Using PHP to GroupBy, looping through results and using count - php

Question

I have search SO and found answers where people GROUP BY but haven't come across an answer where it displays the ability to GROUP BY (in this case date) and then get the user with the highest # of entries (in each of the date results).
I have my table set up as follows.
+----+------+-----------+-----------------+---------------------+
| id | name | reference | email | date |
+----+------+-----------+-----------------+---------------------+
| 1 | dan | 56453 | dan#example.com | 2015-05-01 09:00:01 |
| 2 | bob | 34564 | bob#example.com | 2015-05-01 09:21:03 |
| 3 | dan | 08948 | dan#example.com | 2015-05-01 09:30:08 |
| 4 | bob | 43775 | bob#example.com | 2015-05-02 09:01:43 |
| 5 | bob | 67210 | bob#example.com | 2015-05-02 09:04:13 |
| 6 | dan | 22195 | dan#example.com | 2015-05-02 09:09:11 |
+----+------+-----------+-----------------+---------------------+
Each day users log in and input a reference. Each day they could be logging many entries.
First, what I'm trying to do is GROUP BY the days and output the data.
So $day[0] which should equal "2015-05-01", should output those first 3 rows and who (and their count) has the highest number of entries for that day.
So output should be:
+----+------+-----------+-----------------+---------------------+
| Results of 2015-05-01 |
+----+------+-----------+-----------------+---------------------+
| 1 | dan | 56453 | dan#example.com | 2015-05-01 09:00:01 |
| 2 | bob | 34564 | bob#example.com | 2015-05-01 09:21:03 |
| 3 | dan | 08948 | dan#example.com | 2015-05-01 09:30:08 |
+----+------+-----------+-----------------+---------------------+
+----+------+-----------+-----------------+---------------------+
| Top user/s for 2015-05-01 |
+-----+-----+---------------------------------------------------+
| dan | 2 | |
+-----+-----+---------------------------------------------------+
So I need to be able to output the data for each day including the user with the most entries for that particular day.
My understanding is that I'd need to do something like this but can't quite grasp it.
$query = mysqli_query('SELECT * FROM mydata GROUP BY DATE(mydata.date as $date)');
$groupcount = 0;
while ($row = #mysqli_fetch_array($query))
{
$rowcount = 0;
$date[$groupcount] = $date;
$dayquery = mysqli_query('SELECT * FROM mydata WHERE date = $date)');
while ($entry = #mysqli_fetch_array($dayquery))
$data[] = array($row['$rowcount']['name'], $row['$rowcount']['reference'], $row['$rowcount']['email'], $row['$rowcount']['date']);
foreach ($data as $d) {
echo "Name: " . $d[0] . ", Reference: " . $d[1] . ", Email: " . $d[2] . ", Date: " . $d[3];
}
$rowcount++;
}
So something like that but also having another query that will output the person and the sum of the highest entries for that day. I just don't want to add another query as I think there would be a better way of structuring this to include COUNT in another.
Could potentially be multiple people with the same number of entries for the day so I'm trying to accommodate for that.

Answer 1

This should give you what you are looking for all in one query.
SELECT
max(m.id), -- since this column is unique you can ignore it if it's not important, or using min/max to get one of the values
max(m.reference), -- since this column appears to be unique you can ignore it if it's not important, or using min/max to get one of the values
m.name,
m.email,
cast(m.date as date) date,
count(*) count
FROM
mydata m
GROUP BY
m.name,
m.email,
cast(m.date as date)
ORDER BY
m.date DESC,
m.count DESC,
m.name ASC,
m.email ASC;