Question

我正在尝试按PublishedAt排序。 sql是从json数据插入的，我需要获取最新的，但是此代码获取的最旧。

我的代码有什么问题？

$loop = mysqli_query($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

当您遍历$json['items']时，您将用每个循环的最后一项覆盖$publishedAt。

如果只希望第一项的publishedAt，则应直接获取第一项（无需循环）。但是，如果要插入所有项目，则应将查询放入foreach中。

PHP数组插入不是最后而是首先

1 个答案: