从php远程csv获取json数组

Question

For one of my php code, i want the json array as following from remote csv file in php.

The remote csv file: www.xyz.com/dir/records.csv

NAME    Age ID
Jhon    45  101
Bhil    42  102
Tone    41  103

I want a function which converts this CSV file to a JSON array

Something Like this:

$json = '{"records": 
       [
         {"Name":"Jhon", "Age":"45", "id":"101"},
         {"Name":"Bhil", "Age":"42", "id":"102"},
         {"Name":"Tone", "Age":"41", "id":"103"},
       ]
    }';

Kindly guide me, how to incopporate the above csv file to get the above json array for the follwoing php code.

$myjson = json_decode($json, true);  
  foreach ( $myjson['records'] as $row ) {  
                        if ($row['id'] =='101') { 
                          foreach ( $row as $field => $value ) {  
                               // do the work here
                             }
                   }
         }

Answer 1

Here is the solution, You can modify your output array format accordingly inside csvToJson() function

<?php

     function csvToJson($filename) {
            $handle = fopen($filename,"r");
            $i=0;
            if($handle) {
                while(($line = fgetcsv($handle,1000,",","'")) !== FALSE)
                {
                    if($i == 0) {
                        $c = 0;
                        foreach($line as $col) {
                            $cols[$c] = $col;
                            $c++;
                        }
                    } else if($i > 0) {
                        $c = 0;
                        foreach($line as $col) {
                            $data[$i][$cols[$c]] = $col;
                            $c++;
                        }
                    }
                    $i++;
                }
            }
            $data2['records'] = array($data);       
            fclose($handle);
            // return json_encode($data2);  /*you don't have to convert it into json if you want to use it in your php foreach loop you can directly return this*/ 
            return json_encode($data2);
        }

        $json = csvToJson('records.csv');
        echo "<pre>";
        print_r($json);

    ?>