PHP MySQL：查询疑难解答

Question

I have what I thought was a straight-forward query, but I cant seem to get it to work. All help appreciated!

The Tables:

There are two tables involved: gallery_meta and prm_album . The gallery_meta tbl contains the data for images stored in the file structure. The prm_album tbl simply aligns album ID's with album regular-text Names. For both tbls, the field names used in the query are correct.

The Query:

In the code below, I first get an array of saved album names. I then check the file structure to see which directories actually contain images. At this point I want to query the db again for a thumbnail for each of the albums that return as positives. It is here that the the process fails.

All I get is a whitescreen, no error reporting whatsoever. The array $albums is being filled successfully, and the $id variable is being passed to the $q query. $pattern is simply the file path to the parent directory where the album directories are stored.

            //Get album names from db
            $albums = array(); 
            $q="SELECT ID, Name FROM prm_album WHERE 1";
            $sql=mysql_query($q) or die(mysql_error());
            while($r=mysql_fetch_array($sql)) {
                $album = $r['Name'];
                $album_dir = $pattern.$album.'/';

                //check which albums actually contain images (dividing to account for thumbs, med, lrg.)
                $filecount = count(glob($album_dir.'*'))/3;

                //if images found, put album in array
                if($filecount >= 1) {
                    $albums[ $r['ID'] ] = $filecount;
                }
            }

            foreach($albums as $id => $filecount) {     
                $q="SELECT m.AlbumID, m.FileName, m.FileExt,
                        m.LegacyName, m.IsDefault,
                        p.Name
                        FROM gallery_meta AS m
                            LEFT JOIN prm_album AS p
                                ON m.AlbumID = p.ID

                        WHERE m.AlbumID = $id
                            AND Public = 1
                            ORDER BY IsDefault DESC
                            LIMIT 0,1";

                $sql=mysql_query($q) or die(mysql_error());
                while($r=mysql_fetch_array($sql)) {
                    $album=$r['Name'].'/';
                    $legacyname=$r['LegacyName'];
                    $filename=$r['FileName'].'-sm.';
                    $ext=$r['FileExt'];

                    echo '<img src="'$pattern.$album.$filename.$ext.'" />'.$legacyname.'<br/>';
                }
            }

Answer 1

A couple of points ${id} can be written this way in a string and maybe needs \\"${id}\\" but more likely Put all of the$q string on one line in the PHP code or the parser interpreter could trip over the newline characters in it and certainly the sql will.

$q="SELECT m.AlbumID, m.FileName, m.FileExt, m.LegacyName, m.IsDefault, p.Name FROM gallery_meta AS m LEFT JOIN prm_album AS p ON m.AlbumID = p.ID WHERE m.AlbumID = $id AND Public = 1 ORDER BY IsDefault DESC LIMIT 0,1";

You should be able to test that on a command line into a database first too(bit of wisdom).

Answer 2

Change

echo '<img src="'$pattern.$album.$filename.$ext.'" />'.$legacyname.'<br/>';

to

echo '<img src="'.$pattern.$album.$filename.$ext.'" />'.$legacyname.'<br/>';

The first dot was missing. And always set these lines on top of your script =)

ini_set('display_errors', 1);
error_reporting(E_ALL);