Nonuniform length in table

Similar Questions

• Problems with the length of table/column name

  Hello
  
  I have load data from the xml source in the oracle table. Some of the elements of xml (which ODI inverts like data warehouses) are more than 30 characters long, and it goes the same for columns as well.
  
  In the topology Manager, when I change the XML technology and go to the tab 'OTHER', the 'Maximum length of the Table name' and 'Name of maximum column length' are assigned to the 30. According to the literature, ODI must truncate the table name and the name of the column to the specified maximum length. But when I invert the xml model, with more than 30 characters data warehouses are created, and this causes my interface to fail. Is there a solution for this problem?
  
  Receive your answer.
  
  Thank you!

  In this case, you have 2 options

  (1) change each column in the data to the right length store

  (2) modify the IKM to create columns at the top, 30 characters. The problem is:

  -If the columns 2 to as difference, only the last 2 characters (like: my_column_name_is_really_big_01 and my_column_name_is_really_big_02) you could get a duplicate if column name not the code in the right way.

  I propose option 1.

  This happens because ODI validate the length of column of the tables in model once it doesn't matter of what is not "owned" by him (like C$ E$...)

  Best regards

  Cezar Santos
  http://odiexperts.com

• Lengths of table layout

  I am totally new to the Dreamweaver software and website. I took the online Dreamweaver training, but I'm still learning. I started to build my pages of layout tables and layout cells, because I created my designs in Photoshop and then sliced into cells. When I originally created my layout table, I use an image background, but thought I had a fixed width and height, but I added stuff to the page is longer and repeating the image. I don't see any other cell, which makes it longer and when I try to type in a length fixed it telling me I can't and automatically determines the length. But the background image is repeated below. I tried to learn this for hours, and I'm very, very frustrated. Help, please!
  

  Misfortune and wow. Thanks for the tips. I learned about lynda.com and they do not discuss tables and CSS, but the layout table seemed easier. I'm a designer, so my page layouts are images which are silo-ed and the drawings do not look like your typical grid contained in a box look. I commented on one of my coming soon to splash pages that the slices are visible in Internet explorer, but happening in Foxfire and Safari, Yes... I'm a mac person. So I can't actually test the page on explore until what I publish. This process does not seem easy to use.

• Problem with the function 'length' for table column in a table (fn)

  Hello
  
  I have a piece of javascript code that sets all the values of a column in a table:
  
  function exemptAll (flag)
  {
  Alert (document.wwv_flow. F42. Length);
  for (var i = 0; i < document.wwv_flow.f42.length; i ++)
  {
  var curr_id = document.wwv_flow.f42user.user;
  If (flag == 'Y')
  {
  document.getElementById ("curr_id") .value = "Y";
  }
  on the other
  {
  document.getElementById (curr_id) .value = 'n';
  }
  }
  }
  
  
  It works very well to provide in the form of tables, there are multiple lines, if only one line is the "document.wwv_flow.f42.length" function returns false values - we had '2' and '3' returned to a form of the tab that contains a single row and subsequently code fails with an error "invalid argument".
  
  Is this a recognized bug because we cannot find similar reports of what is happening on the forums. Any help would be appreciated as we have to fake a work around using an element of the page containing the number of lines in the tabular presentation.

  Hello

  I think that "bug" is caused by the dynamic nature of Javascript.

  If there is more than one field called f42, any reference to it will return an array of fields. However, if there is only one instance of it, any reference to it will return the field itself! In the latter case, probably the length function returns the number of characters in the field.

  Therefore, you must check whether what you have is an array or not. Unfortunately, there is no such function only as "isArray" in JS! If you are looking on the net for "isArray" you will find a number of alternatives well...

  I hope this helps.
  Luis

• Length of table name constraint

  Hi all
  
  Today, we have a process that has failed because of a constraint of table name length. The table name is 31 characters long, and we are on Oracle 9. I did some research and realized table name length cannot be more than 30 characters. My question is this:
  
  Is this always the case in newer versions of oracle (10 or 11)?
  
  Thanks for your help!

  Yes, the same limitation of 10g and 11g.

  SQL> select DATA_LENGTH from dba_tab_columns where column_name='TABLE_NAME' and table_name='DBA_TABLES';
  
  DATA_LENGTH-----------         30
  
  SQL> select * from v$version where rownum=1;
  
  BANNER----------------------------------------------------------------Oracle Database 10g Enterprise Edition Release 10.2.0.4.0 - 64bi
  

  Nicolas.

• How to treat only selected tables?

  I am writing a transformation I only applied to certain tables in the model.  I found the "isSelected()' on the table object function, but appears not to do what I thought, he does.  If I selected a table in the user interface and apply my transformation, the table not treated.  Passage through the debugger shows that the isSelected() function returns false for all my paintings.

  Here is the code I use to test...

  tables = model.getTableSet () .toArray ();

  for (t = 0; t < tables.length; t ++) {}
  table = table [t];

  If (table.isSelected ()) {}
  table.setName ("Found_You");
  }
  }

  Is it possible for me to treat only the tables that are currently selected in the user interface?

  Thank you

  Eric

  Hi Eric,.

  Must be something like this

  importPackage(javax.swing);
  mainView = model.getMainView();
  
  selObjects = mainView.getSelectedObjects();
  
  for (var t = 0; t		   

• Transformation to define the schema and user on Tables in relational and physical

  I imported my Designer SDDM 4.0.1.836 deposit.  Now, I want to APP_SYS of subviews allows to define the schema on the tables in the relational model and the user on the physical model.  Our application in designer systems are 90% based scheme.  Here's what I have so far based on the example of OE and HR plans that I imported, built logical, then back to relational.  At this point, I have the tables by subview...  Line 24 is obviously incorrect.  Can someone give me a step by step approach to understand what I need to set the schema of the table and the user?  I looked through the read me file, xml files, the index and searched the known universe and am still confused.

  importPackage(javax.swing);
  // variable to keep a status message for later
  
  var message="";
  var schema="";
  
  // Get all subviews for Relational
      subviews = model.getListOfSubviews().toArray();
  for (var isub = 0; isub<subviews.length;isub++){
      subview = subviews[isub];
      message = message + '\n'+isub+" "+ subviews[isub];
      if(subviews[isub]!=null){
            if(subview == "HR_ERD"){
               shema = "HR";
              }else if(subview == "OE_ERD"){
               schema = "OE";
              }
           tables = model.getTableSet().toArray();
           for (var t = 0; t<tables.length;t++){
               table = tables[t];
               //check for presentation in subview
               tv = table.getFirstViewForDPV(subviews[isub]);
               if(tv!=null){
                   table.setSchemaObject(schema);  
                    message = message + '\n    '+ t +" " +shema+"."+ tables[t] ;
              }
          }
      }
  }  
  
  JOptionPane.showMessageDialog(null, message);
  //Packages.oracle.dbtools.crest.swingui.ApplicationView.log(message);
  

  Hi Marcus,

  First of all, you must make sure that the schema already exists in your relational model.

  You can use the method

  public void setSchema (String schemaName)

  to associate your Table with this scheme.

  (This method is defined on ObjetConteneur, so it can also be used to make and index).

  If you already have a user of your physical model that implements your schema, it is not normally necessary to create a user on the TableProxyOracle in the physical model.

  However if you do not want to set, then the method to use is

  public void setUser (user UserOracle)

  David

• Hoow get you a table name?

  The following is a snippet of code that I use.

  tables = model.getTableSet () .toArray ();
  ...
  for (var i = 0; i < tables.length; i ++)
  {
  table = table [i];

  At this point, I would like to get the name of the table that is held in the table variable.

  However, the documents show no way to get it.

  Is there something that I am missing?

  Hello

  You can get the name of the table by adding the statement as follows:

  name = table.getName ();

  This procedure is inherited from parentclass ModelObject

  Joop

• update of several columns of values in another table

  Hello
  
  I have two tables in a database of 11 GR 2 that contain similar data. There is no FK relationship between the two tables, and for now, I need the data in MASTER to match what is in the SECONDARY. Table data types are the same with the same length.
  
  Table 1 is called MASTER and columns are:
  MTID
  ADDRESS1
  ADDRESS2
  CITY
  
  Table 2 is called SECONDARY and columns are:
  SDID
  ADDRESS1
  ADDRESS2
  CITY
  
  MID and SID are the same on both, but the rest of the columns may be the same or in some cases are different.
  
  I ran the following query that lets me know there are 20 k odd rows that have different data in these lines (I'm not at work at the mo so do not have the exact query I ran or table column names, but it's close enough).
  

  select
  m.address1 as maddress1,
  s.address1 as saddress1,
  m.address2 as maddress2,
  s.address2 as saddress2,
  m.city as mcity,
  s.city as scity
  from master m, secondary s
  where m.mtid = s.sdid
  and m.address1 != s.address1
  or m.address2 != s.address2
  or m.city != s.city
  /

  I need to update the rows in the master of secondary school where these values are different. I tried a few different queries updated with no luck.
  
