Thread: php/mySQL help
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Old 04-28-2005, 08:10 PM   #1
raublekick
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php/mySQL help

hey dudes and dudettes, hopefully someone can help me:

here is where i am stuck now:
i have three sets of queries that insert into seperate tables and then try to view the data that was just inserted. the insertions all work, but only the second SELECT query works. they just don't display the information even though it is in the database. here they are:

not working:
$query = 'INSERT INTO Review (ReviewTitle, ReviewType, Rating, ArtistID, AlbumID, Text, Link, Picture, AuthorID, Date)
VALUES ("'.$reviewtitle.'", "'.$reviewtype.'", "'.$rating.'", "'.$muartist.'", "'.$mualbum.'", "'.$reviewtext.'", "'.$date.'", "'.$url.'", "'.$imageurl.'", "'.$authorname.'")';
$result = mysql_query($query) or die("Query failed: ".mysql_error().$query);
echo "<tr><td><p><strong>Added into Review table<strong></p></td></tr>:";
$query = 'SELECT ReviewTitle, ReviewType, Rating, ArtistID, AlbumID, Text, Link, Picture, AuthorID, Date
FROM Review
WHERE (ArtistID = "'.$muartist.'") AND (AlbumID = "'.$mualbum.'") AND (AuthorID = "'.$authorname.'")';
$result = mysql_query($query) or die("Query failed: ".mysql_error().$query);


working:
$query = 'INSERT INTO Artist (ArtistID, AlbumID) VALUES ("'.$muartist.'", "'.$mualbum.'")';
$result = mysql_query($query) or die("Query failed: ".mysql_error().$query);
echo "<tr><td><p><strong>Added into Artist table<strong></p></td></tr>:";
$query = 'SELECT ArtistID, AlbumID
FROM Artist
WHERE (ArtistID = "'.$muartist.'") AND (AlbumID = "'.$mualbum.'")';
$result = mysql_query($query) or die("Query failed: ".mysql_error().$query);


not working:
$query = 'INSERT INTO Album (AlbumID, ReleaseYear) VALUES ("'.$mualbum.'", "'.$muyear.'")';
$result = mysql_query($query) or die("Query failed: ".mysql_error().$query);
echo "<tr><td><p><strong>Added into Album table<strong></p></td></tr>:</p>:";
$query = 'SELECT AlbumID, ReleaseYear
FROM Album
WHERE (AlbumID = "'.$mualbum.'") AND (ReleaseYear = "'.$muyear.'")';
$result = mysql_query($query) or die("Query failed: ".mysql_error().$query);
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