Parse error: parse error, unexpected $ in /var/www/html/showproducts.php on line 22

coolboarderguy

Hi All,

just testing that the code, so far, is right. The page shows this error,

drew010

its missing one last } at the end to close off the else. generally if you see unexpected $ the dollar sign represents and end of string (file). so if its saying unexpected end of file there is something its looking for that hasnt happened yet, usually no closing braces.

coolboarderguy

Hi All,

thanx. Now I get the following,

Parse error: parse error, unexpected T_ECHO, expecting ')' in /var/www/html/showproducts.php on line 14 after adding the extra }.

drew010

the code above runs fine for me (aside from the db stuff) when i add the last brace. must be some other change you have made since then.

coolboarderguy

Hi All,

no, I made only the 1 change. Now I've changed it some. But get the following error.

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /var/www/html/showproducts.php on line 12

<html>
<head>
<title>Jumbo Status-Show Products</title>
</head>
<body>
<?php
$db = mysql_connect("localhost", "root", "grunger");
mysql_select_db("status", $db);
if (isset($REQUEST[maker_id])){
$maker_id = $REQUEST[maker_id];
$result = mysql_query("SELECT Products.product_name, Products.product_price, Products.product_model FROM Products INNER JOIN Makers ON Products.maker_id = Makers.maker_id WHERE Makers.maker_id = $maker_id");
$num = mysql_num_rows($result); /////Line 12
for ($i=0; $i < $num; $i++){
$myrow=mysql_fetch_array($result);
$product_name = mysql_result($result,$i,"product_name");
$product_price = mysql_result($result, $i, "product_price");
$product_model = mysql_result($result, $i, "product_model");
}
}
else {
if (isset($REQUEST[product_type_id])){
$product_type_id = $REQUEST[product_type_id];
$result = mysql_query("SELECT Products.product_name, Products.product_price, Products.product_model FROM Products INNER JOIN ProductTypes ON Products.product_type_id = ProductTypes.product_type_id WHERE ProductTypes.product_type_id = $product_type_id");
$num = mysql_num_rows($result);
for ($i=0; $i < $num; $i++){
$myrow=mysql_fetch_array($result);
$product_name = mysql_result($result,$i,"product_name");
$product_price = mysql_result($result, $i, "product_price");
$product_model = mysql_result($result, $i, "product_model");
}
}
}
?>
</body>
</html>

Tables:

Products:
product_id
product_name
product_price
product_model
product_type_id
maker_id

ProductTypes:
product_type_id
product_type_detail

Makers:
maker_id
maker_detail

drew010

if it says argument is not a valid mysql resource, it either means the connection to the database was never made successfully, or the query itself failed (usually syntax reasons) and the resource is destroyed. use mysql_error() to check if there is an error after db actions.

i.e.


$conn = mysql_connect("host", "user", "pass");
if (!$conn) {
  echo "no connection because " . mysql_error();
  exit;
}

$res = mysql_query("SHOW TABLES");
if (!$res) {
  echo "Query failed: " . mysql_error();
  exit;
}

//...

coolboarderguy

Hi All,

thanx so much..had a few database errors and syntax in the sql. All is well now. Thanx, again.