how can filter json code - php

Question

i have json file like below:
{
"data": [
{
"id": 8921,
"weather": null,
"status": "TT",
}, {"id": 8922,
"weather": null,
"status": "TF",
},
{
"id": 8923,
"weather": null,
"status": "NT",
}, {"id": 8922,
"weather": null,
"status": "HT",
}
]
}
i read this and show in table
but i want filter that with status and get just status have TF and size depend on.
old code without check status:
$jsonResult = file_get_contents('1.json');
$content_json = json_decode($jsonResult);
$size=sizeof($content_json->data);
the new code (did'nt work true):
$jsonResult = $jsonResult = file_get_contents('1.json');
$content_json = json_decode($jsonResult);
$content_json = array_filter($content_json['data'], function ($data) {
return $data['status'] != 'HT' or $data['status'] != 'NT';
});

Answer 1

You should iterate over the data array to check the status:
$jsonResult = file_get_contents('1.json');
$content_json = json_decode($jsonResult, true);
$data = $content_json['data'];
$filteredData = [];
foreach($data as $entry){
if($entry['status'] == "TF")
$filteredData[] = $entry;
}
$size = sizeof($filteredData);
Also, your JSON file has a syntax error, remove the commas after the status properties like so:
{
"data": [
{
"id": 8921,
"weather": null,
"status": "TT"
}, {"id": 8922,
"weather": null,
"status": "TF"
}
]
}

Answer 2

Using arrays as opposed to objects (pass true as the second parameter to json_decode()), you can use array_filter() to remove any elements your not interested in...
$content_json = json_decode($jsonResult,true);
$content_json = array_filter($content_json['data'], function ($data) {
return $data['status'] == 'TF';
});
echo count($content_json);
Update:
Your code should be...
$content_json = json_decode($jsonResult, true);
$content_json = array_filter($content_json['data'], function ($data) {
return $data['status'] != 'HT' and $data['status'] != 'NT';
});
Note the ,true on the json_decode() and the condition should be and not or.