我想使用服务器上的json文件显示电视节目的列表,包括标题、流派、子流派、描述、开始时间和持续时间,但我得到了以下错误:
E_NOTICE :类型8 --尝试获取非对象的属性--第13行
E_WARNING :类型2 --在第13行为foreach()提供的参数无效
下面是一段json示例:
{"channel":"6280",
"banned":true,
"plan":[
{"id":"-1",
"pid":"0",
"starttime":"00:00",
"dur":"65",
"title":"",
"normalizedtitle": "",
"desc":"",
"genre":"",
"subgenre":"",
"prima":false
},
{"id":"94622386",
"pid":"507461",
"starttime":"01:05",
"dur":"65",
"title":"Sex Researchers",
"normalizedtitle": "sex-researchers",
"desc":"Ep. 2 - Ciclo The Body of...",
"genre":"mondo e tendenze",
"subgenre":"societa",
"prima":false
},下面是我使用的php代码:
<?php
$channel = '6280';
$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'));
//print_r($json);
echo '<ul>';
foreach ($json as $data) {
echo '<li>';
foreach ($data->plan as $prog) {
if ( $current_unix < $prog->starttime ) {
echo $prog->id . '<br>';
echo $prog->starttime . '<br>';
echo $prog->dur . '<br>';
echo $prog->desc . '<br>';
if ( isset($prog->genre)) {
echo $prog->genre . '<br>';
}
}
}
echo '</li>';
}
echo '</ul>';
?>你能帮我解决这个问题吗?谢谢
发布于 2017-07-18 00:16:38
你不能在对象上循环。您的JSON不返回channel数组,而只返回一个channel对象。这是你想要的:
$json = json_decode(file_get_contents('http://guidatv.sky.it/app/guidatv/contenuti/data/grid/'.date('y_m_d').'/ch_'.$channel.'.js'));
echo '<ul>';
foreach ($json->plan as $prog) {
echo "<li>" . $prog->title . '</li>';
}
echo '</ul>';https://stackoverflow.com/questions/45148830
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