Convert MySQL Query into JSON using PHP - php

Question

I have a question about using a MySQL Query to convert my data into a JSON Object. The Query I have is converting to a JSON Object, but it is not working the way I would like.
I have multiple tables in my database that I would like to graph on a chart using the date as the X axis and the values as the Y axis. I am currently joining the tables by date. However, some tables may have multiple submissions per day while others may not have any. Currently, the Query I have is only showing results for dates that data was submitted to all 4 tables.
I would also like to graph the information on a scale of 0-10. Three of the 4 tables only have values from 0-10 so I am taking the average of each value per day. The nutrition table, which holds nf_sugars and nf_total_carbohydrates has larger numbers that I will be using normalization to convert them into a 0-10 scale. For now, I am just attempting to get the SUM per day and will complete the rest of the calculation after this part is working. However, the query I am currently running is giving me results that are much higher than the SUM of the actual numbers in my database.
Any help would be greatly appreciated! Here is the PHP I am currently using to create the JSON Object. As a side note, I did successfully connect to my database, I just did not include that here.
$myquery = "SELECT track_ticseverity.date,
AVG(track_ticseverity.ticnum) as average_ticnum,
track_fatigue.date,
AVG(track_fatigue.fatiguenum) as average_fatiguenum,
track_stress.date,
AVG(track_stress.stressnum) as average_stressnum,
track_nutrition.date,
((SUM(track_nutrition.nf_sugars) ) ) as sum_nf_sugars,
((SUM(track_nutrition.nf_total_carbohydrate) ) ) as sum_nf_total_carbohydrate
FROM track_ticseverity
INNER JOIN track_fatigue
ON track_ticseverity.date=track_fatigue.date
INNER JOIN track_stress
ON track_fatigue.date=track_stress.date
INNER JOIN track_nutrition
ON track_stress.date=track_nutrition.date
WHERE track_ticseverity.user_id=1
AND track_fatigue.user_id=1
AND track_stress.user_id=1
AND track_nutrition.user_id=1
GROUP BY track_ticseverity.date";
$query = mysqli_query($conn, $myquery);
if ( ! $query ) {
echo mysqli_error(s);
die;
}
$data = array();
for ($x = 0; $x < mysqli_num_rows($query); $x++) {
$data[] = mysqli_fetch_assoc($query);
}
echo json_encode($data);
mysqli_close($conn);
EDIT - The Query is successfully returning a JSON object. My issue is that the query I wrote does not output the data in the correct way. I need the query to select information from multiple tables, some with multiple submission per day and others with only one or no submissions.
EDIT2 - I am thinking another way to handle this is to combine multiple SELECT statements into a single JSON Object, but I am not sure how to do this.

Answer 1

The sum is larger than expected because of the joins. Imagine that a certain date
occurs in one track_nutrition record and two track_fatigue records, then the join
will make that the data from the first table is once combined with the first track_fatigue
record, and then again with the second record. Thus the same nf_sugars
value will be counted twice in the sum. This behaviour will also affect the averages.
You should therefore first perform the aggregations, and only then perform the joins.
Secondly, to ensure you catch all data, even if for a certain date not all tables have
values, you should use full outer joins. This will guarantee that each record in each table
will find its way in the result. Now, MySQL does not support such full outer joins, so
I use an extra sub-select to select all different dates from the 4 tables and then
"left join" them with the other aggregated data:
SELECT dates.date,
IFNULL(average_ticnum_n, 0) as average_ticnum
IFNULL(average_fatiguenum_n, 0) as average_fatiguenum
IFNULL(average_stressnum_n, 0) as average_stressnum
IFNULL(sum_nf_sugars_n, 0) as sum_nf_sugars
IFNULL(sum_nf_total_carbohydrate_n, 0) as sum_nf_total_carbohydrate
FROM (
SELECT DISTINCT user_id,
date
FROM (
SELECT user_id,
date
FROM track_ticseverity
UNION
SELECT user_id,
date
FROM track_fatigue
UNION
SELECT user_id,
date
FROM track_stress
UNION
SELECT user_id,
date
FROM track_nutrition
) as combined
) as dates
LEFT JOIN (
SELECT user_id,
date,
AVG(ticnum) as average_ticnum_n
FROM track_ticseverity
GROUP BY user_id,
date) as grp_ticseverity
ON dates.date = grp_ticseverity.date
AND dates.user_id = grp_ticseverity.user_id
LEFT JOIN (
SELECT user_id,
date,
AVG(fatiguenum) as average_fatiguenum_n
FROM track_fatigue
GROUP BY user_id,
date) as grp_fatigue
ON dates.date = grp_fatigue.date
AND dates.user_id = grp_fatigue.user_id
LEFT JOIN (
SELECT user_id,
date,
AVG(stressnum) as average_stressnum_n
FROM track_stress
GROUP BY user_id,
date) as grp_stress
ON dates.date = grp_stress.date
AND dates.user_id = grp_stress.user_id
LEFT JOIN (
SELECT user_id,
date,
SUM(nf_sugars) as sum_nf_sugars_n,
SUM(nf_total_carbohydrate) as sum_nf_total_carbohydrate_n
FROM track_nutrition
GROUP BY user_id,
date) as grp_nutrition
ON dates.date = grp_nutrition.date
AND dates.user_id = grp_nutrition.user_id
WHERE dates.user_id = 1
ORDER BY dates.date;
Note that you will get 0 values in some of the columns when there is no data for that
particular date. If you prefer to get NULL instead, remove the Nvl() from those columns
in the query above.
Then, to normalize all data on a 0 - 10 scale, you could look at the maximum
found for each type of value and use that for a conversion, or if you know beforehand
what the ranges are per type, then it is probably better to use that information, and
maybe code that in the SQL as well.
However, it always looks a bit odd to have values combined in a graph that actually use
different scales. One might easily jump to wrong conclusions with such graphs.

Answer 2

I will prefer to use this way (assuming all other things are working fine e.g query is working fine)
$query = mysqli_query($conn, $myquery);
if ( ! $query ) {
echo mysqli_error($conn); //You need to put $conn here to display error.
die;
}
$data = array();
while($row = mysqli_fetch_assoc($query)) {
$data[] = $row;
}
echo json_encode($data);
mysqli_close($conn);

Answer 3

if you are using PDO (PHP Data Objects) for database operations; then following code can be used to create the JSON from PDO.
$array = $pdo->query("SELECT * FROM employee")->fetchAll(PDO::FETCH_ASSOC);
echo json_encode($array);
For Multiple Select we need to try it like that way :
$array = $pdo->query("SELECT 1; SELECT 2;");
$array->nextRowset();
var_dump( $array->fetchAll(PDO::FETCH_ASSOC) );