Using a function to query a database - php

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.

Answer 1

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>