairs', which means that there are no cases as

  7. '004' | H4

  8. ' 006' | H4

  9. '005' | H4

  Any ideas are much appreciated.

  Thank you

  See if the following can help...

  select id,cm,pm
      from(
          select t.*
                ,count(1) over(partition by cm,pm) cnt
              from testt t
          )
      where cnt>1;
  

• Stuck on a sql query to search for records that have the same parent child records

  Oracle 10 g 2 Enterprise Edition.
  
  Hello
  
  I'm writing a logic to find records in a parent table, who have the same values in a child table.
  This is part of a larger application, but I am stuck on that part for now, so I have mocked some of the below simplified tables to capture the heart of the
  the problem is that I'm stuck.
  Let's say I have a responsible parent, child employee table table and there are a number of many relationships between them.
  The aptly named Join_Table manages the relationship between them. If a manager can manage several employees, an employee can be managed by
  many managers.
  
  I have a feeling it's stupidly easy, but it seems to me having a bad episode of brain freeze today!

  -- parent table
  CREATE TABLE manager (
   id      number primary key,
   name      varchar2(100));
  
  -- child table 
  CREATE TABLE employee (
   id          number primary key,
   name      varchar2(100));
  
  -- link table
  CREATE TABLE join_table (
   manager_id          NUMBER, 
   employee_id      NUMBER,
   CONSTRAINT join_table_pk PRIMARY KEY (manager_id, employee_id),
   CONSTRAINT manager_fk FOREIGN KEY (manager_id) REFERENCES manager(id),
   CONSTRAINT employee_fk FOREIGN KEY (employee_id) REFERENCES employee(id) 
   );
  
  -- Insert some managers
  INSERT INTO manager (id, name) VALUES (1, 'John');
  INSERT INTO manager (id, name) VALUES (2, 'Bob');
  INSERT INTO manager (id, name) VALUES (3, 'Mary');
  INSERT INTO manager (id, name) VALUES (4, 'Sue');
  INSERT INTO manager (id, name) VALUES (5, 'Alan');
  INSERT INTO manager (id, name) VALUES (6, 'Mike');
  
  -- Insert some employees 
  INSERT INTO employee (id, name) VALUES (101, 'Paul');
  INSERT INTO employee (id, name) VALUES (102, 'Simon');
  INSERT INTO employee (id, name) VALUES (103, 'Ken');
  INSERT INTO employee (id, name) VALUES (104, 'Kevin');
  INSERT INTO employee (id, name) VALUES (105, 'Jack');
  INSERT INTO employee (id, name) VALUES (106, 'Jennifer');
  INSERT INTO employee (id, name) VALUES (107, 'Tim');
  
  -- Insert the links
  -- John manages Paul, Simon, Ken
  INSERT INTO join_table (manager_id, employee_id) VALUES (1, 101);
  INSERT INTO join_table (manager_id, employee_id) VALUES (1, 102);
  INSERT INTO join_table (manager_id, employee_id) VALUES (1, 103);
  -- Bob manages Paul, Simon, Kevin, Jack
  INSERT INTO join_table (manager_id, employee_id) VALUES (2, 101);
  INSERT INTO join_table (manager_id, employee_id) VALUES (2, 102);
  INSERT INTO join_table (manager_id, employee_id) VALUES (2, 104);
  INSERT INTO join_table (manager_id, employee_id) VALUES (2, 105);
  -- Mary manages Jennifer, Tim
  INSERT INTO join_table (manager_id, employee_id) VALUES (3, 106);
  INSERT INTO join_table (manager_id, employee_id) VALUES (3, 107);
  -- Sue manages Jennifer, Tim
  INSERT INTO join_table (manager_id, employee_id) VALUES (4, 106);
  INSERT INTO join_table (manager_id, employee_id) VALUES (4, 107);
  -- Alan manages Paul, Simon, Ken, Jennifer, Tim
  INSERT INTO join_table (manager_id, employee_id) VALUES (5, 101);
  INSERT INTO join_table (manager_id, employee_id) VALUES (5, 102);
  INSERT INTO join_table (manager_id, employee_id) VALUES (5, 103);
  INSERT INTO join_table (manager_id, employee_id) VALUES (5, 106);
  INSERT INTO join_table (manager_id, employee_id) VALUES (5, 107);
  -- Mike manages Paul, Simon, Ken
  INSERT INTO join_table (manager_id, employee_id) VALUES (6, 101);
  INSERT INTO join_table (manager_id, employee_id) VALUES (6, 102);
  INSERT INTO join_table (manager_id, employee_id) VALUES (6, 103);
  
