CSV to JSON with PHP?

Question

I need to convert a CSV file to JSON on the server using PHP. I am using this script which works:

function csvToJSON($csv) {
    $rows = explode("\n", $csv);

    $i = 0;
    $len = count($rows);
    $json = "{\n" . '    "data" : [';
    foreach ($rows as $row) {
        $cols = explode(',', $row);
        $json .= "\n        {\n";
        $json .= '            "var0" : "' . $cols[0] . "\",\n";
        $json .= '            "var1" : "' . $cols[1] . "\",\n";
        $json .= '            "var2" : "' . $cols[2] . "\",\n";
        $json .= '            "var3" : "' . $cols[3] . "\",\n";
        $json .= '            "var4" : "' . $cols[4] . "\",\n";
        $json .= '            "var5" : "' . $cols[5] . "\",\n";
        $json .= '            "var6" : "' . $cols[6] . "\",\n";
        $json .= '            "var7" : "' . $cols[7] . "\",\n";
        $json .= '            "var8" : "' . $cols[8] . "\",\n";
        $json .= '            "var9" : "' . $cols[9] . "\",\n";
        $json .= '            "var10" : "' . $cols[10] . '"';
        $json .= "\n        }";

        if ($i !== $len - 1) {
            $json .= ',';
        }

        $i++;
    }
    $json .= "\n    ]\n}";

    return $json;
}

$json = csvToJSON($csv);
$json = preg_replace('/[ \n]/', '', $json);

header('Content-Type: text/plain');
header('Cache-Control: no-cache');
echo $json;

The $csv variable is a string resulting from a cURL request which returns the CSV content.

I am sure this is not the most efficient PHP code to do it because I am a beginner developer and my knowledge of PHP is low. Is there a better, more efficient way to convert CSV to JSON using PHP?

Thanks in advance.

Note. I am aware that I am adding whitespace and then removing it, I do this so I can have the option to return "readable" JSON by removing the line $json = preg_replace('/[ \n]/', '', $json); for testing purposes.

Edit. Thanks for your replies, based on them the new code is like this:

function csvToJson($csv) {
    $rows = explode("\n", trim($csv));
    $csvarr = array_map(function ($row) {
        $keys = array('var0','var1','var2','var3','var4','var5','var6','var7','var8','var9','var10');
        return array_combine($keys, str_getcsv($row));
    }, $rows);
    $json = json_encode($csvarr);

    return $json;
}

$json = csvToJson($csv);

header('Content-Type: application/json');
header('Cache-Control: no-cache');
echo $json;

mario · Accepted Answer · 2011-01-27 00:46:49Z

25

Well there is the json_encode() function, which you should use rather than building up the JSON output yourself. And there is also a function str_getcsv() for parsing CSV:

$array = array_map("str_getcsv", explode("\n", $csv));
print json_encode($array);

You must however adapt the $array if you want the JSON output to hold named fields.

answered Jan 27, 2011 at 0:46

mario

146k20 gold badges243 silver badges293 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

VerizonW Over a year ago

Thank you very much, this is a really clean solution, I edited my post with the code I am using now based on your answer plus I added array_combine to output JSON with property names.

kaiser Over a year ago

+1 That rocks like hell. Didn't know about str_getcsv. If you also know how to prevent breaking multi-line cells ... :D

Joe P · Accepted Answer · 2014-07-07 11:00:48Z

9

I modified the answer in the question to use the first line of the CSV for the array keys. This has the advantage of not having to hard-code the keys in the function allowing it to work for any CSV with column headers and any number of columns.

Here is my modified version:

function csvToJson($csv) {
    $rows = explode("\n", trim($csv));
    $data = array_slice($rows, 1);
    $keys = array_fill(0, count($data), $rows[0]);
    $json = array_map(function ($row, $key) {
        return array_combine(str_getcsv($key), str_getcsv($row));
    }, $data, $keys);

    return json_encode($json);
}

answered Jul 7, 2014 at 11:00

Joe P

3928 silver badges14 bronze badges

Comments

ericjbasti · Accepted Answer · 2015-02-28 01:33:03Z

None of these answers work with multiline cells, because they all assume a row ends with '\n'. The builtin fgetcsv function understands that multiline cells are enclosed in " so it doesn't run into the same problem. The code below instead of relying on '\n' to find each row of a csv lets fgetcsv go row by row and prep our output.

function csv_to_json($file){

    $columns = fgetcsv($file); // first lets get the keys.
    $output = array(); // we will build out an array of arrays here.

    while(!feof($file)){ // until we get to the end of file, we'll pull in a new line
        $line = fgetcsv($file); // gets the next line
        $lineObject = array(); // we build out each line with our $columns keys
        foreach($columns as $key => $value){
            $lineObject[$value] = $line[$key];
        }
        array_push($output, $lineObject);
    }
    return json_encode($output); // encode it as json before sending it back
}

alex · Accepted Answer · 2011-01-27 00:44:44Z

1

Some tips...

If you have URL opening enabled for fopen() and wrappers, you can use fgetscsv().
You can build an array of the CSV, and then convert it with PHP's native json_encode().
The correct mime type for JSON is application/json.

answered Jan 27, 2011 at 0:44

alex

492k205 gold badges891 silver badges992 bronze badges

1 Comment

VerizonW Over a year ago

Thanks, now I am using json_encode() and corrected the mime type.

josefnpat · Accepted Answer · 2011-01-27 00:45:54Z

0

You could probably reduce the overhead by removing all the spaces and \n's. But that's in your note.

You could increase the performance by skipping the preg_replace and passing a boolean that would turn it on and off.

Other than that, the variable unrolling of your var[1-10] actually is good, as long as there are always ten varaibles.

The explode and the foreach approach are just fine.

answered Jan 27, 2011 at 0:45

josefnpat

584 bronze badges

1 Comment

VerizonW Over a year ago

I ended up removing the spaces using a much cleaner solution now, thanks.

John · Accepted Answer · 2012-12-27 18:23:24Z

0

I recommend using Coseva (a csv parsing library) and using the built in toJSON() method.

<?php

// load
require('../src/CSV.php');

// read
$csv = new Coseva\CSV('path/to/my_csv.csv');

// parse
$csv->parse();

// disco
echo $csv->toJSON();

answered Dec 27, 2012 at 18:23

John

3,9166 gold badges35 silver badges39 bronze badges

Collectives™ on Stack Overflow

CSV to JSON with PHP?

6 Answers 6

2 Comments

Comments

Comments

1 Comment

1 Comment

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

6 Answers 6

2 Comments

Comments

Comments

1 Comment

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related