return 0 if a value is null - php

Question

I have a MySQL query in my php script, which is working fine currently. The only issue I have is that some of the columns return null values.
If there is data for those columns then they return the value, but if there is no data or records for the date then they return null. All I want to do is modify this query so that if anything is null it returns '0'.
I'm not sure if I should use IFNULL or coalesce but either way I'm unfamiliar with the best way to apply it to this query.
Any help is much appreciated.
$stmt3 = mysqli_prepare($conn2,
"UPDATE ambition.ambition_totals a
INNER JOIN
(SELECT
c.user AS UserID,
COUNT(*) AS dealers,
ROUND((al.NumberOfDealers / al.NumberOfDealerContacts) * 100 ,2) AS percent
FROM jfi_dealers.contact_events c
JOIN jackson_id.users u
ON c.user = u.id
JOIN jfi_dealers.dealers d
ON c.dealer_num = d.dealer_num
LEFT JOIN (
SELECT user_id, COUNT(*) AS NumberOfDealerContacts,
SUM(CASE WHEN ( d.next_call_date + INTERVAL 7 DAY) THEN 1 ELSE 0 END) AS NumberOfDealers
FROM jackson_id.attr_list AS al
JOIN jfi_dealers.dealers AS d ON d.csr = al.data
WHERE al.attr_id = 14
GROUP BY user_id) AS al
ON al.user_id = c.user
GROUP BY UserID) as cu
on cu.UserID = a.ext_id
SET a.dealers_contacted = cu.dealers,
a.percent_up_to_date = cu.percent;
") or die(mysqli_error($conn2));
UPDATE
Version with IFNULL statment:
UPDATE ambition.ambition_totals a
INNER JOIN
(SELECT
c.user AS UserID,
ifnull(count(*),0) AS dealers,
ifnull(ROUND((al.NumberOfDealers / al.NumberOfDealerContacts) * 100 ,2),0) AS percent
FROM jfi_dealers.contact_events c
JOIN jackson_id.users u
ON c.user = u.id
JOIN jfi_dealers.dealers d
ON c.dealer_num = d.dealer_num
LEFT JOIN (
SELECT user_id, COUNT(*) AS NumberOfDealerContacts,
SUM(CASE WHEN ( d.next_call_date + INTERVAL 7 DAY) THEN 1 ELSE 0 END) AS NumberOfDealers
FROM jackson_id.attr_list AS al
JOIN jfi_dealers.dealers AS d ON d.csr = al.data
WHERE al.attr_id = 14
GROUP BY user_id) AS al
ON al.user_id = c.user
WHERE c.created_at >= CURDATE()
GROUP BY UserID) as cu
on cu.UserID = a.ext_id
SET a.dealers_contacted = cu.dealers,
a.percent_up_to_date = cu.percent;

Answer 1

Yes, you can use IFNULL for this. But be very sure that you actually want this behaviour. PHP is also familiar with NULL values and can handle them just fine. 0 has a very different meaning. But if you do want this behaviour, simply wrap the field or statement that might return null in a IFNULL, for example:
SELECT IFNULL(user_id, 0);
But you can also do this in PHP itself, so you do not have to modify the query:
if (is_null($result['field'])) {
echo 0;
}
Or, if you use PHP 7+, you can also use the null coalescing operator:
echo $result['field'] ?? 0;