Code: Select all
$children = mysql_query("SELECT c.id, c.name, c.age, c.photoName, c.panelColor FROM children as c");
$temp = array();
while ($child = mysql_fetch_assoc($children)) {
// Get the child's reatings
$ratings = mysql_query('
SELECT r.behaviourID, t.points, t.typeName
FROM behaviourRatings as r
JOIN behaviourTypes as t
ON r.behaviourID = t.typeID
WHERE r.childID = ' . $child['id']);
// Loop over the ratings
$totalPoints = 0;
while ($childRatings = mysql_fetch_array($ratings)){
$totalPoints = ($totalPoints + $childRatings['points']);
}
// We can only max out at our set max
if(($totalPoints + $maxPoints) > $maxPoints) {
$total = $maxPoints;
} else if($totalPoints < 0){
$total = ($maxPoints + $totalPoints);
}else{
$total = ($maxPoints - $totalPoints);
}
// Set the child array
$temp[] = $child;
}
$response = array();
$response['timestamp'] = $currentmodif;
$response['children'] = $temp;
echo json_encode($response, JSON_PRETTY_PRINT);
Ich habe versucht, $ temp ['Punkte'] = $ Total < /code> zu machen, aber das hat es außerhalb des Arrays und nicht mit den äußeren Schleifendaten gestellt. < /p>
Dies ist das Ergebnis der Funktion: < /p>
{
"timestamp": 1482918104,
"children": [
{
"id": "1",
"name": "Maya",
"age": "5",
"photoName": "maya.png",
"panelColor": ""
},
{
"id": "2",
"name": "Brynlee",
"age": "3",
"photoName": "brynlee.png",
"panelColor": "green"
}
]
}
< /code>
Ich möchte die Punkte für jedes dieser Kinder zeigen, aber ich bin mir nicht sicher, wie ich es dem Teil des Arrays hinzufügen soll. < /p>