  -- For sanity
  CREATE UNIQUE INDEX employee_name_uidx ON employee(name);

  If I ask for Manager John, so I want to find other managers who manage the exact list and even employees.
  Answer should be Mike.
  If I ask for Manager of Mary, the answer should be Sue.
  
  This query will give me the list of managers who manage some of the same employees as John, but not the same employees accurate...

  SELECT DISTINCT m.name AS manager
  FROM manager m, join_table jt, employee e
  WHERE m.id = jt.manager_id
  AND jt.employee_id = e.id
  AND e.id IN (
       SELECT e.id
       FROM manager m, join_table jt, employee e
       WHERE m.id = jt.manager_id
       AND jt.employee_id = e.id
       AND m.name = 'John')
  ORDER BY 1;

  I thought about using set operations to find managers with a list of employees less than my employees is null and where my employees under their list of employees is null. But there must be an easier way more elegant.
  Any ideas?
  BTW, I need to run as a batch on tables with > 20 million rows so the efficiency of queries is key.

  What about...

  WITH manager_list AS
  (
   SELECT name,
          LTRIM(MAX(SYS_CONNECT_BY_PATH(id,','))
          KEEP (DENSE_RANK LAST ORDER BY curr),',') AS employees
   FROM   (SELECT m.name,
                  e.id,
                  ROW_NUMBER() OVER (PARTITION BY m.name ORDER BY e.id) AS curr,
                  ROW_NUMBER() OVER (PARTITION BY m.name ORDER BY e.id) -1 AS prev
           FROM   manager m,
                  join_table jt,
                  employee e
    WHERE m.id           = jt.manager_id
    AND   jt.employee_id = e.id
    AND   m.name = :P_MANAGER)
    GROUP BY name
    CONNECT BY prev = PRIOR curr AND name = PRIOR name
    START WITH curr = 1
  ), all_list AS
  (
   SELECT name,
          LTRIM(MAX(SYS_CONNECT_BY_PATH(id,','))
          KEEP (DENSE_RANK LAST ORDER BY curr),',') AS employees
   FROM   (SELECT m.name,
                  e.id,
                  ROW_NUMBER() OVER (PARTITION BY m.name ORDER BY e.id) AS curr,
                  ROW_NUMBER() OVER (PARTITION BY m.name ORDER BY e.id) -1 AS prev
           FROM   manager m,
                  join_table jt,
                  employee e
    WHERE m.id           = jt.manager_id
    AND   jt.employee_id = e.id)
    GROUP BY name
    CONNECT BY prev = PRIOR curr AND name = PRIOR name
    START WITH curr = 1
  )
  SELECT a.*
  FROM   manager_list m,
         all_list a
  WHERE  m.employees = a.employees
  

  Would be easier in 11g, but I do not have a facility here so this is based on 10g.

  See you soon

  Ben

• Read the nodes that have the same value as the subnodes - XML

  It is more of a general JAVA / XML problem, but given that it is going into my BlackBerry app I thought I'd see if anyone knows.

  Consider a simple XML document:

                      Whatever 1                   Whatever 2               Whatever 3       
  

  Using the standard org.w3c.dom, I can get the nodes in X by the practice...

  NodeList fullnodelist = doc.getElementsByTagName ("x");

  But if I want to go to the next set of 'e', I try to use something like...

