Using a function to query a database

Question

I have a page and I would really like some help / advice to create a better way / function to call in information from another table.

At the moment, my code looks like: [I do know this is deprecated SQL and would really like to do some nice SQLi for this.]

<?
$menuid = "100";
$imageid = "50";

// ** Talk to 'imagedirectory' table

mysql_select_db($database_db, $BASEdb);
$query_displayimage = "SELECT * FROM imagedirectory WHERE menuid = ".$menuid." AND imageid = ".$imageid."";
$displayimage = mysql_query($query_displayimage, $BASEdb) or die(mysql_error());
$row_displayimage= mysql_fetch_assoc($displayimage);

?>

<img src="/images/assets/<?php echo $menuid; ?>-<?php echo $imageid; ?>-<?php echo $row_displayimage['urlslug']; ?>.jpg" alt="<?php echo $row_displayimage['alttext']; ?>" />

I figure There really has to be a better way because if there is 10 images on a page, this is pretty intense way of doing it.

What do you intend to gain from this question? When you already know that mysql_*() functions are deprecated and using MySQLi is the way to go, you got everything you need to educate yourself. For example here: php.net/mysqli — Marcello Mönkemeyer
– Marcello Mönkemeyer, Commented Nov 12, 2014 at 15:32
hi @icefyre - I appreciate your comments, I am sure I will be to Google and replicate the above code in SQLi without too much trouble I'd imagine. The thing I don't want to do is actually replicate that code. There has to be a better way to do this via a function of some sort, though I don't really know where to begin. What I am after is some help with the logic. — Brandrally
– Brandrally, Commented Nov 12, 2014 at 15:42
I understand. However, you only provided a code without any information about the logic. What is the purpose of your script? — Marcello Mönkemeyer
– Marcello Mönkemeyer, Commented Nov 12, 2014 at 15:45
I might adjust my question a little to be more specific. Though in essence I go through the above code to get the 'urlslug' and 'alttext' fields from a 'imagedirectory' table to display an image. The problem is that if I have 10 images on a page I literally have to repeat the above code 10x, I would like to have it so I have a function where I can easily tell the request the 'menuid' and 'imageid' and it spits out a result: <img src="/images/assets/ $menuid-$imageid-$urlslug.jpg" alt="$alttext" /> — Brandrally
– Brandrally, Commented Nov 12, 2014 at 15:58

Community · Accepted Answer · 2020-06-20 09:12:55Z

Since you seem to know that mysql_* is deprecated, I am assuming you have read up on, and are using mysqli_* instead.

You needn't query the database every time. mysqli_query() returns a mysqli_result, which you can iterate over, and read using functions like mysqli_fetch_assoc(). Here is one way of doing it:

<?php

// store your query in a variable.
$query_displayimage = "SELECT * FROM imagedirectory";

// query the database.
$displayimage = mysqli_query($query_displayimage, $BASEdb);

// check for errors.
$dbErrors = mysqli_error($BASEdb);
if (count($dbErrors))
{
    print_r($dbErrors);
    die();
}

// iterate over the returned resource.
while ($row_displayimage = mysql_fetch_assoc($displayimage))
{
    echo '<img src="/images/assets/' . $menuid . '-' . $imageid . '-' . $row_displayimage['urlslug'] . '.jpg" alt="' . $row_displayimage['alttext'] . '" />';
}

?>

Hope that helped.

EDIT:

You can use this code in a function, too. For example:

<?php

function printImage($menuid, $imageid)
{
    $query_displayimage = "SELECT * FROM imagedirectory";
    $displayimage = mysqli_query($query_displayimage, $BASEdb);
    $dbErrors = mysqli_error($BASEdb);
    if (count($dbErrors))
    {
        print_r($dbErrors);
        die();
    }
    
    if ($row_displayimage = mysql_fetch_assoc($displayimage))
    {
        echo '<img src="/images/assets/' . $menuid . '-' . $imageid . '-' . $row_displayimage['urlslug'] . '.jpg" alt="' . $row_displayimage['alttext'] . '" />';
    }
    else    // if there is a problem getting the image
    {
        echo 'Error getting image.';
    }
}

?>

and elsewhere in your HTML, you would do something like:

<div>
    And here is an image!
    <?php printImage(20, 50); ?>
</div>

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