在PHP中将json解析为html表

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?

Answer 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