All channels to HAVE it have the same value - NI PCI-6221 and NI PCI-6229

Similar Questions

• All channels to HAVE it have the same value

  I use the example for a multichannel AI aiex2.cpp read with mseries NI6280 devices.

  This example works for a lane, but other chains have the same value

  For example: I put 5 Volt on the first string, and then the other channel are 5 Volt too.

  What should I consider in the configuration?

  Hello Beilei,

  I think that you run in theghost of the question.  The other strings that you use are connected to the earth when you connect the first channel to 5V?  If the other channels are floating, they will read the same value as the first string... 5V.

  Steven T.

• I teach online and all my classes have the same user name and password. Now that I clicked "remember me next time", I can connect only in one class. How to unlock my password. Carol in English

  I teach online and all my classes have the same user name and password. Now that I clicked "remember me next time", I can connect only in ONE class. How to unlock my login and my password, so that I can use it for all classes. Carol in English

  "Remember Me" for the site connections automatically when you return to the Web site is done with a Cookie the site in Firefox.

  Try to clear your Cookies for this Web site.
  
  Tools > Options-> life privacy - Cookies = the button show Cookies.
  
  You must use the custom settings for history at the top of this tab to see the View the Cookies button.

  Enter the domain name in the top search bar and all Cookies for this URL will be displayed. Unless you can figure out which is Cookie to "remember me", you will need to delete them all.

  Hold the {Ctrl} key while you click each Cookie in the small window. When this list is all highlighted, click the Cookie delete button at the bottom left.
  When you are finished click Close.

• Clock on windows 8 is wrong time, I tried to set the time of the internet but its still not working, think theres a bug in the clock time, anyone have the same problem for windows 8 and know how to fix?

  Clock on windows 8 is wrong time, I tried to set the time of the internet but its still not working, think theres a bug in the clock time, anyone have the same problem for windows 8 and know how to fix?

  It sets obtained :-) Sorry for the delay

  Thank you very much! has worked perfectly!

• 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;
  

• 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

• GL CCID have the same values for Segment

  Is it possible that two different code_combination_id in the same GL_Code_Combination table has the same values for segment.
  It is, when I run the accounts payable balance report trial of the responsibility of the PA, its follow-up of the transactions ID associated with only 1 and not both.
  But, here, for the same Chart_of_Accounts_Id, we can see that there are two Code_combination_id entries, with the same values for segment.
  Can you please let me know if this is possible?
  
  For example: I'm running the following queries and Segment1, Segment2, Segment3, Segment4, 12855 CCId values Segment5 is the same as Segment1, Segment2, Segment5, Segment4, Segment3 to 13402.
  
  Select CODE_COMBINATION_ID, CHART_OF_ACCOUNTS_ID,
  SEGMENT1, SEGMENT2 SEGMENT3, SEGMENT4, ENABLED_FLAG, SEGMENT5,
  * of the GL_CODE_COMBINATIONS where CODE_COMBINATION_ID = 12855
  
  Select CODE_COMBINATION_ID, CHART_OF_ACCOUNTS_ID,
  SEGMENT1, SEGMENT2 SEGMENT3, SEGMENT4, ENABLED_FLAG, SEGMENT5,
  * of the GL_CODE_COMBINATIONS where CODE_COMBINATION_ID = 13402
  
  CCID - Segment1, Segment2, Segment3, Segment4, Segment5
  -----------------------------------------------------------------------------------------------
  12855 01 000 2220 00000 000
  13402 01 000 2220 00000 000
  
  I hope you understand.
  Thank you
  Bob

  Hi Bob,

  Can you please check the id of chart of accounts for the two combinations...

  Concerning
  Muthu

  Published by: Muthu on April 30, 2013 02:10

• All related Clones have the same SID?

  I played with the persistent linked clones and have noticed all share the same identifier SID like parent VM. Is this normal? Won't this cause of problems with Active Directory?

  Hello

  the quickprep function used with linked clones will create an account active directory for the virtual computer newly created and due to that he gets a unique SID. SOEs not change local SIDS, it's true. When you work in a domain environment, you don't need to change the local SID because it only uses the SID in an environment of the domain. If you work with windows working groups, it is necessary to have unique local SIDS.

