Barand

Received and loaded. Thanks. For each timesheet record you should have up to 7 (1 for each day of the week) records in stafftime. These stafftime records for the week should have a sheet_id value which matches the timesheet id of the parent timesheet record. You stafftime.sheet_id values match none of the timesheet ids.

Not what I asked for. There is no way that enables me to run your query on your data. I wanted something similar to what I posted in your previous topic...

Do you want to post a dump of your data so I can run your query at my end?

Did you read the bit in the JOIN tutorial about LEFT JOIN to a table (user in your case) and then referencing a column from that table in the WHERE clause?

Follow the link in my sig. There's a section in the tutorials dedicated to JOINS

Put this line immediately before your line that creates your mysqli connection mysqli_report(MYSQLI_REPORT_ERROR|MYSQLI_REPORT_STRICT); That will force mysqli to report any erors it encounters.

As @gizmola said, the first problem is missing style= second) You are using obsolete html markup 3) you have specified a couple of classes but not put them inside the class="w3-container" 4) Again inline style without the style= 5) You have broken up the style attributes with a couple of rogue quotes

If you know the cat_id, can't you then determine the category? All you should need to pass is the cat_id, which should be the value in your options.

Why? What's the difference between your category and cat_title?

Looking at the code you posted it looks you have decided to ignore the rules of syntax for inline style definitions and class specifications. Plus a bit of obsolete html markup thrown in for good measure

Haven't we been here before?

Here's my attempt SELECT sale_date , sum(cashpaid) + sum(cashpartpaid) + sum(codpaid) as total_cash_sales , sum(cashdue) + sum(cashpartdue) + sum(carddue) + sum(coddue) as total_credit_sales , sum(codpaid) + sum(coddue) as total_cod_sales , sum(cardpaid) as total_card_sales FROM ( SELECT bad.basket_id , bad.amount_due as due , SUM(p.amount_paid) as paid , p.payment_method_id , b.payment_status , DATE(bad.sale_time) as sale_date , CASE WHEN payment_method_id = 1 AND payment_status = 'paid' THEN sum(p.amount_paid) ELSE 0 END as cashpaid , CASE WHEN payment_method_id = 1 AND payment_status = 'due' THEN bad.amount_due ELSE 0 END as cashdue , CASE WHEN payment_method_id = 1 AND payment_status = 'partial' THEN SUM(p.amount_paid) ELSE 0 END as cashpartpaid , CASE WHEN payment_method_id = 1 AND payment_status = 'partial' THEN bad.amount_due - SUM(p.amount_paid) ELSE 0 END as cashpartdue , CASE WHEN payment_method_id = 2 THEN SUM(p.amount_paid) ELSE 0 END as cardpaid , CASE WHEN payment_method_id = 2 THEN bad.amount_due - SUM(p.amount_paid) ELSE 0 END as carddue , CASE WHEN payment_method_id = 4 AND payment_status <> 'due' THEN SUM(p.amount_paid) ELSE 0 END as codpaid , CASE WHEN payment_method_id = 4 AND payment_status = 'due' THEN bad.amount_due ELSE 0 END as coddue FROM basket_amount_due bad JOIN basket b USING (basket_id) LEFT JOIN basket_payment p USING (basket_id) GROUP BY basket_id, payment_method_id ) detail GROUP BY sale_date;

Or you could do at the source when you query your data. SELECT Firstname , Lastname , Coursename , CASE WHEN DateBooked IS NULL THEN '' ELSE DATE_FORMAT(DateBooked, '%d-%m-%Y') END as DateBooked, , DATE_FORMAT(ExpiryDate, '%d-%m-%Y') as ExpirtyDate , Notes FROM ....

I think you'll find they were mentioned a little earlier than that ... ... and I doubt that is the earliest.

Your payment status values are "Paid/Due/Partial" Do we have a couple of new Schrodinger cats in your latest calculation requirements...

Is it correct that some combinations of method/status should be included in more than one total?

If I were showing totals by "Payment method" I would find it helpful to show the method descriptions in the row output. The main question - is that correct and what you wanted?

Given the adage "Test data is that data for which the code works", here is the data I used for the queries I posted

None of those have a value of both 1 AND 2. That would mean, like Schrodinger's cat, they have two values at the same time, Some, though, do have a value of 1 OR 2

Firstly, can you show me the record where the "payment_method_id" is both 1 and 2 ? You to need to change around the syntax of that first CASE statement , CASE WHEN p.payment_method_id IN (1, 2) THEN sum(amount_due) ELSE 0 END AS total_cash_sales = where date on or after today but before tomorrow. Isn't that just WHERE DATE(bv.sale_time) = CURDATE() Lastly, are you sure you want to GROUP BY sale_time? That gives you a total for each second, which is probably individual sales - not much of a summary.

My money's on "Syntax error". (Error 4)

This fails $j = "{'admin': 1, 'moderator': 1}" ; $a = json_decode($j, 1); echo '<pre> a ' . print_r($a, 1) . '</pre>'; This works $j = '{"admin": 1, "moderator": 1}' ; $b = json_decode($j, 1); echo '<pre> b ' . print_r($b, 1) . '</pre>'; Note the quotes in the JSON string.

Doesn't your console have a "preserve" option? Or, add "return false" to the end of your submitData() function to stop the page refreshing

How do you define "does not work"? Because when I use your form and the first function, my console shows the data entered in the form

$expression2 = [ "type" => "add", 'children' => [ [ "type" => "number", "value" => 100 // $expression2['children'][0]['value'] ], [ "type" => "multiply", "children" => [ [ "type" => "number", "value" => 2 // $expression2['children'][1]['children'][0]['value'] ], [ "type" => "add", "children" => [ [ "type" => "number", "value" => 5 // $expression2['children'][1]['children'][1]['children'][0]['value'] ], [ "type" => "number", "value" => 45 // $expression2['children'][1]['children'][1]['children'][1]['value'] ] ] ] ] ] ] ]; It's easier if you do a print_r() of the array so you see all the indexes echo '<pre>$expression2 = ' . print_r($expression2, 1) . '</pre>'; then follow the indexes down the tree .... showing the path for echo $expression2['children'][1]['children'][1]['children'][1]['value']; // 45