How to insert images from the external file in the block of text using AppleScript

Similar Questions

• How to insert image in the region

  Hello!
  
  How to Insert Image in the region, as I upload image with the name "bar_color.jpg" in my workspace. And in the attributes of the region, I found
  1-static id
  2-region attributes
  View 3-region selector
  image 4-region
  The 5-image tag attributes
  Attributes of cells 6-region HTML table
  
  What to enter in these attributes.
  I entered bar_color.jpg in the column region's image, but the image does not appear
  
  Any help!

  Hello

  You can not simply use an img tag in the Source of the region ? :

  Otherwise, you can view the type of the Image to displayitem, if you want to display the image directly from the BLOB.

• How to insert images in the content pane HTMLEditor?

  How to insert images in the content pane HTMLEditor? or a hyperlink?

  > know the position of the caret

  Well, there is no public API on the HtmlEditor to know that.

  You could create an tweak asks to add such an api.

  In the meantime, I believe that the HtmlEditor is implemented as a contentEditable has WebView below, so I think it's possible to hack together a solution if you want to.

  To do this:

  1. create a program that uses the HtmlEditor.

  2. run the program and use ScenicView to work on the css to display embedded Web search path.

  3. Once your HtmlEditor appears on a scene, use a node search to dig up a reference to the Web view.

  4. Once you have a reference to the Web mode, use executeScript to get the position of the caret by using JavaScript.

  5. Once you know where the caret, use a mechanism similar to my example program above to allow you to insert an image to the location of the caret (get the html text, put the position appropriate, up-to-date then set it again).

  Of course, you don't need to use a HtmlEditor control to do this, you can also just use a plain WebView with contentEditable flag put on it or any number of 3rd party JavaScript html editors have image insert functions.

• How to insert image from mysql into the table using php and create the checkbox in the table?

  How can I insert image from mysql into the table using php and create the checkbox for each data as a vote? Here is my code...

  WELCOME

  connect_error) {die ("connection failed:".)} $conn-> connect_error); } $sql = "SELECT no, Calon, ID, of course, the Image OF THE candidates." $result = $conn-> Query; If ($resultat-> num_rows > 0) {echo ' '; export data of each line while ($row = $result-> fetch_assoc()) {"echo"}}

  NO	Candidate	INFO	Vote
  " . $row ["no"]. "		-" . $row ["Calon"]. " -" . $row ['ID']. " -" . $row ['class']. " "; } ECHO ' ' ;} else {echo '0 results' ;} $conn-> close();?} > hope someone can help me because I am a newbie in this program... need to finish this project... Thank you.

  If you have saved the file name in the database, it's pretty simple.

  echo '';
  

  Is the same for the box:

  echo '';
  

  If you have saved the image file in the database, it is much more complicated. I recommend you store only the file name in the database.

• How to import images from the printer/scanner in Windows 8

  Windows 8 has a program to import images from a printer/scanner?

  Hi TEGKy,

  You can follow the steps described in the connection with the similar question.

  http://productforums.Google.com/Forum/#! 0Qc3joM8ZZU/picasa/subject

  http://forums.MacNN.com/t/299779/how-to-import-photos-from-scanner

  http://answers.Microsoft.com/en-us/Windows/Forum/Windows_7-pictures/import-pictures-and-videos-using-Windows-option/b4a82dce-7b52-4969-8af1-21eb71f160c7

  http://answers.Microsoft.com/en-us/Windows/Forum/Windows_7-windows_programs/import-pictures-from-scanner/995b5888-cd9a-45A3-ACFF-044bec24ec0a

  http://answers.Microsoft.com/en-us/Windows/Forum/windows_vista-pictures/error-in-importing-pictures-into-Photo-Gallery/32e61751-4fb9-41e3-a4fe-8aaab9c1dee1

• OFA: how to insert values from the front end of the OPS to the database page

  Hi friends, I am new OAF, iam using jdeveloper version 9.0.3.5
  
  I created a QueryPG .in this page I HAV set a "create" button on this page when I click that "create" button must navigate to another page (ie) createPersonPG, I have hepatitis has created this createPersonPG in a RegionStyle of = > defaultsinglecolumn, with some of my attribute as "personId, phoneno, e-mail, address" in my createPersonPG I hav also set a button titled "save"...»»» " After you enter the data when I click on save button means a new line should be added to my database table that was my task... for which I wrote the code in my MainCO
  
