Removing Commas in PHP Array for JSON Formatting

Question

I'm outputting a json formatted array. Where I have the address fields and Market fields separated by commas I want to remove commas if a field doesn't exist so I don't end up with empty spaces with commas. I had a go below with implode but that doesn't seem to work... what is the correct way of doing this?

  $data = array(
    'ID' => $Member->memberID(),
    'Name' => $Member->first_name() . ' ' . $Member->last_name(),
    'Job Title' => $Member->expert_job_title(),
    'Organisation' => $Member->expert_org_name(),
    'Organisation Type' => $Member->expert_org_type(),
    'Website' => 'http://' . $Member->expert_org_website(),
    'Phone' => $Member->expert_org_phone(),
    'Expertise' => $Member->expert_org_desc(),
    'Markets' => implode(', ', array($Member->expert_org_market_medical(), $Member->expert_org_market_pharmaceuticals(), $Member->expert_org_market_agriculture(), $Member->expert_org_market_food(), )),
    'Address' => $Member->expert_address_one() . ', ' . $Member->expert_address_two() . ', ' . $Member->expert_address_town_city() . ', ' . $Member->expert_address_county_state() . ', ' . $Member->expert_org_country() . ', ' . $Member->expert_address_code(),
    'Latitude' => $Member->expert_org_latitude(),
    'Longitude' => $Member->expert_org_longitude()
  );



header('Content-Type: application/json');

echo json_encode($data, JSON_PRETTY_PRINT);

UPDATE WITH ANSWER

'Markets' => implode(', ', array_filter(array($Member->expert_org_market_medical(), $Member->expert_org_market_pharmaceuticals(), $Member->expert_org_market_agriculture(), $Member->expert_org_market_food(), ))),
'Address' => implode(', ', array_filter(array($Member->expert_address_one(), $Member->expert_address_two(), $Member->expert_address_town_city(), $Member->expert_address_county_state(), $Member->expert_org_country(), $Member->expert_address_code(), ))),

the quick and dirty way to do it is to trim the string of anything that looks like ', ,' replacing it with ',' — Marshall Tigerus
– Marshall Tigerus, Commented Jun 25, 2018 at 16:59

A l w a y s S u n n y · Accepted Answer · 2018-06-25 17:59:37Z

3

Use array_filter() like this for Address and Markets field

<?php
echo "without array filter = " . implode(', ', array('a', '', 'b', '', ));
echo "<br/>";
echo "with array filter = ". implode(', ', array_filter(array('a', '', 'b', '', )));
?>

DEMO : https://eval.in/1027921

I also urge you to see some example on php.net because it'll help you clearly understand how array_filter() works and what characters are filtered out.

<?php

$entry = array(
             0 => 'foo',
             1 => false,
             2 => -1,
             3 => null,
             4 => ''
          );

print_r(array_filter($entry));
?>

The above example will output:

Array
(
    [0] => foo
    [2] => -1
)

edited Jun 25, 2018 at 17:59

answered Jun 25, 2018 at 17:00

A l w a y s S u n n y

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tadeubarbosa · Accepted Answer · 2018-06-25 17:06:20Z

0

Have some things that you do. But, you can try array_filter, in documentation.

$Markets = array($Member->expert_org_market_medical(), $Member->expert_org_market_pharmaceuticals(), $Member->expert_org_market_agriculture(),  $Member->expert_org_market_food());
$Markets = array_filter($Markets, function($item) {
    // return if that strings isn't null
    // maybe you want to add another restrictions here
    return strlen($item);
}, $Markets);
// all empty items was removed.|
$Markets = implode(',', $Markets);

answered Jun 25, 2018 at 17:06

tadeubarbosa

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Comments

Eric · Accepted Answer · 2018-06-25 21:27:17Z

0

Maybe with preg_replace

 $json = json_encode($data);
 $json = preg_replace('/(^, )|( ,)/mi', '', $json);
 $data = json_decode($json);

edited Jun 25, 2018 at 21:27

answered Jun 25, 2018 at 17:22

Eric

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