Using PHP to loop JSON and generate a new format of nested JSON

Question

In PHP, how do I loop through the following JSON Object's date key, if the date value are the same then merge the time:

[
  {
    date: "27-06-2017",
    time: "1430"
  },
  {
    date: "27-06-2017",
    time: "1500"
  },
  {
    date: "28-06-2017",
    time: "0930"
  },
  {
    date: "28-06-2017",
    time: "0915"
  }
]

Result should looks like this:

[
  {
    date: "27-06-2017",
    time: [{"1430","1500"}]
  },      {
    date: "28-06-2017",
    time: [{"0930, 0915"}]
  }
]

Should I create an empty array, then the looping through the JSON and recreate a new JSON? Is there any better way or any solution to reference?

Thank you!

I think time will be an array, not an array with an object which is not well formated. I would suggest you to read the json manual first. I guess you should have an output like this: [ { date: "27-06-2017", time: ["1430", "1500"] } ] — Marius Bogdan
– Marius Bogdan, Commented May 5, 2017 at 10:11
you can check with if else condition and some find function if you have large data or you can do it simple way like check two column — Ahmed Ginani
– Ahmed Ginani, Commented May 5, 2017 at 10:20

sevavietl · Accepted Answer · 2017-05-05 10:46:08Z

1

This solution is a little overhead, but if you are sure that your data is sequential and sorted, you can use array_count_values and array_map:

$count = array_count_values(array_column($array, 'date'));
$times = array_column($array, 'time');
$grouped = array_map(function ($date, $count) use (&$times) {
    return [
        'date' => $date,
        'time' => array_splice($times, 0, $count)
    ];    
}, array_keys($count), $count);

Here is working demo.

answered May 5, 2017 at 10:46

sevavietl

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1 Comment

dnez Over a year ago

This one looks good! By adding echo json_encode($grouped); I could get the result I am looking for. Let me test it out in 2 days to compare the speed with foreach. Thanks!!

Niklesh Raut · Accepted Answer · 2017-05-05 10:37:05Z

1

Another Idea to do this is

<?php
    $string = '[{"date": "27-06-2017","time": "1430"},{"date": "27-06-2017","time": "1500"},{"date": "28-06-2017","time": "0930"},{"date": "28-06-2017","time": "0915"}]';
    $arrs = json_decode($string);
    $main = array();
    $temp = array();
    foreach($arrs as $arr){
      if(in_array($arr->date,$temp)){
        $main[$arr->date]['time'][] = $arr->time;
      }else{
        $main[$arr->date]['date'] = $arr->date;
        $main[$arr->date]['time'][] = $arr->time;
        $temp[] = $arr->date;
      }
    }
    echo json_encode($main);
?>

live demo : https://eval.in/787695

answered May 5, 2017 at 10:37

Niklesh Raut

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2 Comments

dnez Over a year ago

After gone through the few solutions, i found this one easier to be understand and using only 1 for loop. Thank you so much everyone for the answers!!!

dnez Over a year ago

Sorry, I like the answer but the format is slightly different than what I was asking.

informer · Accepted Answer · 2017-05-05 10:47:01Z

1

Please try this:

$a = [] // Your input array
$op= [];//output array
foreach($a as $key => $val){
    $key = $val['date'];
    $op[$key][] = $val['time'];
}
$op2 = [];
foreach($op as $key => $val){
    $op2[] = ['date' => $key, 'time' => $val]; 
}

edited May 5, 2017 at 10:47

answered May 5, 2017 at 10:12

informer

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4 Comments

Ahmed Ginani Over a year ago

this is not solution

dnez Over a year ago

The answer is good but is there a way to have key value pair like @Marius Bogdan suggested? [ { date: "27-06-2017", time: ["1430", "1500"] } ]

dnez Over a year ago

Thanks for your quick answer, think the $op[] in 2nd foreach is $op2[]. so I have to use 2 foreach loop? if data size is big, it might be not that efficient right?

dnez Over a year ago

The 2nd foreach should be: foreach($op as $key => $val){ $op2[] = ['date' => $key, 'time' => $val]; }

CD001 · Accepted Answer · 2017-05-05 10:52:19Z

// create the "final" array
$final = [];

// loop the JSON (assigned to $l)
foreach($l as $o) {

    // assign $final[{date}] = to be a new object if it doesn't already exist
    if(empty($final[$o->date])) {
        $final[$o->date] = (object) ['date' => $o->date, 'time' => [$o->time]];
    }

    // ... otherwise, if it does exist, just append this time to the array
    else {
        $final[$o->date]->time[] = $o->time;
    }
}

// to get you back to a zero-indexed array
$final = array_values($final);

The "final" array is created with date based indexes initially so that you can determine whether they've been set or not to allow you to manipulate the correct "time" arrays.

They're just removed at the end by dropping $final into array_values() to get the zero-indexed array you're after.

json_encode($final) will give you want you want as a JSON:

[{"date":"27-06-2017","time":["1430","1500"]},{"date":"28-06-2017","time":["0930","0915"]}]

xuma · Accepted Answer · 2017-05-05 10:58:06Z

1

Yet another solution :

$d = [];
$dizi = array_map(function ($value) use (&$d) {
    if (array_key_exists($value->date, $d)) {
        array_push($d[$value->date]['time'], $value->time);
    } else {
        $d[$value->date] = [
            'date' => $value->date,
            'time' => [$value->time]
        ];
    }
}, $array);
echo json_encode(array_values($d));

answered May 5, 2017 at 10:58

xuma

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