我想ID,网址,IMG和在JSON data.My当前代码视频标题不输出任何 可以在任何一个告诉我,我在做什么wrong.Thanks如何使用PHP foreach解析json数组?
$code2 = stripslashes($_POST['outputtext']);
$data = json_decode($code2, true);
$i = 0;
foreach($data->videos as $values)
{
echo $values->id . "\n";
echo $values->url . "\n";
echo $values->img . "\n";
echo $values->title . "\n";
$i++;
}
数据:
{
"cat": {
"id": "1234567",
"source_id": null,
"title_en": "first season",
"description_en": "This is spring category ",
},
"videos": [{
"id": "312412343",
"url": "\/2015-07-17\/1abcd.mp4",
"img": "image\/44\/\/2015-07-17\/1abcd.jpg",
"title": "first",
}, {
"id": "2342343",
"url": "\/2015-07-16\/2dcdeg.mp4",
"img": "images\/44\/\/2015-07-16\/2dcdeg.jpg",
"title": "second",
}];
}
验证JSON数据:
{
"cat":{
"id":"1234567",
"source_id":null,
"title_en":"first season",
"description_en":"This is spring category "
},
"videos":[
{
"id":"312412343",
"url":"\/2015-07-17\/1abcd.mp4",
"img":"image\/44\/\/2015-07-17\/1abcd.jpg",
"title":"first"
},
{
"id":"2342343",
"url":"\/2015-07-16\/2dcdeg.mp4",
"img":"images\/44\/\/2015-07-16\/2dcdeg.jpg",
"title":"second"
}
]
}
可能它不会输出任何内容,因为您尝试像访问对象一样访问数组。 –
$ data = json_decode($ code2,true);返回数组不是对象 –
您的数据无效json http://json.parser.online.fr/ –