How to format json to what I need? - php

Question

ok, Going to get this out of the way right now. I suck at php. I am building an angular app that is going to populate a mobile app with data from the database. I having it pull from the database just fine but I need the json formatted in a special way and I have no idea how to do it.
Using json_encode this is how it is coming from the database:
[
{
"id":"1",
"date":"2014-10-03",
"time":"2014-10-03 10:45:05",
"amount":"5"
},
{
"id":"2",
"date":"2014-10-03",
"time":"2014-10-03 12:21:05",
"amount":"2"
}
]
This is how I need it organized (this is just dummy data im using in on the angular side)
[
{
date: '2014-09-04',
feeding: [
{
id: '1',
time: '1409852728000',
amount: '3'
},
{
id: '2',
time: '1409874328000',
amount: '4'
},
]
},
{
date: '2014-09-05',
feeding: [
{
id: '3',
time: '1409915908000',
amount: '3.5'
},
{
id: '4',
time: '1409957908000',
amount: '5'
},
]
},
]
I needs to be seperated out and grouped by date. How would I go about doing this?

Answer 1

Airtech was just about there. Small update to the function though. The feeding value needs to be an array of objects rather than an object. You then need to push individual objects into that array.
function dateparse($in) {
$in = json_decode($in);
$out = array();
for ($i = 0; $i < sizeof($in); $i++) {
$date = $in[$i]->date;
$isFound = false;
for ($i2 = 0; $i2 < sizeof($out); $i2++) {
if ($date == $out[$i2]["date"]) {
// We've already run into this search value before
// So add the the elements
$isFound = true;
$out[$i2]["feeding"][] = array(
"id" => $in[$i]->id,
"time" => $in[$i]->time,
"amount" => $in[$i]->amount);
break;
}
}
if (!$isFound) {
// No matches yet
// We need to add this one to the array
$feeding = array("id" => $in[$i]->id, "time" => $in[$i]->time, "amount" => $in[$i]->amount);
$out[] = array("date" => $in[$i]->date, "feeding" => array($feeding));
}
}
return json_encode($out);
}

Answer 2

How about the following? I tested it on your json input example and it ran fine.
function parse($in)
{
$in = json_decode($in);
$out = array();
for ($i = 0; $i < sizeof($in); $i++) {
$date = $in[$i]->date;
$isFound = false;
for ($i2 = 0; $i2 < sizeof($out); $i2++) {
if ($date == $out[$i2]["date"]) {
// We've already run into this search value before
// So add the the elements
$isFound = true;
$out[$i2][]["feeding"] = array(
"id" => $in[$i]->id,
"time" => $in[$i]->time,
"amount" => $in[$i]->amount);
break;
}
}
if (!$isFound) {
// No matches yet
// We need to add this one to the array
$out[] = array("date" => $in[$i]->date, "feeding" => array(
"id" => $in[$i]->id,
"time" => $in[$i]->time,
"amount" => $in[$i]->amount));
}
}
return json_encode($out);
}

Answer 3

How about this ? This works fine and works as expected :) :-
function dateparse($var)
{
$counter=0; $var = json_decode($var); $date=array();
foreach($var as $key=>$value){foreach($value as $val1=>$val2)
{if($val1 == "date")
{foreach($value as $val3=>$val4)
{if($val3!="date") //because we don't want date again
{
$date[$val2]["feeding"][$counter][$val3] = $val4; continue;
}}continue;
}
}$counter++;}
return json_encode($date);}