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?

Try category instead of catagory. It won't fix your problem, but it will give you extra marks in your final class report (wink) — Funk Forty Niner
– Funk Forty Niner, Commented Aug 2, 2013 at 13:18
@Fred - Ha thanks! However any idea how i can do what i need to? — Shane
– Shane, Commented Aug 2, 2013 at 13:19
Hint: What's the difference between SELECT * and SELECT cola, colb, colc ? — O. Jones
– O. Jones, Commented Aug 2, 2013 at 13:20
You are using an obsolete database API and should use a modern replacement. You are also vulnerable to SQL injection attacks that a modern API would make it easier to defend yourself from. — Quentin
– Quentin, Commented Aug 2, 2013 at 13:24

klkvsk · Accepted Answer · 2013-08-02 13:22:51Z

2

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 <.

answered Aug 2, 2013 at 13:22

klkvsk

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Comments

AmazingDreams · Accepted Answer · 2013-08-02 13:21:14Z

1

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

answered Aug 2, 2013 at 13:21

AmazingDreams

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1 Comment

Shane Over a year ago

Changed this, first row showing one column, second row 2 columns, 3rd row 1 column?

rink.attendant.6 · Accepted Answer · 2013-08-02 13:26:34Z

<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>

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