  Element element = (Element) fullnodelist.item(0);NodeList nodes = pelement.getElementsByTagName("e");
  

  EXPECTED back '3' nodes (because there are 3 series of 'e'), but it returns '9' - because it gets all entries including the 'e' apperently.

  It would be nice in the above case, because I could probably go through and find what I'm looking for. The problem I have is that when the XML file looks like the following:

              whatever              Something Else                    whatever              Something Else            
  

  When I ask 'e' value, it returns 4, instead of (what I want) 2.

  I am simply not understand how DOM parsing works? Generally, in the past I used my own XML documents so I name never articles like this, but unfortunately this isn't my XML file and I don't have the choice to work like this.

  What I thought I would do, it is write a loop knots "drills down" so that I can combine each node...

    public static NodeList getNodeList(Element pelement, String find)
      {
          String[] nodesfind = Utilities.Split(find, "/");
          NodeList nodeList = null;
  
          for (int i = 0 ; i <= nodesfind.length - 1; i++ )
          {
              nodeList = pelement.getElementsByTagName( nodesfind[i] );
              pelement = (Element)nodeList.item(i);
          }
  
          // value of the nod we are looking for
          return nodeList;
      }
  

  .. While if adopted you ' s/e' in the service, he would return the 2 nodes I'm looking (or elements, perhaps I'm using the wrong terminology?). on the contrary, it returns all the 'e' nodes in this node.

  Anyway, if anyone is still with me and has a suggestion, it would be appreciated.

  Well, there is no doubt that there is a learning curve robust for XML programming. You can take an hour or two and go through one of the tutorials that are circulating on the net. (Like that of w3schools.com.)

  Basically, almost everything in XML is a node, the Document that returns the parser. The API for node tells you that you can test the node type you have by calling getNodeType, which returns one of the constants of type node (Node.ELEMENT_NODE, Node.TEXT_NODE, etc..) If necessary, you can then convert the variable to the corresponding interface (element, text, etc.).

  Similarly, the API documentation say you for any node, calling getChildNodes (or for an element node or Document getElementsByTagName) will give you a NodeList (a little non-types of nodes in the XML API), while calling getFirstChild and getNextSibling to any node will give you another node (or null ).

  Once you learn the API, writing logic of course is not all that hard. For example, if the only 'e' interest tags are those directly under the element root of the document (as shown in your example) you can simply go to them directly:

  Vector getTopENodes(Document doc) {  Vector vec = new Vector();  NodeList nodes = doc.getDocumentElement().getChildNodes();  int n = nodes.getLength();  for (int i = 0; i < n; ++i) {    Node node = nodes.item(i);    if (node.getNodeType() == Node.ELEMENT_NODE &&        "e".equals(node.getNodeName()))    {      vec.addElement(node);    }  }  return vec;}
  

  Note that this example does not assume that all children are nodes of element 'e '. the document could have comments, white space or something else that makes it into the DOM as comment, text or any other type of node.

  On the other hand, if you want to capture every "e" tag which is directly under the ' tag, no matter the level, then you need to do something a little more complicated (it's on the top of my head - no guarantee):

  static class NodeListImp implements NodeList {  private Vector nodes = new Vector();  public int getLength() {    return nodes.size();  }  public Node item(int index) {    return (Node) nodes.elementAt(index);  }  public add(Node node) {    nodes.addElement(node);  }}
  
  NodeList getTargetNodes(Document doc) {  NodeListImp list = new NodeListImp();  getTargetnodes(list, doc.getDocumentElement(), false);  return list;}
  
  void getTargetNodes(NodeListImp list, Node node, boolean parentIsS) {  if (node.getNodeType() == Node.ELEMENT_NODE) {    // node name is tag name for element nodes    String name = node.getNodeName();    if (parentIsS && "e".equals(name)) {      list.add(node);    }    parentIsS = "s".equals(name);    for (Node child = node.getFirstChild();         child != null;         child = child.getNextSibling())    {      getTargetNodes(list, child, parentIsS);    }  }}
  

  I hope that it gets the idea across.

