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Split database output dynamically into two columns for layout

Posted on 2008-10-24
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Medium Priority
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Last Modified: 2013-12-13
I have a simple php script querying a database a listing everything out in one long column. For layout purposes I would like to break it into two columns. Can this be done dynamically so the columns are automatically balanced or do I need to choose an item half way down the table and break there manually?
<?php
$con = mysql_connect("","","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
 
mysql_select_db("", $con);
 
$result = mysql_query("SELECT * FROM vendors ORDER BY vendor ASC");
 
while($row = mysql_fetch_array($result))
  {
  
  print"<div><img src=\"vendorlogos/{$row['logo']}\" height=\"45\" alt=\"\"/><br /><br /><u><strong>{$row['vendor']}</strong></u><br />{$row['description']}<br />Category: {$row['category']}<br /><br /><img src=\"http://www.domain.com/images/mini_arrow.jpg\"/>&nbsp;Product PDFs:<br/><br/><a href=\"http://www.domain.com/vendorpdfs/{$row['pdf']}\">{$row['pdf']}</a><br /><a href=\"http://www.domain.com/vendorpdfs/{$row['pdf2']}\">{$row['pdf2']}</a><br /><a href=\"http://www.domain.com/vendorpdfs/{$row['pdf3']}\">{$row['pdf3']}</a><br /><br /><br />Visit vendor<a href=\"http://{$row['website']}\" target=\"_blank\">Website</a><hr><br /></div>";
  
  }
 
mysql_close($con);
?>

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Question by:guy4graphics
  • 2
  • 2
4 Comments
 
LVL 39

Expert Comment

by:Roger Baklund
ID: 22800656
Maybe something like this could work:
$result = mysql_query("SELECT * FROM ...");
$rowcount = mysql_num_rows($result);
if($rowcount > 0) {
  echo '<table><tr><td>';
  $count = 0;
  while($row = mysql_fetch_array($result)) {
    print "<div>output goes here</div>";
    $count += 1;
    if($count == floor($rowcount/2)) 
      echo '</td></td>';
  }
  echo '</td></tr></table>';
}

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Author Comment

by:guy4graphics
ID: 22800825
This just lists them in one column as before in no particular order. Am I doing something wrong?
 <?php
$con = mysql_connect("","","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
 
mysql_select_db("", $con);
 
             
$result = mysql_query("SELECT * FROM vendors ORDER BY vendor ASC");
 
$rowcount = mysql_num_rows($result);
if($rowcount > 0) {
  echo '<table><tr><td>';
  $count = 0;
  while($row = mysql_fetch_array($result)) {
    
    print"<div><img src=\"vendorlogos/{$row['logo']}\" height=\"45\" alt=\"\"/><br /><br /><u><strong>{$row['vendor']}</strong></u><br />{$row['description']}<br />Category: {$row['category']}<br /><br /><img src=\"http://www.domain.com/images/mini_arrow.jpg\"/>&nbsp;Product PDFs:<br/><br/><a href=\"http://www.domain.com/vendorpdfs/{$row['pdf']}\">{$row['pdf']}</a><br /><a href=\"http://www.domain.com/vendorpdfs/{$row['pdf2']}\">{$row['pdf2']}</a><br /><a href=\"http://www.domain.com/vendorpdfs/{$row['pdf3']}\">{$row['pdf3']}</a><br /><br /><br />Visit vendor<a href=\"http://{$row['website']}\" target=\"_blank\">Website</a><hr><br />
  
  </div>";
    $count += 1;
    if($count == floor($rowcount/2)) 
      echo '</td></td>';
  }
  echo '</td></tr></table>';
}
 
mysql_close($con);
?>

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LVL 39

Accepted Solution

by:
Roger Baklund earned 2000 total points
ID: 22800853
Sorry, there was a bug. Change this:

      echo '</td></td>';

... to this:

      echo '</td><td>';
0
 
LVL 1

Author Comment

by:guy4graphics
ID: 22801399
I should have seen that - thanks for your help! Works great now!
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