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Convert CSV to JSON
* Converts CSV to JSON
* Example uses Google Spreadsheet CSV feed
* csvToArray function I think I found on
header('Content-type: application/json');
// Set your CSV feed
$feed = '';
// Arrays we'll use later
$keys = array();
$newArray = array();
// Function to convert CSV into associative array
function csvToArray($file, $delimiter) {
if (($handle = fopen($file, 'r')) !== FALSE) {
$i = 0;
while (($lineArray = fgetcsv($handle, 4000, $delimiter, '"')) !== FALSE) {
for ($j = 0; $j < count($lineArray); $j++) {
$arr[$i][$j] = $lineArray[$j];
return $arr;
// Do it
$data = csvToArray($feed, ',');
// Set number of elements (minus 1 because we shift off the first row)
$count = count($data) - 1;
//Use first row for names
$labels = array_shift($data);
foreach ($labels as $label) {
$keys[] = $label;
// Add Ids, just in case we want them later
$keys[] = 'id';
for ($i = 0; $i < $count; $i++) {
$data[$i][] = $i;
// Bring it all together
for ($j = 0; $j < $count; $j++) {
$d = array_combine($keys, $data[$j]);
$newArray[$j] = $d;
// Print it out as JSON
echo json_encode($newArray);

aendrew commented Apr 27, 2012

Great script, saved me a bunch of time for an interactive I was working on for the Guardian.

Couple quick things --
a. I was getting PHP warnings on ln. 54. Adding an @ in from of array_combine(); suppresses that.
b. If you're wanting the JSON to be consumed on a different domain than the one hosting the PHP (i.e., through jQuery's getJSON(); function), you'll need a way around AJAX's same domain origin policy. I've done so by changing the header to script/javascript and replacing line 59 with echo $_GET['callback']. '(' . json_encode($newArray) . ');';. This allows you to consume the output through JSONP by attaching a callback variable to the path (e.g., "?callback=mycallback").


robflaherty commented Apr 30, 2012

Thanks! Regarding the PHP warnings, what version of PHP are you using?

aendrew commented Jun 1, 2012

Sorry for the slow response -- I was using 5.2.17.

Great script, thanks.

On a MAC add

ini_set("auto_detect_line_endings", true);

Great script. Thanks, In my fork I made it for command line use.

Thank you.

Really nice!

Licensing? am I free to use this or are there any restrictions?

We executed this for CSV dataset that we got and the results were incredible. Thank you!!!!!!

ram-you commented Jun 13, 2014

Thank you. Saved me.

Thank you. I make my portfolio based on your script.

This is exactly what I needed.
It may be obvious but for anyone who needs to access the PHP created json via javascript, just use this:

<?php $json = json_encode($newArray); ?>
var $json = <?php echo $json; ?>;
// do your magic here

Thanks a million Rob! :)

afenix commented Jun 19, 2015

So awesome. Thanks much for the share! I was going to have to refactor a bunch of code, and this worked like a charm. I hope I can repay the favor one day.

I have a nested/setted JSON format I need to produce... Can anyone suggest how to format a spreadsheet to produce this:

Or would that require a set of spreadsheets/CSVs?

ur92 commented Feb 25, 2016

Great Gist!

I forked it and shortened the csvToArray function to this:
$data = array_map('str_getcsv', file($feed));
as this is more compact and performance should be better.

Tested => works!

This is exactly what I was looking for! Thanks a million!!!!

g4spow commented Jun 9, 2016

My file path for my images is folder/imagename.jpg.

When I run this great script, this is the only thing throwing an error. The path come out as folder/imagename.jpg

Here is the screenshot of the issue:

Also, can the output be saved to a .json file rather than within the PHP?

Really useful. Thanks.

imsabin commented Jul 2, 2017

hi, my server has disabled fopen. so is there any alternative like using curl to get the same result??

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