php - 如何从准备好的pdo中输出数组

Question

我正在尝试编写一个准备好的语句，该语句从一组绑定(bind)字段收集数据并将它们输出到名为 $data 的数组变量。

global $db;
$user_id = (int)$user_id;
$fields = 'field1, field2, field3';
$stmt = $db->prepare("SELECT :fields FROM users WHERE user_id = :user_id");
$stmt->bindValue(':user_id', $user_id);
$stmt->bindValue(':fields', $fields);
$stmt->execute();
$result = $stmt->fetchall();

foreach ($result as $row) {
    $data[] = array(
        'field1' => $row['field1'], 
        'field2' => $row['field2'], 
        'field3' => $row['field3']
    );
}

print_r($data);

然而我得到的只是......

Notice: Undefined index: field1 in C:\\\\\\.php on line #
Notice: Undefined index: field2 in C:\\\\\\.php on line #
Notice: Undefined index: field3 in C:\\\\\\.php on line #
Array ( [field1] => [field2] => [field3] => )

这是 vardump(注意，这里的代码是我的代码的副本。转储是删除了字段名称的副本和帖子。)

array(1) { [0]=> array(2) { ["field1, field2, field3"]=> string(22) "field1, field2, field3" [0]=> string(22) "field1, field2, field3" } } Array ( [field1] => [field2] => [field3] => )

这可能非常简单..但我现在正在从错误中学习。

此外，我想将 $field 变量从字符串更改为数组。如何绑定(bind)一个数组来处理每条数据，然后将其输出到 $data 数组中？

Answer 1

假设 user_id 是主键，则不需要 foreach 循环

$fields = 'field1, field2, field3';
$stmt = $db->prepare("SELECT $fields FROM users WHERE user_id = ?");
$stmt->execute(array(1));
$result = $stmt->fetch(PDO::FETCH_ASSOC); 

$data = array(
    'field1' => $result['field1'], 
    'field2' => $result['field2'], 
    'field3' => $result['field3']
);

print_r($data);

我的意思是，如果两个用户可以有相同的 id，为什么还要有 user_id