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php-general | 2001032

Re: [PHP] Catagory list From: Joe Stump (joe <email protected>)
Date: 03/31/01

This is what I'd do:

  $sql = "SELECT disinct category FROM article ORDER BY category";
  $r = mysql_query($sql);
  while($cat = mysql_fetch_array($r))
  {
    $id = $cat['id'];
        $sql = "SELECT * FROM article WHERE category='$id'";
        $x = mysql_query($sql);
        while($row = mysql_fetch_array($x))
        {
          // echo articles here
        }
  }

The permanent solution is to make a separate table called "categories" and then
do a join on the categoryID from the two tables.

--Joe

On Fri, Mar 30, 2001 at 10:33:11PM +1000, Mark Bayfield wrote:
> I am trying to make a catagory list that displays a distinct catagory, and
> then listing titles under that catagory.
> Unfortunatly all the data is in one table, and the catagory field is not set
> catagories, so there is duplicates in that field.
> The code so far is (but I am probably going about it the wrong way, if you
> can see whats wrong with my code plase help, or point me in some direction),
>
> $query = mysql_query("SELECT DISTINCT catagory from article order by
> catagory");
> $query2 = mysql_query ("SELECT * from article ORDER BY catagory");
>
> for ($index = 0; $index < mysql_num_rows($query); $index++) {
> $Catagory = mysql_fetch_row ($query) or die (mysql_error());
> $name1 = mysql_fetch_row ($query2) or die (mysql_error());
> $x = $name1;
> $result1 = array_unique($Catagory);
> $result = array_merge($Catagory, $name1);
>
> print ("$result1[0]<BR>");
> for ($x=0; $x == $result1[0]; $x++) {
> print ("&nbsp;&nbsp;&nbsp;&nbsp;<a
> href='update.php?passed=$name1[0]'>$name1[2]</a><br>\n");
> }
> }
>
> I am trying everything to get it do just print the following:
>
> Catagory
> title1
> title2
> Catagory1
> title3
> title4
>
> Please help me out....
>
> thanks
>
>
>
> --
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