Replies

• allan Posts: 63,498Questions: 1Answers: 10,471 Site admin

  March 2011

  You can just add HTML to the JSON return for your column which includes the link tag you want. If you have a look at line 162 in this example: http://www.datatables.net/development/server-side/php_mysql - you can see how you can add special formatting for a given column.
  
  Allan
• BiggestBadDaddy Posts: 6Questions: 0Answers: 0

  March 2011

  Thanks for the response Allan. I apologize, but I'm not following how to add the html to the code.
  
  I want both the listing ID and the listing TITLE to be hyperlinks to the listing page for SEO purposes, as well as the entire row to be click-able, which, to my knowledge, doesn't appear to the spiders like the text links do. This is what have now on the data table I wrote:
  
  [code]
  >
  <?php echo $id; ?>
  <?php echo $title_one; ?>, <?php echo $title_two; ?>
  <?php echo $photoYes; ?>
  
  [/code]
• allan Posts: 63,498Questions: 1Answers: 10,471 Site admin

  March 2011

  To modify my example to add a link to Google for the version column you can just do this (again at line 162):
  
  [code]
  $row[] = ''. $aRow[ $aColumns[$i] ] .'';
  [/code]
  That's the basic idea of how you can format the string for the return.
  
  Allan
• BiggestBadDaddy Posts: 6Questions: 0Answers: 0

  March 2011

  You are super cool, thank you so much. I'll give this a try this afternoon.
  Thanks.
• BiggestBadDaddy Posts: 6Questions: 0Answers: 0

  March 2011

  Hi Allan
  I tried the example you provided, unfortunately I kept getting the JavaScript error. I even tried to copy and paste the entire code and replace my database with the one in the example and still got the error.
  Not sure how to fix this.
  Thanks
• allan Posts: 63,498Questions: 1Answers: 10,471 Site admin

  March 2011

  What is the error?
• BiggestBadDaddy Posts: 6Questions: 0Answers: 0

  March 2011

  [code]
  while ( $aRow = mysql_fetch_array( $rResult ) )
  {
  $sOutput .= "[";
  for ( $i=0 ; $i
• allan Posts: 63,498Questions: 1Answers: 10,471 Site admin

  March 2011

  I think both examples are saying what the issues is and where to start debugging. So with your own code (which is probably what you want more than mine :-) ) have a look at the JSON return from the server. The error is saying that it is not valid, so it stands to reason that there is an issue with the return (you can run it through http://jsonlint.com to check it). Looking at the return in Firebug will help.
  
  Allan
• BiggestBadDaddy Posts: 6Questions: 0Answers: 0

  March 2011

  Will do, thanks for the help.