【问题标题】:PHP store array into seperate columnPHP将数组存储到单独的列中
【发布时间】:2018-04-18 08:17:40
【问题描述】:

目前我有一个数组:

$movie_seat = "seat1, seat2, seat3, seat4";

foreach($movie_seat as $keys =>$values)
        {
        $movie_id = $values["movie_id"];
        $movie_seats = $values["movie_seats"];

            $query = "INSERT INTO `testing` (`movie_id`,`movie_seat`) VALUES ('$movie_id ','$movie_seats')" ;

            $addmoviequery = mysqli_query($conn,$query);
            if(!$addmoviequery){
                die("QUERY FAILED".mysqli_error($conn));

            }

从数组中,我将得到以下结果存储在数据库中。它将整个数组存储到movie_seat列中。

+-----------------+-------------------------------+
| movie_id        | movie_seat                    |
+-----------------+-------------------------------+
| movie01         | seat1, seat2, seat3           |
| movie02         | seat5,seat6                   |
+-----------------+-------------------------------+

但是,我想记录到数据库中的不同列

我想要的结果:

 +-----------------+-------+-------+-------+-------+-------+-------+
 | movie_id        | seat1 | seat2 | seat3 | seat4 | seat5 | seat6 |
 +-----------------+-------+-------+-------+-------+-------+-------+
 | movie01         | booked|booked |booked |       |       |       |
 | movie02         |       |       |       |       |booked |booked |
 +-----------------+-------+-------+-------+-------+-------+-------+

我尝试使用 explode 功能,但它会显示

警告:在 C:\

中为 foreach() 提供的参数无效

【问题讨论】:

  • 首先在foreach 中使用$movie_seat - 而不是movie_seat
  • 是的..我错过了 $..

标签: php sql arrays mysqli


【解决方案1】:

您正在执行的操作存在多个错误。

$movie_seat = "seat1, seat2, seat3, seat4"; 

这不是array。这是string

你必须把explode它放到一个数组中,像这样:

$seats = explode(",", $movie_seat);

这将创建一个名为$seats 的字符串array,其中$seats[0] 等于“seat1”等...

然后您需要使用您创建的新数组而不是旧变量来使用您的foreach。像这样:

foreach($seats as $key => $value)
{
    //Your code goes here, 
    //First iteration should give you $key = 0, $value = "seat1", etc...
}

【讨论】:

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