Select all customers and each customer's total orders - php

Question

For each customer, I want to return: id, name, address, city, cap, country and total of orders. Table names are: customers (10 records) and orders (20 records)
I've tried this code, but it doesn't works
<?php
$conn = mysqli_connect('127.0.0.1','...','...','...') or die("Connection failed: " . $conn->connect_error);
$select = mysqli_query($conn,"SELECT customers.id,customers.name,customers.address,customers.city,customers.cap,customers.country COUNT(customerid) as TotalOrders FROM customers LEFT JOIN orders GROUP BY customer.id");
while($row = mysqli_fetch_array($select,MYSQLI_ASSOC))
{
$tb = <<<table
...{$row['totalorders']}...
table;
echo $tb;
}
mysqli_free_result($select);
echo $num_record = mysqli_affected_rows($conn);
$conn->close();
?>

Answer 1

So something like...
<?php
include('../path/to/connection/stateme.nts');
$query =
"
SELECT c.id
, c.name
, c.address
, c.city
, c.cap
, c.country
COUNT(o.customerid) TotalOrders
FROM customers c
LEFT
JOIN orders o
ON o.customerid = c.id
GROUP
BY c.id;
";
$result = mysqli_query($db,$query); ** where $db is something like mysqli_connect("myhost", "myusername", "mypassword");
while($row = mysqli_fetch_assoc($result))
{
$tb = <<<table
...{$row['totalorders']}...
table;
echo $tb;
}
mysqli_free_result($result);
echo $num_record = mysqli_affected_rows($db);
$db->close();
?>
(I'm no php coder)

Answer 2

You have a comma missing -
customers.country COUNT(customerid)
-----------------^
and you have left out an ON condition -
FROM customers LEFT JOIN orders //ON would be here
One thing you should always do is to create and run your queries against your database before you put those queries in your code. And always, always include error checking.