Get data from JSON file in php

Question

I am trying to get data from json file , parse it and add data from it to database. I am using as database, oracle. I am getting the errors :

This is my JSON file

{
"data": {
    "gps": [
        { "location": "45.778123,24.151165"     
        },
        { "location": "45.7875116,24.1549801"               
        }
    ]
}

This is my php file

    <?php
        $file = 'JSONdata.json';
        $json = file_get_contents($file,0,null,null);
        $obj = json_decode($json,true);
        $data = $obj->data->gps;

        include 'connection.php';

       function insert_data($connname, $conn)
       {
          foreach ($data as $post) {
             $stmt = oci_parse($conn, "insert into GPSLOGS
                   values('".$post->location."')");
            oci_execute($stmt, OCI_DEFAULT);
            echo "$connname inserted row without committing<br>\n";
            }
       }

        insert_data('c1', $conn);
   ?>

Do you know what am I doing wrong? And why it keeps getting those errors? I also tried using another format of JSON file and parse it differently , but it keeps getting the same errors.

.json
      [{ "gps": "45.778123,24.151165"},{"gps": "45.7875116,24.1549801"}]

 .php
     $file = 'JSONdata.json';
     $json = file_get_contents($file,0,null,null);
     $obj = json_decode($json,true);    
       foreach($obj as $item)
       {      echo $item['gps'];}

Dimi · Accepted Answer · 2017-02-21 21:29:09Z

2

$obj = json_decode($json,true);
$data = $obj->data->gps;

returns data in a form of array, not object. Remove the 2nd argument from json decode and everything should work.

make it

$obj = json_decode($json);
$data = $obj->data->gps;

answered Feb 21, 2017 at 21:29

Dimi

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