Parsing json to html table in PHP

Question

I have an array and when I do

$array = json_decode($response, true);
print_r($array);

This is what I get from an array.

Array ( [data] => Array ( [0] => Array ( [id] => accenture [type] => companies [attributes] => Array ( [name] => Accenture [description] => Accenture is a global management consulting, technology services and outsourcing company, with more than 293,000 people serving clients in more than 120 countries. Combining unparalleled experience, comprehensive capabilities across all industries and business functions, and extensive research on the world’s most successful companies, Accenture collaborates with clients to help them become high-performance businesses and governments. The company generated net revenues of US$xxx for the fiscal year ended Aug. 31, 2013. Its home page is www.some-url.com. [employee_count_range] => 10001+ [founded_year] => 1989 [industries] => Array ( [0] => Information Technology & Services ) [website_url] => www.some-url.com [logo] => https://some-url/logo.png [square_logo] => https://some-url/square_logo.png [followed] => [claimed_status] => 1 [last_reviewed_at] => ) ) [1] => Array ( [id] => accenture-banglore [type] => companies [attributes] => Array ( [name] => Accenture, Banglore [description] => [employee_count_range] => [founded_year] => [industries] => [website_url] => [logo] => [square_logo] => [followed] => [claimed_status] => 1 [last_reviewed_at] => ) ) [2] => Array ( [id] => accenture-gmbh [type] => companies [attributes] => Array ( [name] => Accenture GmbH [description] => [employee_count_range] => [founded_year] => [industries] => [website_url] => [logo] => [square_logo] => [followed] => [claimed_status] => 1 [last_reviewed_at] => ) )

I tried with foreach loop to create a table and show the data in the table. This is how I try but I get an error "Invalid argument supplied for foreach()"

echo '<table>';
foreach($array as $result){
  echo '<tr>';
echo '<td>'.$result->id.'</td>';
echo '<td>'.$result->name.'</td>';
echo '<td>'.$result->description.'</td>';
echo '<td>'.$result->funded_year.'</td>';
  echo '</tr>';
}
echo '</table>';

Can anyone help me show the data in the table?

Matko Đipalo · Accepted Answer · 2017-09-03 17:14:55Z

1

Your data for HTML table is inside data array key. You should write:

foreach($array['data'] as $result){

or even better:

$array = json_decode($response, true);
$resultArray = isset($array['data']) ? $array['data'] : [];

echo '<table>';
foreach($resultArray as $result){
        echo '<tr>';
        echo '<td>'.(isset($result['id']) ? $result['id'] : '-') .'</td>';
        echo '<td>'.(isset($result['attributes']['name']) ? $result['attributes']['name'] : '-').'</td>';
        echo '<td>'.(isset($result['attributes']['description']) ? $result['attributes']['description'] : '-').'</td>';
        echo '<td>'.(isset($result['attributes']['funded_year']) ? $result['attributes']['funded_year'] : '-').'</td>';
        echo '</tr>';
}
echo '</table>';

http://php.net/manual/en/function.json-decode.php

edited Sep 3, 2017 at 17:14

answered Sep 3, 2017 at 15:41

Matko Đipalo

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2 Comments

Barlet Over a year ago

thanks... now the error is gone. But I see only ID's. On name I see "-" same as on description and funded_year...

Matko Đipalo Over a year ago

It is because name and founded_year are one level deeper. They are in attributes key. You can access them with $result['attributes'][name'] and $result['attributes']['funded_year']

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