shows warning that suplied argument is not a valid

Question

while performing inner join when i add data in mysql it shows a error mysql_fetch_array is not a valid argument. code is below:

    echo $sql="SELECT * FROM info
 INNER JOIN item ON info.info_id=item.info_id where item.info_id=".$info_id;
        $query=mysql_query($sql);
         while($result=mysql_fetch_array($query))
        { 
             echo "<tr>";
             echo "<td>" .$result['Name']. "</td>";
        echo "<td>" .$result['Item']. "</td>";
         echo "<td>" .$result['Price']. "</td>";
        echo "<td>" .$result['info_id']. "</td>";
             echo "<td><a href='item.php?act=edit&id=".$result['id']."'>Edit</a></td>";
             echo "<td><a href='item.php?act=delete&id=".$result['id']."'>Delete</a></td>";
             echo "<td><a href='item.php?act=item&id=".$result['id']."'>Item</a></td>";
echo "</tr>";

while running the above code it shows the following warning:

warning:mysql_fetch_array(): supplied argument is not a valid MySQL result resourc

Check your query that returns anything. I think it returns false — Edwin Alex, Apr 08 '13 at 10:39
You are using [an **obsolete** database API](http://stackoverflow.com/q/12859942/19068) and should use a [modern replacement](http://php.net/manual/en/mysqlinfo.api.choosing.php). You are probably also **vulnerable to [SQL injection attacks](http://bobby-tables.com/)** that a modern API would make it easier to [defend](http://stackoverflow.com/questions/60174/best-way-to-prevent-sql-injection-in-php) yourself from. — Quentin, Apr 08 '13 at 10:40
Please share the creating connection code, are you selecting the database? One thing important is `mysql` api is deprecated use `mysqli` instead. — Muhammad Ummar, Apr 08 '13 at 10:40
check `echo mysql_num_rows($query);` before ur `fetch` statement — Suhel Meman, Apr 08 '13 at 10:41
@Quentin is right, you should probably not be using mysql_query as it is deprecated ([starting from PHP 5.5.0](http://php.net/manual/en/function.mysql-query.php)) — Erik Schierboom, Apr 08 '13 at 10:46
try checking error `$query=mysql_query($sql) or die(mysql_error())`..I think it should return you with some thing useful..... and I suggest you start using mysqli_* functions — alwaysLearn, Apr 08 '13 at 10:47
@Ummar i have created connection with the database, and the connection is successfully established. **mysql_connect('localhost','root','root') or die('oops error in connection'); mysql_select_db('friend_list') or die('error in selecting db');** — user39891, Apr 09 '13 at 10:06

score 0 · Answer 1 · answered Apr 08 '13 at 10:48

0

The mysql_query method probably fails silently. If it does, it returns FALSE so you should be doing this:

$query=mysql_query($sql);
if (!$query) {
    die('MSSQL error: ' . mssql_get_last_message());
}

answered Apr 08 '13 at 10:48

Erik Schierboom

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shows warning that suplied argument is not a valid

1 Answers1