  Successful work means:
  --------------------------------------
  I can able to query the data from the database
  From my QyeryPG, I can access createPersonPG
  
  MY PROBLEM
  -------------------
  When I enter values in the createPersonPG wen I click on "Save" button means that the data that I enter are not save to database, but in my QueryPG I get confirmation that saved successfully in my QueryPG. " I want data to be saved in the database... when I get select * from table_name average, should be a rank more
  
  can anyone help me with this task please I'll be greateful to you.. _ I was doing this job for more than 20 days Please helpme I complete this task
  
  Thank you
  Thilib
  
  my code for in MainCO
  
  in LICS I HAV written query command and navigate to the code "createPersonPG" like this:
  
  ' Public Sub processFormRequest (pageContext OAPageContext, OAWebBean webBean)
  {
  super.processFormRequest (pageContext, webBean);
  String actionInMainProgramScreen = pageContext.getParameter (EVENT_PARAM);
  String personId = pageContext.getParameter ("PersonId");
  If (pageContext.getParameter ("GO")! = null)
  {
  System.out.println ("Submit button");
  OAQueryUtils.checkSelectiveSearchCriteria (pageContext, webBean);
  String firstName = pageContext.getParameter ("FirstName");
  String phoneNO = pageContext.getParameter ("PhoneNO");
  [Serializable] parameters = {personId, firstName, phoneNO};
  System.out.println ("AM method is called");
  OAApplicationModule am = pageContext.getApplicationModule (webBean);
  am.invokeMethod ("initQuery", parameters);
  }
  
  
  If (actionInMainProgramScreen.equals ("Create"))
  {
  System.out.println ("create key");
  () pageContext.setForwardURL
  'OA.jsp?page=/gwp/oracle/apps/xxi/aaaPersonDetails/webui/createPersonPG '.
  null
  OAWebBeanConstants.KEEP_MENU_CONTEXT
  null
  null
  true
  OAWebBeanConstants.ADD_BREAD_CRUMB_YES
  OAWebBeanConstants.IGNORE_MESSAGES);
  }
  }------------------------------------------------------------------------------
  IN MY createPersonCO I HAV wrote this code
  
  ' Public Sub processRequest (pageContext OAPageContext, OAWebBean webBean)
  {
  super.processRequest (pageContext, webBean);
  String paramPersonAction = pageContext.getParameter ("xxParamActionOnPerson");
  OAApplicationModule am = pageContext.getApplicationModule (webBean);
  PersonParamList serializable [] = {paramPersonAction};
  
  If (paramPersonAction! = null & & paramPersonAction.equals ("Create"))
  {
  am.invokeMethod ("createPerson");
  }
  }
  
  /**
  * How to manage remittances form for form elements in
  * a region.
  @param pageContext OA page context
  @param webBean the grain of web for the region
  */
  
  ' Public Sub processFormRequest (pageContext OAPageContext, OAWebBean webBean)
  {
  super.processFormRequest (pageContext, webBean);
  OAApplicationModule am = pageContext.getApplicationModule (webBean);
  
  If (pageContext.getParameter ('Save')! = null)
  {
  System.out.println ("first line button Save");
  am.invokeMethod ("saveProgramToDatabase");
  System.out.println ("navigate to home page');
  String message = "saved successfully ';
  OAException exp = new OAException (message, OAException.CONFIRMATION);
  pageContext.putDialogMessage (exp);
  () pageContext.forwardImmediately
  'OA.jsp?page=/gwp/oracle/apps/xxi/aaaPersonDetails/webui/aaaPersonDetailsMainPG '.
  null
  OAWebBeanConstants.KEEP_MENU_CONTEXT
  null
  null
  true
  ADD_BREAD_CRUMB_YES);
  }
  }-------------------------------------------------------
  IN my AM I wrote this code
  
  ' public void initQuery (String personId, String firstName, String phoneNo)
  {
  System.out.println ("Am initQuery");
  aaaPersonDetailsVOImpl vo = (aaaPersonDetailsVOImpl) getaaaPersonDetailsVO1 ();
  System.out.println ("aaaPersonDetailsVO initQuery method called");
  vo.initQuery (personId, firstName, phoneNo);
  }
  
