Question

我正在从JSON链接解析数据，然后我想将它插入到SQL Server 2012中的一个表中。解析运行良好，我唯一的问题是将数据插入SQL Server。

示例JSON数据：

"results": [   {
   "name": "The JPT Look On Do More - Commercial Transformation",
   "metrics": {
    "video": {
     "playthrough_75": "2981",
     "plays": "5466",
     "time_watched": "4030502288",
     "playthrough_100": "2288",
     "uniq_plays": {
      "monthly_uniqs": "4731",
      "daily_uniqs": "5136",
      "weekly_uniqs": "4991"
     },

脚本：

$context = stream_context_create($opts);
$content = file_get_contents($url, false, $context);
$json = json_decode($content, true);

foreach($json['results'] as $item)
{
$VideoName = $item['name'];
$playthrough_75 = $item['metrics']['video']['playthrough_75'];
$plays = $item['metrics']['video']['plays'];
$time_watched = $item['metrics']['video']['time_watched'];
$playthrough_100 = $item['metrics']['video']['playthrough_100'];
$plays_monthly_uniqs = $item['metrics']['video']['uniq_plays']['monthly_uniqs'];
$plays_daily_uniqs = $item['metrics']['video']['uniq_plays']['daily_uniqs'];
$plays_weekly_uniqs = $item['metrics']['video']['uniq_plays']['weekly_uniqs'];

$serverName = "SERVER_NAME";
$connectionInfo = array("Database"=>"Web_Analytics");
$conn = sqlsrv_connect($serverName, $connectionInfo);

*INSERT SQL SCRIPT HERE* 
}

因此，从上面的脚本中，我将每个字段存储到一个变量中，然后我想要做的是在下面执行此查询。我怎么能用PHP执行它？

$sql = "INSERT INTO Ooyala_Analytics([VideoName]
      ,[VideoId]
      ,[playthrough_75]
      ,[plays]
      ,[time_watched]
      ,[playthrough_100]
      ,[plays_monthly_uniqs]
      ,[plays_daily_uniqs]
      ,[plays_weekly_uniqs]
      ,)
      VALUES('$VideoName'
        ,'$VideoId'
        ,'$playthrough_75'
        ,'$plays'
        ,'$time_watched'
        ,'$playthrough_100'
        ,'$plays_monthly_uniqs'
        ,'$plays_daily_uniqs'
        ,'$plays_weekly_uniqs' 
        ')";
    }

使用PHP将Json数据插入SQL Server 2012

0 个答案: