Help me with this error please

[romioaa]

I am getting the following error:

Warning: mysql_fetch_array() expects parameter 1 to be resource, string given in C:\wamp\www\accounting\members.php on line 10

This is the code: The connection to the database is working

<?php include_once 'include/mysql.php';?>

<?php
echo "Welcome to the Memebers Area";



$result1 = mysql_query("SELECT * FROM trans, $connection");

while($row1 = mysql_fetch_array('$result1'))
  {
  	echo $row1['product_name'];

  }


mysql_close()

?>

[traq]

don't quote the variable:

mysql_fetch_array('$result1')
// _wrong_, should be:
mysql_fetch_array($result1)

[johanafm]

romioaa;10986063 wrote:

$result1 = mysql_query("SELECT * FROM trans, $connection");

Also, unless $connection contains a table identifier, the you have a SQL syntax error. To test if the query went ok, you should always

$query = "SELECT * FROM trans, $connection"
$result1 = mysql_query($query);

if (!$result)
{
   echo 'An error occured. Please try again later';
   error_log(mysql_error.' ('.mysql_errno().'): ' . $query);
   # possibly: exit;
}
# If you terminate program execution with exit above, then you don't need "else"
else
{
   # if the query was ok...
   $row = mysql_fetch_assoc($result1);

   # etc...
}

And the most likely error is that you should have terminated the string literal before , $connection, as in

$result1 = mysql_query("SELECT * FROM trans", $connection);