  Public Sub createPerson()
  {
  System.out.println ("first createPersonMethod line");
  aaaPersonDetailsVOImpl vo = (aaaPersonDetailsVOImpl) getaaaPersonDetailsVO1 ();
  System.out.println ("Line before if condition createPersonMethod");
  If (!) VO.isPreparedForExecution ())
  {
  System.out.println ("first line If condition createPersonMethod");
  vo.setMaxFetchSize (0);
  }
  System.out.println ("creating New Row");
  Line r = vo.createRow ();
  System.out.println ("new insertion line");
  vo.insertRow (r);
  System.out.println ("inserted row");
  r.setNewRowState (r.STATUS_INITIALIZED);
  System.out.println ("before"sequence initialized");
  Number personId = this.getOADBTransaction () .getSequenceValue "(XXI. PER_DETAIL");
  System.out.println ("after"sequence initialized");
  vo.getCurrentRow () .setAttribute ("Person_Id", personId);
  System.out.println ("sequence number =" + personId);
  System.out.println ("last line of the Create method");
  }
  
  Public Sub saveProgramToDatabase()
  {
  
  System.out.println ("before validation.");
  getOADBTransaction () .commit ();
  System.out.println ("new program created.");
  }
  
  Published by: user13307444 on July 26, 2010 10:20

  Hello

  This could be the problem if there is that data validation is there that does not back up the data on the page. Because for example I had this problem when my database column has been primary key while on the EO page it wasn't. & When I tried to record data EO was allowing me to record, but the database did not allow. Please check your OA

  AJ

• How to insert images without the audio 1 and the soundtrack

  I'm new to first 10 items and I have a problem when I move clips such as images in the middle of the timeline between video clips.

  When I move a photo directly in the middle of the time line it clips and separates the music I have on the soundtrack.   I found through Lynda.com and Adobe help if you hold down the "alt" key while inserting the new image you will not affect the soundtrack or other securities.   This problem I have with the alt key is now the rest of the clips in the timeline panel have audio track 1 and track the offset of 1 video by the duration of the image that was inserted.  My project is more than an hour for a long time then it is difficult to go back and that straighten.

  What is the best way to handle this?

  Adding additional video clips in the middle of the project works very well with the alt key as the video 1 and audio 1 move together.

  Unfortunately, you have to decide whether you want the new clip to insert. You can set the ripple to insert on a timeline and not other chronologies of ripple.

  If you don't hold down the CTRL key, the clip will insert you place - but it will break your piece of music.

  If you hold the Ctrl key, nothing on the timeline will go - but, if you try to place it where there is already a clip, it will overlap.

• How can I read from the InputStream when I use StreamConnectionNotifier

  I want to make a program that will listen on port 100 and exchange with PC message.

  And the program uses wifi to connect with the PC.

  I have StreamConnectionNotifier choice and I can receive request for socket of PC, but when I open imputStream, Exception throw.

  My code:

  import java.io.IOException;
  import java.io.InputStream;
  import java.io.OutputStream;

  Import javax.microedition.io.Connector;
  Import javax.microedition.io.StreamConnection;
  Import javax.microedition.io.StreamConnectionNotifier;

  SerializableAttribute public class ServerThread extends Thread {}
  constants
  private public static final String URL = ""http://:100;interface=wifi;deviceside=false "; "
  fields
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  private InputStream _in;
  private OutputStream _out;
  private StreamConnection stream = null;
  private HelloWorld _app;
     
  Private boolean isStop = false;
   
  public ServerThread() {}
    
  }
   
  {} public void setApp (HelloWorld app)
  _APP = app;
  }
   
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  try {}
  {while(!isStop)}
  {Synchronized (this)}
  _notify = Connector.open (URL) (StreamConnectionNotifier);
  }
  Stream = _notify.acceptAndOpen ();
  _in = stream.openDataInputStream ();  //<-- i="" think="" this="" line throw="" excepiton,="" excepiotn="" 's="" message="" is:request line="" is="">
  _out = stream.openDataOutputStream ();
  extract the data from the input stream
  StringBuffer sb = new StringBuffer();
  Reference int = - 1;
  {While (-1! = (reference = {_in.read ()))}
  Reference of the SB. Append ((Char));
  }
      
  _APP.updateText (SB. ToString());
      
