Formatting a MySQL query in PHP/HTML

Question

This is my output code so far. It currently displays everything underneath each other, as expected.

<table id="dslrTable">

                <?php 

                while($row = mysqli_fetch_assoc($result)) { ?>

                            <?php echo "<tr><td> <a href=model.php?id=" . $row["ID"] . ">" . "<img src=images/" . $row["Image URL"] . " </a></td></tr>"; ?>
                            <?php echo "<tr><td> <a href=model.php?id=" . $row["ID"] . ">" . $row["Make"] . " " . $row["Model"] . " </a></td><tr>"; ?>
                            <?php echo "<tr><td> <a href=#.php>[add to compare] </a> </td></tr>"; ?>

                <?php } ?>

            </table>

What I am trying to achieve is three tr's per row in the database. The image, make/model and then a hyperlink. After 4 "results", then make a new tr. Something like this.

<table>
                <tr>
                    <td>Image 1</td>
                    <td>Image 2</td>
                    <td>Image 3</td>
                    <td>Image 4</td>
                </tr>
                <tr>
                    <td>Make and Model 1</td>
                    <td>Make and Model 2</td>
                    <td>Make and Model 3</td>
                    <td>Make and Model 4</td>
                </tr>
                <tr>
                    <td>Hyperlink 1</td>
                    <td>Hyperlink 2</td>
                    <td>Hyperlink 3</td>
                    <td>Hyperlink 4</td>
                </tr>
                <tr>
                    <td>Image 5</td>
                    <td>Image 6</td>
                    <td>Image 7</td>
                    <td>Image 8</td>
                </tr>
                <tr>
                    <td>Make and Model 5</td>
                    <td>Make and Model 6</td>
                    <td>Make and Model 7</td>
                    <td>Make and Model 8</td>
                </tr>
                <tr>
                    <td>Hyperlink 5</td>
                    <td>Hyperlink 6</td>
                    <td>Hyperlink 7</td>
                    <td>Hyperlink 8</td>
                </tr>
            </table>

And for this to keep going until my while loop stops. I know I need to use modulo division as well as a counter variable, but I can't get it to work.

Any help is appreciated, thanks.

Amarnasan · Accepted Answer · 2017-03-08 13:33:51Z

1

    <?php
         $i=0;
         while($row = mysqli_fetch_assoc($result)) {
            $image[$i] = $row["Image URL"];
            $model[$i] = $row["Model"];
            $hyper_link[$i] = $row["Hyperlink"];
            $i++;
        }

    ?>


    <table border="1">
         <?php 
         $k=0;
         $l=1;
         $m=2;
         $n=3;
         for($j=0;$j<count($image)/4;$j++){
             echo "<tr>";
             echo "  <td> {$image[$k]}</td>";
             echo "  <td> $image[$l]</td>";
             echo "  <td> $image[$m]</td>";
             echo "  <td> $image[$n]</td>";
             echo "</tr>";
             echo "<tr>";
             echo "  <td> $model[$k] </td>";
             echo "  <td> $model[$l]</td>";
             echo "  <td> $model[$m]</td>";
             echo "  <td> $model[$n]</td>";     
             echo "</tr>";
             echo "<tr>";
             echo "  <td> $hyper_link[$k] </td>";
             echo "  <td> $hyper_link[$l]</td>";
             echo "  <td> $hyper_link[$m]</td>";
             echo "  <td> $hyper_link[$n]</td>";
             echo "</tr>";
             $k +=4;
             $l +=4;
             $m +=4;
             $n +=4;
         } 
         ?>
    </table>

edited Mar 8, 2017 at 13:33

Amarnasan

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answered Mar 7, 2017 at 17:29

Resheil Agarwal

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1 Comment

Jay Blanchard Over a year ago

While this code may answer the question, providing additional context regarding why and/or how this code answers the question improves its long-term value.

J. Pichardo · Accepted Answer · 2017-03-07 17:58:19Z

0

I pretty sure that's not the appropriate use of the function mysqli_fetch_assoc(). You really should store your results in an array variable, count the result, and pass it all through a for loop.

$row = mysqli_fetch_assoc($result);

$result_count = count($row);

for($i = 0; $i <= $result_count; $i++){

}

edited Mar 7, 2017 at 17:58

J. Pichardo

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answered Mar 7, 2017 at 17:19

Joshua White

1

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