• 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.

• (3) all my computers have the same problem with installing KB953297. Microsoft must provide an easier way to solve this problem. Obviously a lot of people is affected.

  End users should not have to be remove and reinstall anything. Get it together!

  Can you guys try the following steps:

  1 uninstall .net Framework 1.1 from Add/Remove programs
  2 run the .net Framework Cleanup Utility and remove all .net Framework 1.1 related Articles (located here: http://blogs.msdn.com/astebner/archive/2006/05/30/611355.aspx)
  3 Windows Update to download updates.
  4 reboot and redo the update of Windows

  Can you let me know if you find any success?

  Thank you! If you find my answer was what you're looking for, remember to click on the box "mark as answer" below!

• How to determine how many times result set have the same value

  HI -.
  
  I do a report that will be used for the analysis of trends in payment.
  
  My initial result set looks like this:
  

  HOUSEHOLD_ID     JAN_PMT     FEB_PMT     MAR_PMT     APR_PMT     MAY_PMT     JUN_PMT     JUL_PMT     AUG_PMT     SEP_PMT     OCT_PMT     NOV_PMT     DEC_PMT
  90026845409     1     1     1     1     2     1     1     1     1     0     1     0

  (number of lines, of course, pivot table result set)
  
  I need to determine the households that have a value > 0 in three or more consecutive months.
  
  I hope that someone will have some suggestions because that the only solutions I come up with right now would be a nightmare of coding (lots of "GOLD"), and I assume (hope) there's a better solution out there.
  
  Thank you!
  Christine

  If you are limited to only 12 months I would use GOLD in conjunction with the sign function.

  sign of a positive number returns 1
  sign of 0 returns 0
  sign of a negative number returns - 1

  Really, it doesn't look like bad lto for me. Obviously if you have a numbe rof months variable must be a different solution.

  with the data
  as
  (
  Select hh '90026845409', 1 jan, Feb 1, 1 March, 1 April, 2 may, 1 June, 1 July, August 1, 1 sep, oct, 1 nov, dec double 0 0
  )
  Select * from data
  where
  (sign (jan) sign (feb) + sign (mar)) = 3
  or
  Sign (Feb) + Sign (Tue) + Sign (APR) = 3
  ...
  or
  Sign (Oct) + Sign (Nov) + Sign (DEC) = 3

• How can I know what power adapter is mine since we both have the same for macbook pro 2013?

  I mixed my adapter upward with my friend and we both have the same adapter for macbook pro and mine is almost more recent than its adapter. I need to know how can I know who is mine?

  I doubt that you can unless it adapt serial numbers or other identification on the cards.

• 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

• Photos with the same values of position...

  I have a picture with the edited/customized position values (scale, rotation, inclination, perspective, disort, warp)... I have some photos that will be on the different layers and I want that they have the same values as the first (it's stupid to do the same thing with 10 photos...) I tried to duplicate the layers and replacement is contained, but values are lost, so the photos appear in the normal position...

  So how can guys, I do this?

  For this to work properly, all images must be of the same size and resolution.

  1. Integrate the file of the first image.
  2. Layer > Smart object > convert to smart object
  3. Edit > free transform and establish the size, position and orientation... .OK
  4. Cmd + J to create a layer of dupe
  5. Layer > Smart Object > replace content
  6. Choose the following image > replace

  Repeat steps 4-6

• rows with the same value

  Hello
  I have this request:
  Select b.db_name, a.handle from bp a, dbinc b where a.db_key = b.db_key;
  
  By request I see the lines where the handle columns have the same values?
  Thank you.

  Hello

  First thing is that your question is very abstract, we can not know which data table you is central to the present moment.

  Select b.db_name, a.handle from bp a, dbinc b where a.db_key = b.db_key;
  By request I see the lines where the handle columns have the same values?

  What you mean you 'same' values with what data need to be compared to 'the same '.
  Fact it is same or some values.

  -Pavan Kumar N