  I want to do something like
  

  update m.address1
  set m.address1=s.address1
  where m.address1 != s.address1
  and m.mtid = s.stid;

  But I want to update columns address1 and address2, city of master.
  
  Any ideas on how I can do this? After you run the update results of my select query should return no rows.
  
  Thanks in advance and I hope that the question made sense.

  1008991 wrote:
  I got a similar error as well:

  update master m
  set (sm_user_addr_1, sm_user_addr_2, sm_user_city, sm_user_postal_cde)
  = (select sm_user_addr_1, sm_user_addr_2, sm_user_city, sm_user_postal_cde
  from secondary s
  where s.sd_key = m.sm_key
  )
  where exists (
  select 1 from secondary s
  where s.sd_key = m.sm_key
  and (s.sm_user_addr_1     != m.sm_user_addr_1 or
  s.sm_user_addr_2     != m.sm_user_addr_2 or
  s.sm_user_city       != m.sm_user_city   or
            s.sm_user_postal_cde != m.sm_user_postal_cde));
  
  ERROR at line 3:
  ORA-01427: single-row subquery returns more than one row
  

  It simply means that you have several lines with the same sd_key for a given sm_key.
  Is that you have secondaries for one master. You must choose only one line
  If you need additional terms in your where clause

  select sm_user_addr_1, sm_user_addr_2, sm_user_city, sm_user_postal_cde
             from secondary s
           where s.sd_key = m.sm_key
  

• AS3 getting a random multiple number of table

  Hello:

  I have a hard time finding ways to manage the division with random numbers of tables using AS3.

  On my stage, I have two dynamic boxes; Box1 and box2 and a button, btn.

  in the Assembly, I have the following berries:

  var myDivisor:Array = [3,6,9,12,15,18,21,24,27,30];

  var myDivider:Array = [1,3,6];

  The idea is to divide 1 box per box of 2, so myDivisor should be a random number greater than a random number between myDivider

  WITH THE HELP OF BTN, HOW CAN I GET A RANDOM OF EACH TABLE, SO THIS DIVISOR NUMBER IS HIGHER THAN THE DIVISOR?

  In the timeline for button, here is the code:

  btn.addEventListerner (MouseEvent, CLICK, rNum);

  function rNum (MouseEvent): void

  {

  Box1. Math.Floor (Math.random) * myDivisor.length) / box2. Math.Floor (Math.random) * myDivider.leng th);

  }

  Please suggest ways to make this work.

  German

  var myDivisor:Array = new Array (3,6,9,12,15,18,21,24,27,30,23,89,90,1,34,56,78,99,345,456);
  var myDivider:Array = new Array (1,3,6);
  var divisor: int;
  var divisor: int;

  getNumbers();
  function getNumbers() {}
  divisor = getRandom (myDivisor);
  divisor = getRandom (myDivider);
  checkValues();
  }
  function getRandom(array:Array):int {}

  var NombrAl = Math.floor (Math.random () * array.length);
  return table [NombrAl];

  }
  function checkValues() {}
  If (getRandom (myDivisor) > getRandom (myDivider)) {}
  trace (getRandom (myDivisor));
  trace (getRandom (myDivider));

  } else {}
  getNumbers();
  }
  }

• Tables not set

  I'm having a problem with the animation of the text loaded from an XML document. The info is loaded, and when the trace is run on the individual attributes the content is there. However, the tables in which the data are pushed an error undefined:
  = set output =.
  Map title: 0
  Length of table legends: 3
  RandomText: 3
  Title: undefined, Description: undefined
  =================================================
  The actionscript code is included below. Can anyone help? TX!
  

  your cardTitles bays and the descriptions are declared as local inside your onLoad handler variables, which means that they are not available outside this function.

  put these two lines:
  var cardTitles:Array = new Array();
  descriptions of the var: Array = new Array();

  below this line:
  Legends of the var: Array = new Array();

• Help with the query to return the results of the group in the table?

  Hello
  Im a total noob, so please be nice...!
  
  I'm looking around a table contains a column of WORD and a WORD_TYPE_ID column.
  Words can have the same type_id.
  
  What I'm trying to do is to write a query that will return the longest word for each word_type_id.
  I tried for hours to get it and everything seems to get is two error messages or just the longest word in the whole WORD column.
  
  Is the furthest I can get before things break down...
  
  Select Word
  table
  where
  Length (Word) =
  (
  Select
  Max (length (Word))
  table
  )
  
  Any help on this or if I could be pointed in the right direction it would be greatly appreciated.
  
