使用PHP解碼多個JSON URL使用PHP

Question

我使用YouTube API嘗試將來自多個URL的JSON數據放入mySQL數據庫。我試圖在多個JSON鏈接上使用json解碼器，此刻我的代碼刪除了第一個數組，並用第二個數組覆蓋它。請注意，我不想使用array_merge函數，因爲我想使用大約6-7個json URL。如果你可能非常具體，那也會很棒，因爲我對PHP很陌生。

$linkone = 'https://www.googleapis.com/youtube/v3/playlistItems?part=snippet&maxResults=25&playlistId={playlistID}&key={Key}'; 
$linktwo = 'https://www.googleapis.com/youtube/v3/playlistItems?part=snippet&maxResults=25&playlistId={PlaylistID}&key={Key}'; 


$content = file_get_contents($linkone); 
$content2 = file_get_contents($linktwo); 

$json = json_decode($content, true); 


foreach($json['items'] as $row) 
{ 
$title = $row['snippet']['title']; 
$title = mysql_real_escape_string($title); 
$description = $row['snippet']['description']; 
$description = mysql_real_escape_string($description); 

$publishedAt = $row['snippet']['publishedAt']; 
$publishedAt = mysql_real_escape_string($publishedAt); 
$high = $row['snippet']['thumbnails']['high']['url']; 
$high = mysql_real_escape_string($high); 

$sql = "INSERT INTO table(title, description, publishedAt, high) VALUES('".$title."','".$description."','".$publishedAt."','".$high."')"; 

if(!mysql_query($sql,$conn)) 
{ 
die('Error : ' . mysql_error()); 
} 
} 

Answer 1

$linkone = 'https://www.googleapis.com/youtube/v3/playlistItems?part=snippet&maxResults=25&playlistId={playlistID}&key={Key}'; 
$linktwo = 'https://www.googleapis.com/youtube/v3/playlistItems?part=snippet&maxResults=25&playlistId={PlaylistID}&key={Key}'; 

$jsonArray = array(
    json_decode(file_get_contents($linkone)), 
    json_decode(file_get_contents($linktwo)) 
); 

foreach($jsonArray as $json) { 
    foreach($json['items'] as $row) 
    { 
    $title = $row['snippet']['title']; 
    $title = mysql_real_escape_string($title); 
    $description = $row['snippet']['description']; 
    $description = mysql_real_escape_string($description); 

    $publishedAt = $row['snippet']['publishedAt']; 
    $publishedAt = mysql_real_escape_string($publishedAt); 
    $high = $row['snippet']['thumbnails']['high']['url']; 
    $high = mysql_real_escape_string($high); 

    $sql = "INSERT INTO table(title, description, publishedAt, high) 
     VALUES('".$title."','".$description."','".$publishedAt."','".$high."')"; 

    if(!mysql_query($sql,$conn)) 
    { 
     die('Error : ' . mysql_error()); 
    } 
    } 
}