我想从获取数据到我的iTunes文件。在搜索表单中,你可以填写歌曲的艺术家和标题。我只想展示这首歌的数据。这是我的密码:
$title = $_POST["title"];
$artist = $_POST["artist"];
$query_string = urlencode($title);
$query_string2 = urlencode($artist);
$json = file_get_contents('https://itunes.apple.com/search?term=$query_string&term=$query_string2'); // this WILL do an http request for you
$data = json_decode($json);
echo "<div class='container' id='movies'>";
echo "<div class='col-md-4' id='movie'>";
echo "<img src='" .$data->artworkUrl60. "' width='200'>";
echo "<h4>" .$data->artistName. " - " .$data->trackName. "</h4>"; 我得到了这个错误:注意:未定义的属性: stdClass::$artworkUrl60 60 in.我的密码怎么了?
发布于 2017-02-12 14:00:00
因为您的$data对象不包含名为$artworkUrl60的属性。您的HTTP查询不能正常工作,您应该使用双引号而不是单引号。
// For limit you can add 'limit' parameter
$json = file_get_contents("https://itunes.apple.com/search?term=$query_string&term=$query_string2&limit=20");
// OR concatenation
// $json = file_get_contents('https://itunes.apple.com/search?term='.$query_string.'&term='.$query_string2);
$data = json_decode($json);
if (0 == $data->resultCount) {
echo 'No result found ! ';
}
else {
foreach ($data->results as $row){
echo "<div class='container' id='movies'>";
echo "<div class='col-md-4' id='movie'>";
echo "<img src='" .$row->artworkUrl60. "' width='200'>";
echo "<h4>" .$row->artistName. " - " .$row->trackName. "</h4>";
}
}https://stackoverflow.com/questions/42188133
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