  Thank you

  Hello

  Welcome to the forum!

  Here's one way:

  SELECT    word_type_id
  ,       MIN (word) KEEP (DENSE_RANK LAST ORDDER BY LENGTH (word))     AS longest_word
  FROM       table_x
  GROUP BY  word_type_id;
  

  If there be a tie in some word_type_id (i.e. 2 or more words have exactly the same length, which is the longest in this group) the expression above will return the first one alphabetically. (This is what means here MIN.)

  Published by: Frank Kulash, 11 August 2009 13:56

  You almost had it.
  As you have noticed, you get the longest line across the table. This is because your subquery was watching the entire table.
  If you to correlate the subquery to the row in the main table, as shown below, you can get the longest word in each group:

  select  word
  from     table     m                              -- m for main
  where      length (word) = (
                              select  max (length(word))
                 from     table
                 where     word_type_id  = m.word_type_id     -- New
                 );
  

• Form required if condition of meeting

  I hope that someone can help me with a formula allowing several possible answers based on criteria in the worksheet 3.

  The first sheet has the details of the vehicle.

  Veh Reg		front of cabin at the rear of the 27 ft. trailer	Front of the cabin at the rear of the trailer 35 ft	Front of the cabin at the rear of the trailer 45 feet
  ABC		10,73 ft	10.50 ft	9.5
  CAD		10,73 ft	10.50 ft	9.5

  Second sheet

  Trailer No.		length
  2231		27 ft
  3359		35 ft
  4456		45 feet
  4423		45 feet
  2232		27 ft
  3357		35 ft
  3358		35 ft

  Third worksheet, the formula is required by the regulation. vehicle to find the first sheet and trailer No. conclusion. in the second sheet and calculating these 2 length to give me a length of harvesters.

  Vehicle Reg	Trailer No.	Combined length
  ABC	4456	= 9.5 + 45
  CAD	2231	= 10,73 + 27

  I hope that this is explained properly

  Thank you very much, David

  Hi David,

  Here's a possible solution, using MATCH and INDEX

  

  Note that each Table has a different name. It is important that it allows formulas designate each Table without having to include the name of the worksheet. Rename the tables of your document to match these, and you will be able to use the same formulas in table 3

  Table 1 and table 2 contains none of these options. All data is entered directly.

  A second header line has been added to table 1 to allow using a number instead of a string of text at the top of the columns C, D and E, which allows to MATCH search value, he can get from table 2.

  There is NO merged cell in row 1 of table 1. The text is entered in cell D1, lined up to be centered within the cell and with "Wrap text" unchecked in the Inspector to Format (text), empties into empty cells in C1 and E1.

  Table 3A that one formula, is entered in C2 and filled down from there until the end of the column. The basic formula has two parts, joined by a +. Each party gets one of the two numbers in the sum, the FCtoBT length for table 1, the length of the trailer of table 2.

  INDEX (table 1::A:E, MATCH (Table a, 1::A, 0), GAME (B, 2::A, 0) + 1) + INDEX (table 2::A:C, MATCH B, table 2::A, 0, 3)

  The heart of the formula will return an error "can't find it" for rows where column A or B contain empty cells, the 'switch' below (in bold) is thus added to check the two cells for data and prevent the heart of the formula to be called if one is found empty.

  IF (GOLD (A = "", B =""), "",INDEX (1::A:E, GAME table (Table a, 1::A, 0), GAME (B, 2::A, 0) + 1) + INDEX (table 2::A:C, MATCH B, table 2::A, 0, 3)")

  Copy the formula cell, select C2 of table 3, type a =, and then paste.

  Fill the formula giving the rest of the column.

  Kind regards

  Barry

• text with a larger font for printing on paper of reflow

  I visit the http://paragraf1.cappelendamm.no/c165513/sammendrag/vis.html?tid=205854 page and try to print it.

  The result is "scaling to fit page', and has a very small font.
  I can print in landscape mode, which solves the problem in this particular case, but I would like to be able to print in portrait mode.

  If I feel 100% nationally, much of the text disappears to the right of the edge of paper.

  I wish that it is a control to rearrange the text using large fonts without increasing the total width of the layout.

  Is there a way?

  This site cleverly created which removes the first column of the table of three columns when you print a style sheet, but they neglected the fact that the width of the table is fixed in pixels for width too wide for portrait printing.

  In this area and many other cases, you can manually 'hack' of the page to print better, but unfortunately, there is no single solution to this problem.

