PHP array into HTML table

Question

I am trying to show all products in a database with a certain category in a HTML table. However I'm not sure how to limit the table to three columns only.

Here is my code:

<table>
    <?php
  $catagory=$_GET["q"];


$con = mysql_connect("localhost","cl49-XXX","XXX");
if (!$con) 
  {
   die('Could not connect: ' . mysql_error());
   }

mysql_select_db("cl49-XXX", $con)or die( "Unable to select database");


$result=mysql_query("SELECT * FROM products WHERE catagory  = '$catagory' ")or die('You need enter a catagory ' );

  for ($i = 0; $i < mysql_num_rows($result); $i++)
    {
        $row = mysql_fetch_array($result);

        $prodname = $row['prodname'];
        $prodID = $row['prodID'];
        if ($i % 5 == 0 || $i == 0) {
            echo "<tr>";
        }
        echo "
        <td>
            <b>$prodname </b><br />
            Product ID: $prodID<br />
            <img src='/userpics/$prodID.jpg' height='200' width='200'>
        </td>";

        if ($i % 3 == 0 || $i == (mysql_num_rows($result)-1)) {
            echo "</tr>";
        }
    }
    ?>
<table>

I am waiting to show prodID, prodtitle and image all in the same "cell" but only have three columns (three products per row).

How do I do this?

Answer 1

echo "<tr>"; // first row beginning
for ($i = 1; $i <= mysql_num_rows($result); $i++)
{
    $row = mysql_fetch_array($result);

    $prodname = $row['prodname'];
    $prodID = $row['prodID'];

    echo "
    <td>
        <b>$prodname </b><br />
        Product ID: $prodID<br />
        <img src='/userpics/$prodID.jpg' height='200' width='200'>
    </td>";

    if ($i % 3 == 0) {
        echo "</tr> <tr>"; // it's time no move to next row
    }
}
echo "</tr>"; // last row ending

Note that $i is now starting from 1 and it loops while <= of num_rows, not < .

Answer 2

This does not look good.

if ($i % 5 == 0 || $i == 0) {

I think it should be

if ($i % 3 == 0 || $i == 0) {

Otherwise a new <tr> Won't be opened when you close the old one.

Expanding on this, you could make this a whole lot easier for yourself.

echo "<table><tr>"; // Open the first row
for ($i ..... etc.) {

    -- SNIP --

   if ($i % 3 == 0) {
        echo "</tr><tr>"; // imediately open new row
   }
}
echo "</tr></table>"; // Close last row as well as table

Answer 3

<table>
<?php    
    $con = mysql_connect("localhost","cl49-XXX","XXX");
    if (!$con) {
       die('Could not connect: ' . mysql_error());
    }

    mysql_select_db("cl49-XXX", $con)or die( "Unable to select database");

    $catagory = mysql_real_escape_string($_GET['q']); // prevent SQL injections

    if(empty($catagory)) {
        die('You need to enter a catagory');
    }

    $result = mysql_query("SELECT * FROM products WHERE catagory = '$catagory';") or die(mysql_error());
    // a general error might have occurred,
    // may or may not be related to not entering a catagory

    $rows = mysql_num_rows($result); // cache the number for performance

    for ($i = 0; $i < $rows; ++$i) {
        $row = mysql_fetch_array($result);

        $prodname = $row['prodname'];
        $prodID = $row['prodID'];
        if ($i % 3 === 0) {
            echo "<tr>";
        }
        echo "
        <td>
            <b>$prodname</b><br />
            Product ID: $prodID<br />
            <img src='/userpics/$prodID.jpg' height='200' width='200'>
        </td>";

        if ($i % 3 === 0) {
            echo "</tr>";
        }
    }

    if($i % 3 > 0) {
        // last row was not full
        echo '</tr>';
    }
?>
</table>