显示数据库中的数组

Question

我有这个数据库

id | domain    | whois_server         |
---------------------------------------
1  | ac        | whois.nic.ac         |
2  | ae        | whois.nic.ae         |
3  | id        | whois.pandi.or.id    |

如何显示没有id的数组？

array (
    'ac' => 'whois.nic.ac',
    'ae' => 'whois.nic.ae',
    'id' => 'whois.pandi.or.id',
)

我完成了以下代码：

$sql = "SELECT * FROM whois_server ";
$result = $db->prepare($sql);
$result->execute();
while ($row = $result->fetch(PDO::FETCH_ASSOC)) {
    $whoisservers =  array_merge($whoisservers, array_map('trim', explode(",", $row[1])));
}

但出现的是：

array (
    [0] => whois.nic.ac
    [1] => whois.nic.ae
    [2] => whois.pandi.or.id
)

Answer 1

$sql = "SELECT * FROM whois_server ";
$result = $db->prepare($sql);
$result->execute();
$dataReturn = [];
while ($row = $result->fetch(PDO::FETCH_ASSOC)) {
      $dataReturn[$row['domain']] = $row['whois_server'];
}
print_r($dataReturn);

Answer 2

在您的循环中，您可以将键添加到每个值的$whoisseervers变量中：

$sql = "SELECT domain, whois_server FROM whois_server";
$result = $db->prepare($sql);
$result->execute();

$whoisservers = [];
while ($row = $result->fetch(PDO::FETCH_ASSOC)) {
    $whoisservers[$row['domain']] =  $row['whois_server'];
}

print_r($whoisservers);

Answer 3

$whoisservers = array();
while ($row = $result->fetch(PDO::FETCH_ASSOC)) {
  $whoisservers[$row["domain"]] = $row["whois_server"];
}

会更简单(也更清晰)。

正如注释中提到的，在SQL中不要选择不需要的字段。