  For example, in this case, you can remove the width setting of the table of three columns with a snippet of JavaScript code.

  (1) select the following script, and then copy it to the Clipboard:

  
  var tables=document.querySelectorAll("table"); for (var num=0; num<tables.length; num++){tables[num].removeAttribute("width");}
  

  (2) open the web console by pressing Ctrl + Shift + k (it's also in the Web Developer menu).

  (3) next to the circumflex accent (>) paste the line script and press ENTER to run it

  (4), try to preview before printing to confirm that it worked. You can close the web console using the small "x" in the upper right corner of the pane, or by pressing Ctrl + Shift + k new.

  I've always wanted an add-on that could attach to the top of any page to print, but because of the wide variety of page layouts, it is difficult to automate. For now, you can try one that allows you to add the page to the preview screen: Print Edit - Add - ons for Firefox. It takes some practice to select the correct element, but if you prefer really clean and usable prints, it might be worth investing a bit of time to learn it. Regarding the ' "smart printing, maybe one day!

• Help automate and functions by the numbers

  Hello!

  I need assistance with numbers 3.6.1 for OS X El Capitan. My question is really complicated, and I couldn't even begin to understand what to search using search engines, that is why I turn to the experts in communities for help.

  What I'm trying to do is set up a system where when an a table on a single sheet is changed, the information on a second table on a second sheet is updated to match. Specifically, on the first table on the first worksheet, I keep a spreadsheet of all the video games I have in my collection, and with each game video I save several data as platform, Publisher, genre and ESRB rating points, and so on, as follows:

  

  On the second picture, I have an assortment of tables of data that takes into account the number of games I have they fall into each of these same points data in separate tables, so, how many games I have for every platform, every Editor, in every kind and with each ESRB rating, as follows:

  

  I was able to set up all the tables on the second sheet automatic update of the charges for each individual in each category element, so if I add, say, a new game of Nintendo 3DS to the first sheet on the first table, the corresponding to "Nintendo 3DS" number on the table of the second sheet platform is automatically updated. I did this by manually applying the COUNTIF function to each individual B cell in each table and was pointing at the corresponding column of table on the first worksheet with the correspondent of expression of the cell [for Game Boy Advance, for example, it looks like this: COUNTIF (Collection::Table 1: $Platform, "Game Boy Advance")]. Part of what I want to accomplish is to have all the B cells on the second act of sheet as "smart cells", if you want to, where they are all automatically use the COUNTIF function using the expression in each corresponding to a cell, instead of the manual set up I have now. Is something like this? If so, I want finally to do is to have the paintings on the second sheet automatically be updated to take account of the new factors of table on the first worksheet. That is if I add a game to a new platform, we will say Virtual Boy, to the table on the table of the first sheet, is it possible to have the picture of the platform on the second sheet automatically add a new entry for Virtual Boy, and should the COUNTIF automatic formatting for the work of B lymphocytes as well, automatically count the number of entries of Virtual Boy in the platform column the first sheet as well?

  Here is a link to my spreadsheet in iCloud for reference (remember, it's 'View Only'): https://www.icloud.com/numbers/000AvaqQSb1iRb3-5XbyaoNww#My_Collection

  I hope all this makes sense to someone out there, I tried to explain as clearly as I could. I'm still sort of novice in numbers, if I appreciate any amount of teaching or imparting wisdom!

  Thank you

  -Kill

  Kill them salvation.

  for all B cells on the second act of sheet as "smart cells", if you want to, where they are all automatically use the COUNTIF function using the expression in each corresponding to a cell, instead of the manual set up I have now.

  

  Formula in B2 (and fill by dragging the handle downwards to fill yellow)

  = COUNTIF(Collection::Table 1::$B,A2)

  B2 in the platform table and paste in the length of Table B2. It will show zero because he seeks the wrong column in the table on the sheet Collection.

  

  Click on the "Collection::Table 1: $B" diamond in the formula editor to select.

  Go to the Map Collection and click on the label C.

  AHA! It always displays zero because does not match the description of length:

  "Very short (less than 5 hours)" on the card product and "Very short" on the Collection sheet.

  The descriptions must match exactly.

  The easy way to change them in the Collection journal is to sort the table. But first of all insert another column, and then type 1 then 2 and fill down. Then you can sort to the original order.

  

  Now sort the column Length.

  Change the description (copy and paste into each "block")

  Funny looks because the line height are different

  

  The COUNTIF works because the descriptions are now (the formula now displays column D because of the new column of original order of Collection).

  I'll post this before the time-out annoys me.

  

  Kind regards

  Ian.