passing a $variable; then calling it's row from database

I have built a dropdown menu pulling all the info found under the "product" column from my table, myproducts ( I am using MySQL for database).

"On change", I am posting the form, which passes the variable $product to another .php page. This is successful as my new .php page is recognizing that the variable $product = "Cool Product" ("Cool Product" being the line chosen in dropdown menu).

I used

echo "$product";

to test that the variable is there and has a value of "Cool Product"

My problem is:

How do I query all the other information associated with just that product and only that product?

ie: how do I call up the row for the selected variable $product?

Do a query on the database seeking that product only:

select * from PRODUCTS where NAME='Cool Product'

Put your actual table name in place of PRODUCTS and your actual column name in place of NAME.

I will not know the actual name of the product. (ie: "Cool Product") The dropdown menu was populated automatically from database using php script.

I have named the drop down menu on the first page "product" which passes the variable $product (and it's value) to next page.

That said, should I write:

select * from PRODUCTS where NAME='$product'

I think I have tried that, but maybe my syntax is wrong.

thanks for your help.

this is what produced the results I needed on my second page. I passed my $product_id instead of $product. Not sure if that is what made the difference:

$result = mysql_query("SELECT * FROM products WHERE product_id='$product_id'") or exit;


if ($myrow = mysql_fetch_array($result)); {

do {

printf("<tr><td>%s</td><td>%s</td></tr>\n", $myrow["product_id"], $myrow["product"]);



} while ($myrow = mysql_fetch_array($result));

echo "</table>\n";

mysql_free_result($result);