PHP Variable table size with restricted number of columns

Question

I'm trying to make an image preview in a table, just for checking the submitted files. The table can't have more than 3 cols, and the number of rows and cels is variable, because I skip the "not found" images in DB. I did the code below, but couldn't solve the logic by myself. The table shows the same image for row, and jumps 2 results for the next one.

    <?php
        $dados      = mysql_fetch_array (mysql_query("SELECT id,placa,usuario FROM dados WHERE id='".$_SESSION['novaOS']."'"));
        $itens      = mysql_result (mysql_query("SELECT itens_acessorios FROM vist_aval WHERE idp='".$_SESSION['novaOS']."'"),0);

        $sqlFotos   = "SELECT * FROM imagens WHERE idp='".$_SESSION['novaOS']."'";
        $qrFotos    = mysql_query($sqlFotos) or die();
        $rowFotos   = mysql_fetch_array($qrFotos) or die();
    ?>

    (...)

    <div>
        <table>
            <?php
                $dir_img = "../uploads/fotos/";
                for($f=2;$f<=33;){
                    $foto = $rowFotos[$f];
                    $td = '<td><img src="'.$dir_img.''.$foto.'" style="margin:0;width:100%;height:auto"></td>';
                    $tr = '<tr id="gridpreview"></tr>';
                    if ($foto != false){
                        for ($r=0;$r<=3;$r++){
                            if ($r>0){
                                echo $td;
                                $f++;
                            }
                            else{
                                echo $tr;
                            }
                        }
                    }
                }
            ?>
        </table>
    </div>

In addiction, the message "Fatal error: Maximum execution time of 30 seconds exceeded in C:\xampp\htdocs\app\Finalizar.php on line 88" appears on loading. The line 88 refers to if ($foto != false).

WARNING: If you're just learning PHP, please, do not use the mysql_query interface. It’s so awful and dangerous that it was removed in PHP 7. A replacement like PDO is not hard to learn and a guide like PHP The Right Way explains best practices. Your user parameters are not properly escaped and there are SQL injection bugs that can be exploited. — tadman
– tadman, Commented Sep 23, 2016 at 18:43
@tadman Yes, I'm dealing with PHP not long ago. I'm just using it this way because of the rush for something functional. After that, I'll have more time for reading and applicate best practices. So I'll probably change a lot of things. Thank you very much for your warning. — Diogo Muniz
– Diogo Muniz, Commented Sep 23, 2016 at 18:52

Poiz · Accepted Answer · 2016-09-23 17:49:41Z

1

That's not how you do a columnar output. It should be more like

$col = 0;
echo '<tr>';
while($row = ... fetch row from result ...) {
    echo "<td>$row[whatever]</td>";
    $col++;
    if ($col > 2) {
       echo '</tr><tr>';
       $col = 0;
    }
}
echo '</tr>';

You don't need two loops, just the one loop, which keeps repeating until there's more no results left in the DB.

edited Sep 23, 2016 at 17:49

Poiz

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answered Sep 23, 2016 at 17:48

Marc B

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2 Comments

Diogo Muniz Over a year ago

In this case, isn't it creating <tr>'s one inside another?

Diogo Muniz Over a year ago

I made same changes in this code and it works. Thanks for your help!

Diogo Muniz · Accepted Answer · 2016-09-23 18:55:28Z

0

Just to register the functional code:

<table>
<?php
$dir_img = "../uploads/fotos/";
$col=0;
$f=2;
echo "<tr>";
while ($f<=33){
$foto = $rowFotos[$f];
if ($foto != false){
if ($col>2){
echo "</tr><tr>";
$col = 0;
}
else{
echo "<td><img src='".$dir_img."".$foto."' style='margin:0;width:100%;height:auto'></td>";
$col++;
$f++;
}
}
else{
    $f++;
}
}
echo '<td>';
?>
</table>

answered Sep 23, 2016 at 18:55

Diogo Muniz

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