  StringBuffer res = new StringBuffer();
  Res. Append ("HTTP/1.0 200 OK \r\n");
  Res. Append ("Content-type: text/html \r\n");
  Res. Append ("Content-Length: 1 \r\n");
  Res. Append ("1");
  _out. Write (res. ToString(). GetBytes());
      
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  _notify. Close();
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  } catch (IOException e) {}
  probably the stream has been closed.
  System.Err.println (try ());
  _APP.updateText (try ());
  } {Finally
  try {}
  If (stream! = null) {}
  Stream.Close ();
  }
  } catch (Exception ex) {}
  }
  try {}
  If (_notify! = null) {}
  _notify. Close();
  }
  } catch (Exception ex) {}
  }
  }
  }
   
  {} public void stop()
  isStop = true;
  try {}
  If (stream! = null) {}
  Stream.Close ();
  }
  } catch (Exception ex) {}
  }
  try {}
  If (_notify! = null) {}
  _notify. Close();
  }
  } catch (Exception ex) {}
  }
  }

  }

  When I use IE application this port the program can block reading of inputStream.

  What's wrong? Please help me.

  What could happen is the following (Please check if this is the case however). Your code side device receives an incoming connection, but is then stuck on waiting for InputStream.read return anything. This translates your code aside PC it also blocks its BufferedReader.readLine, because the device-side code is not send anything (yet). So, you have a classic network impasse. If it is exactly what is happening, then you can try to fix the problem in several ways.

  1. change the code aside device to read byte-by-byte (int InputStream.read () instead of void InputStream.read (byte [],...)). It might help because on a few BlackBerrys the TCP stack returns the data received so that the stream is closed by a peer until you start viewing byte-by-byte.

  AND/OR

  2. change the PC-side code to close the output stream before reading the input stream (that's how works HTTP/1.0, for example). It might help because on a few BlackBerrys the TCP stack does not return data received until the stream is closed by a peer.

• How to send images from the photo library

  don't know how to do for the form by e-mail photos picture library

  This depends on your e-mail program. Open your program and create a new email. Select an item called "Send a file" or similar. Access your photos, select those that you want to send. Now your photos will be sent as an attachment when you send this e-mail.

  If not, right-click on the image you in the library, select ' send to ' and then 'recipient '.

  All the answers and suggestions are provided by an enthusiastic amateur and are therefore no explicit or implicit guarantee. Basically, you use my suggestions at your own risk.

• How to insert rows from an xml file in a table

  IM Using Oracle 11g, is show below error when I run...
  
  
  
  declare
  charString varchar2 (80);
  finalStr varchar2 (400): = null;
  whole rowsp;
  v_FileHandle UTL_FILE. TYPE_DE_FICHIER;
  Start
  -the name of the table specified in our DTD
  xmlgen.setRowsetTag ('Zipcodes');
  -the name of the data set as shown in our DTD
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  dbms_output. Enable (100000);
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  v_FileHandle: = utl_file.fopen (TMP_DIR1,'XML_NEW_CITIES.) XML', 'r');
  loop
  BEGIN
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  exception
  When no_data_found then
  UTL_FILE.fclose (v_FileHandle);
  "exit";
  END;
  dbms_output.put_line (charstring);
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  finalStr: = finalStr | charString;
  on the other
  finalStr: = charString.
  end if;
  end loop;
  -for the insertion of XML data in the table
  rowsp: = xmlgen.insertXML('SYS.) Codes POSTAL, finalStr);
  dbms_output.put_line ('INSERT FACT' |) To_char (rowsp));
  xmlgen.resetOptions;
  end;
  /
  
  
  
  
  XML_NEW_CITIES. XML is
  
  <? XML version = "1.0"? >
  < zip codes >
  < Record mappings = "4" >
  CA < STATE_ABBREVIATION > < / STATE_ABBREVIATION >
  < ZIPCODE > 94301 < / code >
  < CITY > Palo Alto < / CITY >
  < / maps >
  < Record mappings = "5" >
  < STATE_ABBREVIATION > CO < / STATE_ABBREVIATION >
  < ZIPCODE > 80323 < / code >
  < ZIP_CODE_EXTN > 9277 < / ZIP_CODE_EXTN >
  Boulder < CITY > < / CITY >
  < / maps >
  < / zip codes >
  
  
  
