Php foreach echo (if/else)

Question

It prints on the screen when it is correctly entered, but I do not want to do anything when it is entered incorrectly

How can I do that?

<?php
header('Content-type: text/html; charset=utf8');

$api_key = 'local';
$keyword = 'test';

$url = 'test.json' . $api_key . '&' .'keyword' .'=' . $GLOBALS['q'] ;

$open = file_get_contents($url);
$data = json_decode($open, true);

$istatistikler = $data['data'];
if ($data) {
foreach ( $istatistikler as $istatistik ){

    echo '<div class="right">';
    echo 'Oyun modu: ' . $istatistik['title'] . '<br />' .
    'Kazanma: ' . $istatistik['content'] . '<br />' .
    'Kazanma: ' . $istatistik['image'] . '<br />' .
    'Kazanma: ' . $istatistik['category'] . '<br />' .
    '<br />' .
    '<hr/>';
 $karakter_simge = 'http://google.com' . $istatistik['image'] . '';
 echo "<img src=".$karakter_simge." >" ;
 echo '</div>';
 }
  }
?>

Successful output

Failed output

Warning: file_get_contents(http://localhost/api/detail?X-Api-Key=local&keyword=a): failed to open stream: HTTP request failed! HTTP/1.1 406 Not Acceptable in /opt/lampp/htdocs/weather-master/php/php-api.php on line 10

"I do not want to print unsuccessfully"

thank you for your help!

You should do that at your PHP configuration level. Disable the error displaying, that is best practice. You should check your logs for errors. — chris85, Jun 17 '17 at 23:38

score 1 · Answer 1 · edited Jun 17 '17 at 23:41

1

This may be helpful:

$open = @file_get_contents($url);

@ sign before a function name (in a call) prevents from showing any warnings (It's a bad practice though).

Good luck!

edited Jun 17 '17 at 23:41

chris85

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answered Jun 17 '17 at 23:38

kourouma_coder

1,078
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Thank you so much for your reply! – İlkin Yıldız Jun 17 '17 at 23:59

score 0 · Accepted Answer · answered Jun 17 '17 at 23:38

Change

$open = file_get_contents($url);

into

$open = @file_get_contents($url);
if ($open === false)
    die("wrong");

The @ suppresses the error message. Using die() will abort the script completely with the given message.

Alternatively, change the condition to !== false and wrap the rest of your "successful" code in its body:

$open = @file_get_contents($url);
if ($open !== false)
{
    $data = json_decode...
    ...
    ...
}

I guess I overshot the goal here a little, but not even running into code that won't work properly without its data isn't a bad idea at all.

Thank you so much for your reply, Problem solved! – İlkin Yıldız Jun 17 '17 at 23:59 — İlkin Yıldız, Jun 17 '17 at 23:59

Php foreach echo (if/else)

2 Answers2