MySql_fetch_array() error in PHP

[dslade]

I got the following message when trying to do a mysql php query:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\Program Files\Apache Group\Apa....

Can anyone help offer suggestions?

Here is the code:
$query = "SELECT title, role_name, CONCAT(first_name, ' ', last_name) AS name FROM actors, roles, movies WHERE actors.act_id = roles.act_id AND roles.movie_id = movies.movie_id AND year > 1999 ORDER BY year;";

$result = @mysql_query ($query);
// Print the results
echo '<table><tr><td align="left"><b>Movie Title </b></td><td align="left"><b>Role Name </b></td><td align="left"><b>Actor</b></td></tr>';
while ($row = mysql_fetch_array
($result, MYSQL_ASSOC))

[ak1dnar]

Next time please enclose your php lines with PHP tags :)

Since i don't have your Table structure i didn't check the SQL Query.
If this is not working post the SQL script with some sample values.
then i can check it in my end.

[PHP]<?php
$query = "SELECT title, role_name, CONCAT(first_name, ' ', last_name) AS name FROM actors, roles, movies WHERE actors.act_id = roles.act_id AND roles.movie_id = movies.movie_id AND year > 1999 ORDER BY year;";

$result = mysql_query($query);
// Print the results
echo '<table><tr><td align="left"><b>Movie Title </b></td><td align="left"><b>Role Name </b></td><td align="left"><b>Actor</b></td></tr>';
while ($row = mysql_fetch_array($result))
{
echo '<tr>
<td align="left">'.$row['title'].'</td><td align="left">'.$row['role_name'].'</td><td align="left">'.$row['name'].'</td>
</tr>';
}
echo '<table>';
?>[/PHP]

[code green]

Get rid of the idiotic '@' symbol then you might see why it is failing. As the error occurs here

$result = @mysql_query ($query);

. Your query is wrong. We don't know why without your table

[ronverdonk]

You suffer from the same shortcoming many developers have: you must always check the results of any database or file related command!

In this case the problem is that your query returns an error, so you must capture that error before you continue. Like this:

$result = mysql_query ($query) 
    or die("Query error: " . mysql_error());

Ronald :cool: