CSV to JSON with PHP? - 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;

Answer 1

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.

Answer 2

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);
}

Answer 3

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
}

Answer 4

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.

Answer 5

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.

Answer 6

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();