How to skip json object if it has a empty value?

Question

I'm stuck on this since many days... I'm using this code to parse a json EPG guide stored on a server:

<?php
$channel = '899';
$current_unix = time();
$json = json_decode(file_get_contents('http://guidatv.sky.it/app/guidatv/contenuti/data/grid/'.date('y_m_d').'/ch_'.$channel.'.js', true));
echo '<ul>';
foreach ($json->plan as $prog) {
echo "<li>" . $prog->title . "</li>";
echo '</ul>';  
}
?>

I would like to skip this part of the json from showing because it has empty fields

"banned":true,
"plan":[
{"id":"-1",
"pid":"0",
"starttime":"00:00",
"dur":"65",
"title":"",
"normalizedtitle": "",
"desc":"",
"genre":"",
"subgenre":"",
"prima":false 
},

How can I do that?

based on which empty element basically you want to skip? title, desc, or any-one else? — Death-is-the-real-truth
– Death-is-the-real-truth, Commented Jul 22, 2017 at 17:46
@YvesLeBorg What about $plan->title = "" ? that will pass the condition — Accountant م
– Accountant م, Commented Jul 22, 2017 at 18:02
Get used to good habits. do echo "<li>" . htmlspecialchars($prog->title) . "</li>"; instead of echo "<li>" . $prog->title . "</li>"; . to protect yourself from HTML injection. — Accountant م
– Accountant م, Commented Jul 22, 2017 at 18:07

Godiez · Accepted Answer · 2017-07-22 18:05:04Z

1

You can add in your foreach a condition like:

foreach ($json->plan as $prog) {
    if(empty($json->plan->title)) continue;
    echo "<li>" . $prog->title . "</li>";
}
echo '</ul>';

edited Jul 22, 2017 at 18:05

answered Jul 22, 2017 at 17:53

Godiez

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3 Comments

Accountant م Over a year ago

Be careful with the "0" problem, if the title is "0" it will not be echoed. also remove the </ul> from the loop

fredix11 Over a year ago

it works but the first entry has a strange indentation that I can't remove: phpfiddle.org/lite/code/i765-2ajn

Godiez Over a year ago

As Accountant said, </ul> must be outside, I updated my answer, check it out

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