Convert mysql_result to mysqli
I am not a programmer and I know very little PHP, but I am trying to fix a code in the osCommerce store that gives the following error on PHP 5.4:
mysql_result (): The supplied argument is not a valid MySQL result resource
This is the code:
$products = $cart->get_products();
for ($i=0, $n=sizeof($products); $i<$n; $i++) {
$id_produto = (INT)$products[$i]['id'];
$sql = tep_db_query("SELECT p.manufacturers_id,m.manufacturers_cep,m.manufacturers_name FROM products p
LEFT JOIN manufacturers m ON m.manufacturers_id = p.manufacturers_id
WHERE p.products_id = '$id_produto'")OR DIE(mysql_error());
$id_fabricante = mysql_result($sql,'0','manufacturers_id');
$cep_fabricante = mysql_result($sql,'0','manufacturers_cep');
$nome_fabricante = mysql_result($sql,'0','manufacturers_name');
$id_fabricantes[$id_fabricante]['peso'] += $products[$i]['quantity']*$products[$i]['weight'];
$id_fabricantes[$id_fabricante]['cep'] = $cep_fabricante;
$id_fabricantes[$id_fabricante]['nome'] = $nome_fabricante;
}
I tried to change it and there are no more errors, but it still doesn't work. Is this the correct way to do it?
$products = $cart->get_products();
for ($i=0, $n=sizeof($products); $i<$n; $i++) {
$id_produto = (INT)$products[$i]['id'];
$sql = tep_db_query("SELECT p.manufacturers_id,m.manufacturers_cep,m.manufacturers_name FROM products p
LEFT JOIN manufacturers m ON m.manufacturers_id = p.manufacturers_id
WHERE p.products_id = '$id_produto'")OR DIE(mysql_error());
$row = mysqli_fetch_assoc($sql);
$id_fabricante = $row['manufacturers_id'];
$row = mysqli_fetch_assoc($sql);
$cep_fabricante = $row['manufacturers_cep'];
$row = mysqli_fetch_assoc($sql);
$nome_fabricante = $row['manufacturers_name'];
$id_fabricantes[$id_fabricante]['peso'] += $products[$i]['quantity']*$products[$i]['weight'];
$id_fabricantes[$id_fabricante]['cep'] = $cep_fabricante;
$id_fabricantes[$id_fabricante]['nome'] = $nome_fabricante;
}
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No, this is not true. If you check manual , you can see that the second parameter is the string to retrieve in your result set. In the original example, you only retrieve data from the first row - 0
- and nothing else.
In your mysqli code, you select a new row before each assignment so that the data is a combination of different field values from different rows.
The correct way would look something like this:
// fetch the first row in the result set
$row = mysqli_fetch_assoc($sql);
$id_fabricante = $row['manufacturers_id'];
$cep_fabricante = $row['manufacturers_cep'];
$nome_fabricante = $row['manufacturers_name'];
Also, you will need to add error handling to make sure there is a line.
You should also avoid running sql queries in a loop. You could probably get all the rows in 1 query using for example mysql IN
, after which you can iterate over that result set.
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