• Block 10 FireFox accesses local files the same source in Ajax? Chrome for this, but the workaround is - disable-web-security.

  I have a few Ajax in a test program that is accessing a local file. It worked well until I updated to Firefox 10. 10 Firefox blocks access to local files, even if they have the same source location? Chrome has this problem but I can overcome it by launching Chrome with the "-disable-web-security" option. Is there a similar to Firefox 10 option?

  A good place to ask for advice on web development is to the 'Web Standards Development/evangelism' MozillaZine forum.

  • http://forums.mozillazine.org/viewforum.php?f=25

  Aid to this forum are better informed on issues related to web development.
  
  You must register on MozillaZine forum site to post in this forum.

• Consultation problem Mac Photoshop Elements 11. Unable to open the program. He tells me that it is damaged. To remove the program and reinstall it from the app store. Again, I have install my purchases in the app store, and yet, I have the same problem. T

  Consultation problem Mac Photoshop Elements 11. Unable to open the program. He tells me that it is damaged. To remove the program and reinstall it from the app store. Again, I have install my purchases in the app store, and yet, I have the same problem. Thank you.

  I just wanted to make account - it seems that the problem has been resolved. After another remove & reinstall it is operational!

• My husband and I have the same phone.  We see the two of you on each of our phones calls him, and that missed us a call from (the phone number is in red when it's a missed call).  Thanks for your help.

  My husband and I have the same phone.  We see them entering and fact phone outgoing calls and missed calls (phone # is in red) from the other phone.  Thank you for your help.

  This happens when you both share the same Apple ID. It is recommended - for a number of reasons - that each have their own Apple ID and then uses the sharing of the family for things you want to share. http://www.Apple.com/icloud/family-sharing/

• Why Apple has the code of two factor authentication on the same device that I log in with?

  I just installed Sierra and chose to use the two factor authentication with my iPhone chosen as a device to receive the code.

  But then, Apple displays a digit code 6 on my Mac itself and then asked me to go on my Mac.

  What sense does that make?

  A wild guess - were you log into your account in Safari on Mac when he showed you the digit code 6 on Mac? And you had already completed the sign-in icloud in System Preferences?

  If so, the macOS has been approved, but Safari wasn't. If macOS was able to show the code. It seems strange to first have the same computer application and provide the code, but really it is two layers of security and you had gotten through the first layer already.

• Can someone explain why these two paintings 'look' different even if they are from the same source?

  Hi all!

  I'm just curious, why these two tables are different and are they really different in terms of structure, even if they are from the same source? The left is the exit of tdms get properties and the other is a power of arry auto indexed.

  Nilesh-

  No doubt, you are referring to the visual aspect, since the data are the same. One of them uses a different type of indicator of the chain. You probably click right and choose 'create-> Indicator '. When you perform this LabVIEW will use the same style of indicator as what sub-VI/uses. In case you made directly from the PDM function, but the other was created between the autoindex loop tunnel for, in the second he used everything that was default in your preferences.

• New Hardware Wizard two elements that show in detail: the first watch Root\legacy_cd20xrnt\software... and the second is Root\legacy_hpn\software

  I hope I am posting this in the right space... on windows xp sp3.

  At the start of the installation Assistant starts... want to install and unknown device... (not tried to install anything in a long time) Looking in the Device Manager, there are two elements that show in detail...

  The first watch Root\legacy_cd20xrnt\software... and the second is Root\legacy_hpn\software.

  have not clicked on install only cancel...

  ran a total analysis of AVG and total bytes Malware scan and they found nothing.

  What are and how...

  Check these:

  http://www.cybertechhelp.com/forums/showthread.php?t=211599&page=19

  http://www.computing.NET/answers/security/root-legacy-npf0000/22753.html

  http://www.malwareremoval.com/forum/viewtopic.php?f=11&t=30570&p=299717