Displaying data from database in table

Question

I am trying to display data in table form with 3 columns. Each should have a main category with some drop down lists. I get all the information to display but all is in one column and the drop down information does not display with the correct heading.

echo "<table>";

while ($row = mysql_fetch_array($result)) {
    $count = 1;

    if ($count = 1) {
        $sCatID = ($row['CatID']);
        echo "<tr valign='top'><td><b><a href='#" . $sCatID . "'>" . $sCatID . "</a></b><br>";
        // column 1 categories

        $result2 = mysql_query("SELECT * FROM test_prefixSubCat WHERE CatID=$sCatID");
        // sub-cats
        while ($row2 = mysql_fetch_array($result2)) {
            $sSub = ($row2['CatID']);
            $sSubID = ($row2['SubID']);
            echo "<dd><a href='#'>" . $sSub . "</a><br>";
            echo "</td>";
        }
        $count = 2;
    } elseif ($count = 2) {

        $sCatID = ($row['CatID']);
        echo "<td><b><a href='.$sCatID.'>" . $sCatID . "</a></b><br>";
        // column 1 categories

        $result2 = mysql_query("SELECT * FROM test_prefixSubCat WHERE CatID=$sCatID");
        // sub-cats
        while ($row2 = mysql_fetch_array($result2)) {
            $sSub = ($row2['CatID']);
            $sSubID = ($row2['SubID']);
            echo "<dd><a href='#'>" . $row2['Sub'] . "</a><br>";
            echo "</td>";
        }
        $count = 3;
    } elseif ($count = 3) {

        $sCatID = ($row['CatID']);
        echo "<td><b><a href='.$sCatID.'>" . $sCatID . "</a></b><br>";
        // column 1 categories

        $result2 = mysql_query("SELECT * FROM test_prefixSubCat WHERE CatID=$sCatID");
        // sub-cats
        while ($row2 = mysql_fetch_array($result2)) {
            $sSub = ($row2['CatID']);
            $sSubID = ($row2['SubID']);
            echo "<dd><a href='.$sSub.'>" . $sSub . "</a><br>";
            echo "</td></tr>";
        }

        $count = 1;
    }
}
if ($count = 2) {
    echo "<td>&nbsp;</td><td>&nbsp;</td></tr>";
} elseif ($count = 3) {
    echo "<td>&nbsp;</td></tr>";
}
echo "</table>";

It doesn't seem to close the rows and table correctly... And it is also putting some of the drop down items before it displays the first heading.

If i display it in only one column it is working fine.

Not sure for what you mean Radu, all I know its is not working the way I would like it to. Somehow the loop is not working correctly and Sorry! but i don't know why... — Marinus
– Marinus, Commented May 21, 2012 at 9:16
Your if statements should be using == rather than = as using a single = will reassign the value rather than compare against it — lethalMango
– lethalMango, Commented May 21, 2012 at 9:17
I have tried that lethalmango, but now it displays the numbers of the CATID rather than the value. — Marinus
– Marinus, Commented May 21, 2012 at 9:45
Are you trying to display the same data into three different table columns? what's the point? — Jetoox
– Jetoox, Commented May 21, 2012 at 9:59
I Have decided to create a table for each of the categories with it own sub headings. It works fine so I have removed the the $count=2 and $count=3 completely. Thanks anyway for anyone that had a look at it. More than one way to do something I guess. — Marinus
– Marinus, Commented May 21, 2012 at 10:22

saji89 · Accepted Answer · 2012-05-21 10:41:49Z

1

You should use == instead of single = in your if statements. Else it would execute everytime as that condition is always true.

answered May 21, 2012 at 10:41

saji89

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