Php数组插入不是最后一个，而是先

Question

I am trying to sort by publishedAt. The sql is inserting from json data and I need to get the newest, but this code getting the oldest.

What is wrong with my code?

$loop = mysqli_query($conn, "SELECT channelid, videoId FROM users ORDER BY id")
    or die (mysqli_error($conn));

    while ($row = mysqli_fetch_array($loop))
    {
        $channelid = $row['channelid'];
        $url = 'https://www.googleapis.com/youtube/v3/search?order=date&part=snippet&channelId='.$channelid .'&maxResults=5&key=KEY';
        $content = file_get_contents($url); 
        $json = json_decode($content, true);    

        if(!isset($json['items'])) continue; //skip if no items

        $videos = ['videoId'=>'','videoId1'=>'','videoId2'=>'','videoId3'=>'','videoId4'=>''];
        $i = 0;

        foreach($json['items'] as $items)
        {
            $publishedAt = $items['snippet']['publishedAt'];
            $key  = 0==$i ? 'videoId' : 'videoId'.$i;
            $videos[$key] = $items['id']['videoId']; 
            ++$i;
        }
        $ii = 0;


        $sql = "INSERT INTO users(channelid, publishedAt, videoId, videoId1, videoId2, videoId3, videoId4) VALUES('$channelid', '$publishedAt', '{$videos['videoId']}', '{$videos['videoId1']}', '{$videos['videoId2']}', '{$videos['videoId3']}', '{$videos['videoId4']}')    
            ON DUPLICATE KEY UPDATE channelid=VALUES(channelid), publishedAt=VALUES(publishedAt), videoId=VALUES(videoId), videoId1=VALUES(videoId1), videoId2=VALUES(videoId2), videoId3=VALUES(videoId3), videoId4=VALUES(videoId4);
            ";
        mysqli_query($conn, "SET NAMES utf8");
        if(!mysqli_query($conn,$sql))
        {
            die('Error : ' . mysqli_error($conn));
        }
    }

Answer 1

When you're looping through $json['items'] you're overwriting $publishedAt with the last item for each loop.

If you only want the publishedAt of your first item you should get the first item directly (no loop necessary). But if you want to insert every items you should put your query inside the foreach.