php - Using a function to query a database -

php - Using a function to query a database -

i have page , help / advice create improve way / function phone call in info table.

at moment, code looks like: [i know deprecated sql , nice sqli this.]

<? $menuid = "100"; $imageid = "50"; // ** talk 'imagedirectory' table mysql_select_db($database_db, $basedb); $query_displayimage = "select * imagedirectory menuid = ".$menuid." , 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 has improve way because if there 10 images on page, pretty intense way of doing it.

since seem know mysql_* deprecated, assuming have read on, , using mysqli_* instead.

you needn't query database every time. mysqli_query() returns mysqli_result, can iterate over, , read using functions mysqli_fetch_assoc(). here 1 way of doing it:

<?php // store query in variable. $query_displayimage = "select * imagedirectory"; // query database. $displayimage = mysqli_query($query_displayimage, $basedb); // check errors. $dberrors = mysqli_error($basedb); if (count($dberrors)) { print_r($dberrors); die(); } // iterate on 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 helped.

edit:

you can utilize code in function, too. example:

<?php function printimage($menuid, $imageid) { $query_displayimage = "select * 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 problem getting image { echo 'error getting image.'; } } ?>

and elsewhere in html, like:

<div> , here image! <?php printimage(20, 50); ?> </div>

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