Filtering JSON data with php

Question

I am trying to iterate over this json file and filter out the unwanted elements I would like to split the result so I have have a list of Customers or a list of Suppliers json File:

{
  "$descriptor": "Test",
  "$resources": [
    {
      "$uuid": "281d393c-7c32-4640-aca2-c286f6467bb1",
      "$descriptor": "",
      "customerSupplierFlag": "Customer"
    },
    {
      "$uuid": "87a132a9-95d3-47fd-8667-77cca9c78436",
      "$descriptor": "",
      "customerSupplierFlag": "Customer"
    },
    {
      "$uuid": "8a75345c-73c7-4852-865c-f468c93c8306",
      "$descriptor": "",
      "customerSupplierFlag": "Supplier"
    },
    {
      "$uuid": "a2705e38-18d8-4669-a2fb-f18a87cd1cc6",
      "$descriptor": "",
      "customerSupplierFlag": "Supplier"
    }
  ]
}

for example

    {
      "$uuid": "281d393c-7c32-4640-aca2-c286f6467bb1",
      "$descriptor": "",
      "customerSupplierFlag": "Customer"
    },
    {
      "$uuid": "87a132a9-95d3-47fd-8667-77cca9c78436",
      "$descriptor": "",
      "customerSupplierFlag": "Customer"
    },

my php code is

$array = json_decode($result, true);


for ($i = 0; $i < count($array['$resources']); $i++){
    foreach ($array['$resources'][$i]['customerSupplierFlag'][Customer] as $item)
    {
        // do what you want
        echo $item['customerSupplierFlag' ]. '<br>';
        echo $item['$uuid'] . '<br>';
    }
}

I have been struggling with this for a few days now an appear to be going over the same articles and getting now were any suggestions would be appreciated.

You don't have $array['$resources'][$i]['customerSupplierFlag'][Customer] elements in your json. Only $array['$resources'][$i]['customerSupplierFlag'] — u_mulder
– u_mulder, Commented Aug 1, 2015 at 16:18
Customer is the is the value held in customerSupplierFlag filed that I would like to filter I believe it is a similar expression to ('//$resources[$uuid>=87a132a9-95d3-47fd-8667-77cca9c78436][customerSupplierFlag="Customer"]'); that is used in xpath — Artful_dodger
– Artful_dodger, Commented Aug 1, 2015 at 16:44
It's not xpath here, it's a plain array, received from JSON. — u_mulder
– u_mulder, Commented Aug 1, 2015 at 16:50

myte · Accepted Answer · 2015-08-01 16:38:21Z

13

$data = file_get_contents('./your_json_data.txt');

$json = json_decode($data, 1);

$resources = $json['$resources'];

$suppliers = array();
$customers = array();

foreach ($resources as $rkey => $resource){

    if ($resource['customerSupplierFlag'] == 'Customer'){

        $customers[] = $resource;

    } else if ($resource['customerSupplierFlag'] == 'Supplier') {

        $suppliers[] = $resource;

    }

}

header('Content-Type: application/json');

echo json_encode($customers);
echo json_encode($suppliers);

die();

answered Aug 1, 2015 at 16:38

myte

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1 Comment

JamesWilson Over a year ago

depending on where the json data is coming from and how often it is updated, you might consider storing the filtered data in a json file itself, and then you can just pass-through the json directly through the webserver instead of loading up php and filtering the data on each request.

jai3232 · Accepted Answer · 2020-07-26 14:45:45Z

5

$array = json_decode($json, true);
$flag = "Supplier";
$resources = array_filter($array, function ($var) use ($flag) {
    return ($var['customerSupplierFlag'] == $flag);
});
print_r($resources);

answered Jul 26, 2020 at 14:45

jai3232

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4 Comments

Mark Rotteveel Over a year ago

Please don't post only code as answer, but also provide an explanation what your code does and how it solves the problem of the question. Answers with an explanation are usually more helpful and of better quality, and are more likely to attract upvotes.

jai3232 Over a year ago

Nothing complex about the code. It's quite straight forward. Just use two basic builtin PHP functions. Refer to PHP documentation: 1. json_decode at php.net/manual/en/function.json-decode.php 2. array_filter at php.net/manual/en/function.array-filter.php

Mark Rotteveel Over a year ago

The point is not whether or not code is complex or not, an explanation will help the OP and future visitors to your question to understand your solution. Answers shouldn't be about blindly copy/paste a piece of code and pray it works.

Frank Over a year ago

Thanks for adding this code, even if you don't have time for an explanation, I still am grateful.

Jose A. de los Santos · Accepted Answer · 2022-11-24 14:59:47Z

1

$json_a = json_decode($string, true);
            
 foreach($json_a as $jsonDataKey => $jsonDataValue){
    foreach($jsonDataValue as $jsonArrayKey => $jsonArrayValue){
                print_r($jsonArrayValue['id']);
                print_r($jsonArrayValue['name']);
        }
          }

answered Nov 24, 2022 at 14:59

Jose A. de los Santos

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