如何从 While 循环 PHP 中正确填充数组

Question

我正在从 While 循环填充一个数组，但无法生成正确的数组格式。我遇到了许多不同的变化，但不是我正在寻找的，包括在下面。目的是用于第 3 方 SMTP，如变量 $to 中所示

所需的数组：

array(3) { 
    ["email1@email1.com"]=> array(1) { ["userid"]=> string(1) "6" } 
    ["email2@email2.com"]=> array(1) { ["userid"]=> string(2) "64" } 
    ["email3@email3.com"]=> array(1) { ["userid"]=> string(3) "503" } 
}

尝试 1：

$str = array();
while($userEmails = $query->fetch(PDO::FETCH_ASSOC)){

    $str[] = '"'.$userEmails['user_email'].'"  => array( 
       "userid" => "'.$userEmails['user_id'].'" 
    )';
}
$to = $str;

尝试 1 不正确的数组：

array(3) {
     [0]=> string(54) ""email1@email1.com" => array( "userid" => "6" )"       
     [1]=> string(54) ""email2@email2.com" => array( "userid" => "64" )" 
     [2]=> string(54) ""email3@email3.com" => array( "userid" => "503" )" 
}

尝试 2：

$str = "";
while($userEmails = $query->fetch(PDO::FETCH_ASSOC)){

    $str .= '"'.$userEmails['user_email'].'"  => array( "userid" => "'.$userEmails['user_id'].'" )';
}

$emails = rtrim($str, ", ");

$to = array($emails);

尝试 2 不正确的数组：

array(1) { 
    [0]=> string(162) ""email1@email1.com" => array( "userid" => "6" )"email2@email2.com" => array( "userid" => "64" )"email3@email3.com" => array( "userid" => "503" )" }

Answer 1

你想要这个，我想：

$str = array();
while($userEmails = $query->fetch(PDO::FETCH_ASSOC)){
    $str[$userEmails['user_email']] = array(
        'user_id' => $userEmails['user_id']
    );
}

Answer 2

我个人认为将while 与fetch 一起使用会迫使您接受丑陋的代码。我通常有几个实用函数来编写更多的声明性代码

function query_reduce(callable $f, $initialValue, $query) {
  $result = $initialValue;
  while($x = $query->fetch(PDO::FETCH_ASSOC))
    $z = call_user_func($f, $x);
  return $result;
}

现在你可以写这个了

$result = query_reduce(function ($acc, $x) {
  $acc[$x['user_email']] = [['user_id'] => $x['user_id']];
  return $acc;
}, [], $query);

var_dump($result);

---

query_reduce 只是我使用的助手之一。这是一个类似于foreach

// use like foreach
function query_foreach(callable $f, $query) {
  query_reduce(function($i, $x) {
    call_user_func($f, $i++);
  }, 0, $query);
}

// $query returns [{id:'a'}, {id:'b'}, {id:'c'}]
query_foreach(function ($x, $i) {
  echo "{$x['id']} is in position {$i}";
});

// output
// a is in position 0
// b is in position 1
// c is in position 2

或用于映射查询的一种

// use like array_map
function query_map(callable $f, $query) {
  $i = 0;
  return query_reduce(function ($acc, $x) use ($i) {
    return array_merge($acc, [call_user_func($f, $x, $i++)]);
  }, [], $query);
}

// $query returns [{id:'a'}, {id:'b'}, {id:'c'}]
$ids = query_foreach(function ($x, $i) {
  return ['id'=> $x['id'], 'position'=> $i];
});

print_r($ids);

// output...
// Array
// (
//     Array
//     (
//         [id] => a
//         [position] => 0
//     )
//     Array
//     (
//         [id] => b
//         [position] => 1
//     )
//     Array
//     (
//         [id] => c
//         [position] => 2
//     )
// )

Answer 3

试试这个方法：

$str = array();
while($userEmails = $query->fetch(PDO::FETCH_ASSOC)){
    $str[$userEmails['user_email']] = array('userid' => $userEmails['user_id']);
}

$to = $str;