MYSQL Query Fill Gaps with Data with LEFT JOIN - mysql

Question

I am querying a MYSQL database which has a table named PRICE with the following fields: user, price_date, morning, afternoon. I am grabbing all the morning prices for the current week for the current user to be used as an array in a JS.Charts line chart. The issue I am having is that there may be missing days so the array that I am creating from the query is creating date gaps in the chart.
I created another table named calendar with three fields: datefield, morning, afternoon. I filled it with this years dates (YYYY-MM-DD) along with 0's for the morning and afternoon values.
I now have been trying to write a LEFT JOIN query to select all morning values for the current week and join it with the second table to fill in the date gaps with zeros but I can not get it to work. Any help would be greatly appreciated.
Query:
$sql = "SELECT p.morning, c.morning
FROM price p
LEFT JOIN calendar c ON p.price_date = c.datefield
WHERE p.user = '$user' AND YEARWEEK(p.price_date) = YEARWEEK(NOW())
ORDER BY p.price_date";
The data being used and what is being outputted:
Table Price:
user
price_date
morning
afternoon
lpepper
2021-03-15
23
35
lpepper
2021-03-17
43
52
lpepper
2021-03-18
24
35
lpepper
2021-03-19
78
85
Table Calendar (Partial - goes for whole year)
datefield
morning
afternoon
2021-03-15
0
0
2021-03-16
0
0
2021-03-17
0
0
2021-03-18
0
0
2021-03-19
0
0
2021-03-20
0
0
2021-03-21
0
0
I need the query to return the morning prices for this week (M to SUN) with zeros filled in for the missing dates:
Array should be: 23,0,43,24,78,0,0
My query above returns: ,0,0,0
To make the array I am doing:
$result = mysqli_query($conn, $sql);
$morning = array();
while ($row = mysqli_fetch_assoc($result))
{
$morning[] = $row["morning"];
}
Then when used in the graph:
<?php echo implode(", ", $morning); ?>

Answer 1

In a LEFT JOIN, the table with all the rows that should be in the output should be first.
If you have conditions on the other table, you need to put them in the ON clause. If there's no match to c.datefield, you'll get NULL for the p columns, and testing them in WHERE will filter those rows out.
The date should be filtered from the calendar table, not price. And you should return the date so you know what the rows are for.
Give aliases to p.morning and c.morning so you can distinguish them when getting the columns from the results.
You should order by the column in c, since p.price_date will be NULL for the missing dates.
$sql = "SELECT c.datefield, IFNULL(p.morning, 0) AS price_morning, c.morning AS cal_morning
FROM calendar c
LEFT JOIN price p ON p.price_date = c.datefield
AND p.user = '$user'
WHERE YEARWEEK(c.datefield) = YEARWEEK(NOW())
ORDER BY c.datefield";
DEMO
On an unrelated note, you should use a prepared statement rather than substituting a variable into the SQL.