Cannot find other rows from db in PHP Mysqli

Question

Before anything else this is my PHP Code

<?php
    $conn = mysqli_connect("localhost", "root", "", "mydb");
    if (!$conn) {
        die("Connection failed: " . mysqli_connect_error());
    }
    $promo1 = "";
    $promo2 = ""; 
    $promo3 = "";

    $sql = "SELECT prmo_itemDescription FROM promos";
    $result = mysqli_query($conn, $sql);

    $row = mysqli_fetch_array($result);
          $promo1 = $row[0];
          $promo2 = $row[1];
          $promo3 = $row[2];
    mysqli_free_result($result);
    mysqli_close($conn);
    ?>

and this is where I want the description to be echoed

      <div class="col-md-3">
          <p><?php echo $promo1?></p>
            <button type="button" class="btn btn-success" data-toggle="modal" data-target="#details-1">Buy Now!</button>
        </div>

        <div class="col-md-3">
          <p><?php echo $promo2?></p>
            <button type="button" class="btn btn-success" data-toggle="modal" data-target="#details-2">Buy Now!</button>
        </div>

        <div class="col-md-3">
          <p><?php echo $promo3?></p>
            <button type="button" class="btn btn-success" data-toggle="modal" data-target="#details-3">Buy Now!</button>
        </div>

and this is my database in phpmyadmin My problem is I managed to display the $promo1 but the other two(promo2 and promo3) doesnt seem to display, it looks like it cannot find the other two rows. There's an error like unidentified offset. am I missing something?

php.net/manual/en/mysqli-result.fetch-array.php - mysqli_fetch_array — Fetch a result row as an associative, a numeric array, or both — splash58
– splash58, Commented Oct 1, 2018 at 10:47
You've selected only prmo_itemDescription column in your SQL query, that's why you can't display the other ones. — AnTrakS
– AnTrakS, Commented Oct 1, 2018 at 10:49
You are only reading one single row of the result set, and pretend like that contained all the descriptions in multiple columns - which it doesn’t. You need to loop over the result, to output the data from each single row in the result set. — misorude
– misorude, Commented Oct 1, 2018 at 10:51
youve fetched one row, and assigned columns not rows to each var — delboy1978uk
– delboy1978uk, Commented Oct 1, 2018 at 11:13

Gonzalo · Accepted Answer · 2018-10-01 10:59:41Z

1

It looks like you are trying to iterate over your results. You could do it with a loop, something like this

<?php
    //Create Connection
    $conn = new mysqli("localhost", "root", "", "mydb");
    if (!$conn) {
        die("Connection failed: " . mysqli_connect_error());
    }

    $sql = "SELECT prmo_itemDescription as pname FROM promos";
    if ($promos = $conn->query($sql)) {
        $index = 1;
        while ($promo = $promos->fetch_assoc()) {
?>
    <div class="col-md-3">
        <p><?php echo $promo['pname']; ?></p>
        <button type="button" class="btn btn-success" data-toggle="modal" data-target="#details-<?php echo $index; ?>">Buy Now!</button>
    </div>
<?php
        $index += 1;
        } //END WHILE

        $promos->free();

    } //END IF

    $conn->close();
?>

Instead of index you could use promo id and use it on your html.

answered Oct 1, 2018 at 10:59

Gonzalo

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2 Comments

Nyotnoti Over a year ago

at first I didn't understand what you did then I read it carefully. Not only you accurately solved my problem you also shortened the code, appreciate it.

Gonzalo Over a year ago

The idea is to use the power of programming to don't repeat code, today you may need just 3 promos but tomorrow you could have 5 or much more, with a loop you don't have to worry about it.

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