[RESOLVED] Sort of advanced database output

Maggan

Having problems putting it down in Notepad just how to confront the following problem.

I have a database table called tablename that I want to output into a html table just one of the columns called name and I want to skip the rows that has another column called other_name as NULL

Query looks like this.

$result=mysql_query("SELECT * FROM tablename WHERE other_name IS NOT NULL");

Right, I know how to set up loops and such and have done so but not gonna take up space here to show it as it seems unnecessary. What I want is a html table to display three records of the name in one single table row or column(suspecting table row will be easier, but would prefer column.), one in each field and if there are more, create another table row/column and continue with new 3 fields.

Any takers on this?

Thank you in advance.

leatherback

$i=1;
$table = "<table><tr>";
$result=mysql_query("SELECT * FROM tablename WHERE other_name IS NOT NULL");  

while($row = mysql_fetch_array($result))
  {
   $table .="<td>".$row[name]."</td>";
    if($i == 3)
      {
       $table .= "</tr><tr>";
       $i=0;
      }
     $i++;
   }

for($j = $i, $j<3, $j++)
  {
   $table .= "<td></td>"
  }
$table .= "</tr></table>"

echo $table

May need some tinkering, but this is the general drill

maxpup979

something like this?

function draw_table($arr,$len){
print "<table><tr>";
for($x=0;$x<count($arr);$x++){
	if($x%$len==0){print "</tr><tr>";}
	print "<td>".$arr[$x]."</td>";
}
print "</tr></table>";
}

Maggan

Thank you very much. This, I can work with.