Merge results of single JSON output with PHP - php

Question

I have a JSON array as per below:
{
"aaData": [
{
"Date_time": "23",
"traffic": "22",
"direction": "sent"
},
{
"Date_time": "24",
"traffic": "55",
"direction": "sent"
},
{
"Date_time": "25",
"traffic": "60",
"direction": "sent"
},
{
"Date_time": "26",
"traffic": "43",
"direction": "sent"
},
{
"Date_time": "27",
"traffic": "50",
"direction": "sent"
},
{
"Date_time": "23",
"traffic": "50",
"direction": "received"
},
{
"Date_time": "24",
"traffic": "42",
"direction": "received"
},
{
"Date_time": "25",
"traffic": "52",
"direction": "received"
},
{
"Date_time": "26",
"traffic": "47",
"direction": "received"
},
{
"Date_time": "27",
"traffic": "36",
"direction": "received"
}
]
}
What I'd like to do with it is combine all the results with the same date into a single entry - so for date_time 23 I want it to appear like this
"Date_time": "23",
"traffic-sent": "22",
"traffic-received": "50"
I'd like to do this with PHP if possible? The data is coming from two separate mySQL queries, coming from to different mySQL databases. I've tried combining the output of the query to do what I need (tried Joins and Unions) but can't get past the separation of the results as per my first example.
The part of the SQL query creating the JSON looks like this:
while($row = mysqli_fetch_assoc($result)) {
$model[$i]['Date_time'] = $row['the_day'];
$model[$i]['traffic'] = $row['traffic'];
$model[$i]['direction'] = $row['TABLE_NAME'];
$i++;
}
And the SQL looks like this:
(SELECT
DAY(`Time`) AS the_day,
count(accounts.accName) AS traffic,
"sent" AS TABLE_NAME
FROM
bss.ss_sent LEFT JOIN bss.accounts ON ss_sent.Customer = accounts.accName
WHERE
YEARWEEK(`Time`) = YEARWEEK(CURRENT_DATE)
AND
Customer != " "
AND
accShortName = "QRR"
GROUP BY
the_day)
UNION
(SELECT
DAY(Date_time) AS the_day,
count(AS_Task) AS traffic,
"received" AS TABLE_NAME
FROM
im_stats.as_counter
WHERE
AS_Task = "QRR3 Incoming"
AND
YEARWEEK(Date_time) = YEARWEEK(CURRENT_DATE)
GROUP BY
the_day
Order by the_day)
IF anyone can advise of a way to combine the results I'd very much appreciate it.
UPDATE:
This is how I've entered Populus's code:
$i = 0;
while($row = mysqli_fetch_assoc($result)) {
$model[$i]['Date_time'] = $row['the_day'];
$model[$i]['traffic'] = $row['traffic'];
$model[$i]['direction'] = $row['TABLE_NAME'];
$i++;
}
$combined = array();
foreach ($model as $val) {
$date_time = $val['Date_time'];
if (!isset($combined[$date_time)) {
$combined[$date_time] = array(
'Date_time' => $date_time,
'traffic_sent' => 0,
'traffic_received' => 0,
);
}
if ('received' == $val['direction']) {
$combined[$date_time]['traffic_received'] += $val['traffic'];
} else {
$combined[$date_time]['traffic_sent'] += $val['traffic'];
}
}
header('Content-type: application/json');
print json_encode(array('aaData' => $combined), JSON_PRETTY_PRINT);

Answer 1

This could probably be done using SQL (which you haven't provided), but if you really want PHP:
$combined = array();
foreach ($model as $val) {
$date_time = $val['Date_time'];
if (!isset($combined[$date_time])) {
$combined[$date_time] = array(
'Date_time' => $date_time,
'traffic_sent' => 0,
'traffic_received' => 0,
);
}
if ('received' == $val['direction']) {
$combined[$date_time]['traffic_received'] += $val['traffic'];
} else {
$combined[$date_time]['traffic_sent'] += $val['traffic'];
}
}
Your desired array is now in $combined. If you don't want the keys, you can remove it:
$result = array_values($combined);

Answer 2

Try it this way:
while($row = mysqli_fetch_assoc($result)) {
if ($row['direction'] == 'sent')
$dt[$row['Date_time']]['traffic-sent'] += $row['traffic'];
elseif ($row['direction'] == 'recieved')
$dt[$row['Date_time']]['traffic-recieved'] += $row['traffic'];
}
foreach ($dt as $date) {
echo "Date_time: " . key($date) . ",<br/>" .
"traffic_sent: " . $date['traffic-sent'] . ",<br/>" .
"traffic-recieved: " . $date['traffic-recieved'] . "<br/><br/>";
}