php - 如何根据表中的特定字段回显变量？

Question

这是我想要实现的一个非常简化的示例：

GIFT
sender  |  type
phil    |  1
phil    |  2

在这里，1 = 蛋糕，2 = 松饼。

$table = query("SELECT sender, type FROM users WHERE sender = Phil");
foreach ($table as $row) {
$sender = $row['sender'];
$type = $row['type'];

echo '$sender has bought a $type';
}

这将输出：

Phil has bought a 1
Phil has bought a 2 

我怎样才能得到下面的输出呢？

Phil has bought a cake
Phil has bought a muffin

我应该使用数组吗？

 $type = array(
 1 => 'cake',
 2 => 'muffin'
 );

问题是 $type 已经定义为 $row['type']。

Answer 1

使用将数字与名称相关联的关联数组：

$types = array(1 => 'cake', 2 => 'muffin');

然后将其用作：

echo "$sender has bought a {$types[$type]}";

请注意，对于要在字符串内展开的变量，您必须使用双引号，而不是代码中的单引号。

Answer 2

或者，如果您想从数据库中返回它：

$table = query("SELECT sender, 
                case when type = '1' then 'cake' 
                     when type = '2' then 'muffin'
                end as type 
                FROM users WHERE sender = Phil");

foreach ($table as $row) {
  $sender = $row['sender'];
  $type = $row['type'];

  echo "$sender has bought a $type"; //Use double quotes for $sender and $type to be interpolated.
}

Answer 3

鉴于您的示例，我将使用这样的数组：

$items = array(
    "1" => "cake",
    "2" => "muffin"
);

然后这样做：

echo "$sender has bought a {$items[$type]}";