  
  Error report:
  ORA-06550: line 8, column 3:
  PLS-00201: identifier ' XMLGEN. SETROWSETTAG' must be declared
  ORA-06550: line 8, column 3:
  PL/SQL: Statement ignored
  ORA-06550: line 10, column 3:
  PLS-00201: identifier ' XMLGEN. SETROWTAG' must be declared
  ORA-06550: line 10, column 3:
  PL/SQL: Statement ignored
  ORA-06550: line 14, column 34:
  PLS-00201: identifier 'TMP_DIR1' must be declared.
  ORA-06550: line 14, column 3:
  PL/SQL: Statement ignored
  ORA-06550: line 31, column 12:
  PLS-00201: identifier ' XMLGEN. INSERTXML' must be declared
  ORA-06550: line 31, column 3:
  PL/SQL: Statement ignored
  ORA-06550: line 33, column 3:
  PLS-00201: identifier ' XMLGEN. RESETOPTIONS' must be reported
  ORA-06550: line 33, column 3:
  PL/SQL: Statement ignored
  06550 00000 - "line %s, column % s:\n%s".
  * Cause: Usually a PL/SQL compilation error
  
  Published by: 898235 on November 20, 2011 23:02

  OK, so you tried the XMLTable example, or wait for someone write at the bottom for you? :)

  You must first create an object of directory Oracle pointing to the physical location of the file (must be somewhere on the db server or at a location that the db server can access on the network).
  Then, create the procedure, in this case a simple INSERT SELECT:

  SQL> create directory tmp_dir as 'c:\temp';
  
  Directory created
  
  SQL> create table zipcodes (
    2   state_abbreviation  varchar2(2),
    3   zipcode             varchar2(5),
    4   zip_code_extn       varchar2(10),
    5   city                varchar2(80)
    6  );
  
  Table created
  
  SQL> CREATE OR REPLACE PROCEDURE insertZipcodes (
    2    p_directory IN VARCHAR2
    3  , p_filename  IN VARCHAR2
    4  )
    5  IS
    6  BEGIN
    7
    8    INSERT INTO zipcodes (state_abbreviation, zipcode, zip_code_extn, city)
    9    SELECT state_abbreviation
   10         , zipcode
   11         , zip_code_extn
   12         , city
   13    FROM XMLTable('/Zipcodes/mappings'
   14           passing xmltype(bfilename(p_directory, p_filename), nls_charset_id('AL32UTF8'))
   15           columns state_abbreviation varchar2(2)  path 'STATE_ABBREVIATION'
   16                 , zipcode            varchar2(5)  path 'ZIPCODE'
   17                 , zip_code_extn      varchar2(10) path 'ZIP_CODE_EXTN'
   18                 , city               varchar2(80) path 'CITY'
   19         )
   20    ;
   21
   22  END;
   23  /
  
  Procedure created
  
  SQL> exec insertZipcodes('TMP_DIR', 'xml_new_cities.xml');
  
  PL/SQL procedure successfully completed
  
  SQL> commit;
  
  Commit complete
  
  SQL> select * from zipcodes;
  
  STATE_ABBREVIATION ZIPCODE ZIP_CODE_EXTN CITY
  ------------------ ------- ------------- ---------------
  CA                 94301                 Palo Alto
  CO                 80323   9277          Boulder
   
  

• How to insert images in the composition of body

  Hello
  All of a sudden I cannot insert emoticons Hotmail images cannot insert anything of clip art images, BING or Smiley Central in my emails. It is also difficult to copy and paste images exactly where I want to be..., the cursor goes in all directions? I have never had a problem like this before and do not know what has happened or what caused it. I can't understand why I'm having a problem with everything else? Anyone else having problems? Can someone help me please? Thank you.

  Sudhakar-

  Hello

  1. is this problem limited to only Hotmail?

  2. do you use internet explore to surf internet browser?

  Try to update the chipset drivers as the cursor moves around.

  Please see the link below: update a hardware driver that is not working properly: http://windows.microsoft.com/en-US/windows-vista/Update-a-driver-for-hardware-that-isn ' t-work properly

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  The question you have posted is related to Windows Live Hotmail and would be better suited to the community of Windows Live Hotmail. Please visit the link below to find a community that will provide the best support.

  http://windowslivehelp.com/product.aspx?ProductID=1

• How to download images from the camera at the top of the Tower

  I just need to upload photos of a family on my counter top party

  Just connect the camera and syncronise it then copy all the images you want and paste it to any folder on the PC.

• How to insert image in the dashboard of BIEE

  If it is a dynamic state image will be good?

  Hello

  1. download the images on D:\ORAHome\j2ee\home\applications\analytics\analytics\res\s_oracle10\images (installation of the Oracle application server) and
  D:\OracleBI\web\app\res\s_oracle10\images (installation OBIEE)
  2 then after restart presentation services.
  3. in the responses, change the view title and your image.

  Reference: http://addsrikanth.wordpress.com/2008/06/10/displaying-images-in-obiee/

  Kind regards
  Suresh

• How to insert images as .jpg... gif attach file in email without any accessories

  Hi, this is Terry

  I would to know how to insert images on the part of email without an attachmenk.

  Thank you

  Tanveer Rauf

  An inserted picture is an attachment.  The embedded image is sent as an attachment and then it is returned in the message in an HTML message, but it is still an attachment to the message.

  In Outlook Express, you can embed a picture compose a new message in HTML then using Insert | Photo.  Who will integrate the image in the email, but it always comes as an image as an attachment.  It's just the way OE manages which shows as embedded.

  Alternatively, you can just link to an image by referencing a URL to the image on a website and this way no image will actually be embedded or attached.

  Steve

• How to display an image from the file in the database

  I want to be able to display an image from the file that I have stored in a database. In the database, the image is store as a file with numbers. When I go to display only the sign of an image appears. I want to be able to display the image not the symbol for numbers of the image or file. I know I'm close I just can't understand what I'm messing, to get the image to display.

  Here is the code:

  <? PHP require Connections/Connections.php"?" >

  <? PHP

  session_start();

  {if (isset($_SESSION["fname"]))}

  }

  else {}

  Header('Location:NewRequest.php');

  }

  ? >

  <? PHP

  $User = $_SESSION ['Pnom"];

  $result = $con-> query ("SELECT ALL * from newrequest where Fname ="$User"" ");

  ? >

  <! doctype html >

  < html >

  < head >

  < link href = "CSS/Master.css" rel = "stylesheet" type = "text/css" / > "

  < link href = "CSS/Menu.css" rel = "stylesheet" type = "text/css" / > "

  < meta charset = 'utf-8 '.

  ReviewRequest < title > < / title >

  < / head >

  < body >

  < div class = "Container" >

  < div class = "Header" > < / div >

  < div class = "Menu" >

  < div id = 'Menu' >

  < nav >

  < ul class = "cssmenu" >

  < li > < a href = "Account.php" > account < /a > < /li > ""

  < li > < a href = "UpdateAccount.php" > update account < /a > < /li > ""

  < li > < a href = "NewRequest.php" > new request < /a > < /li > ""

  < li > < a href = "ReviewRequest.php" > Review Request < /a > < /li > ""

  < li > < a href = "LogOut.php" > sign out < /a > < /li > ""

  < /ul >

  < / nav >

  < / div >

  < / div >

  < div class = "LeftBody" > < / div >

  < div class = "RightBody" >

  < name of the form = "form1" method = "post" action = "" >

  < table border = "1" >

  < b >

  < td = '26' height > name < table >

  Location < td > < table >

  Description < td > < table >

  Image of < td > < table >

  < Td > status < table >

  < /tr >

  <? PHP while ($row = $result-> fetch_assoc()) {? >}

  < b >

  < class td = "GFied" > <? PHP echo $row ["Fname"];? > < table >

  < class td = "GFied" > <? PHP echo $row ['Location'];? > < table >

  < class td = "GFied" > <? PHP echo $row ['Description'];? > < table >

  < class td = "GFied" > < img src = "<?" PHP echo $row ["numero_imagenom_image"];? > "alt =" "> < table >"

  < class td = "GFied" > <? PHP echo $row ['Status'];? > < table >

  <? PHP}? >

  < /tr >

  < /table >

  < / make >

  < / div >

  < div class = "Footer" > < / div >

  < / div >

  < / body >

  < / html >

  This is what it looks like when it displays:

  

  How can I actually picture to display?

  emrenilo32520834 wrote:

  I added the .jpg but the image still does not display, the images are saved in a jpg file.

  If the images in the database to follow this format - 1234567.jpg, 9897654.jpg

  Then your original code must have worked:

  

  Are the images in their own folder, if so you need to add this folder for code name: