您使用的是什么数据库?在PostgreSQL中，我喜欢使用ANY(数组)。重复一下你的例子:

<?php
$ids=array(1,2,3,7,8,9);
$db = new PDO(...);
$stmt = $db->prepare(
    'SELECT *
     FROM table
     WHERE id = ANY (:an_array)'
);
$stmt->bindParam('an_array',$ids);
$stmt->execute();
?>

不幸的是，这是非常不可移植的。

在其他数据库中，您将需要像其他人提到的那样，创建自己的魔法。当然，您需要将该逻辑放入类/函数中，以使其在整个程序中可重用。看看PHP mysql_query页面上的注释。关于这个主题的更多想法和这个场景的示例。

如果列只能包含整数，则可以不使用占位符，直接将id放入查询中。您只需将数组的所有值强制转换为整数。是这样的:

$listOfIds = implode(',',array_map('intval', $ids));
$stmt = $db->prepare(
    "SELECT *
     FROM table
     WHERE id IN($listOfIds)"
);
$stmt->execute();

这应该不会受到任何SQL注入的攻击。

当你有其他参数时，你可以这样做:

$ids = array(1,2,3,7,8,9);
$db = new PDO(...);
$query = 'SELECT *
            FROM table
           WHERE X = :x
             AND id IN(';
$comma = '';
for($i=0; $i<count($ids); $i++){
  $query .= $comma.':p'.$i;       // :p0, :p1, ...
  $comma = ',';
}
$query .= ')';

$stmt = $db->prepare($query);
$stmt->bindValue(':x', 123);  // some value
for($i=0; $i<count($ids); $i++){
  $stmt->bindValue(':p'.$i, $ids[$i]);
}
$stmt->execute();

你可以这样转换:

$stmt = $db->prepare('SELECT * FROM table WHERE id IN('.$in.')');

在此:

$stmt = $db->prepare('SELECT * FROM table WHERE id IN(:id1, :id2, :id3, :id7, :id8, :id9)');

然后用这个数组执行它:

$stmt->execute(array(
        :id1 =>1, :id2 =>2, :id3 =>3, :id7 =>7, :id8 =>8, :id9 => 9
    )
);

因此:

$in = array();
$consultaParam = array();
foreach($ids as $k => $v){
    $in[] = ':id'.$v;
    $consultaParam[':id'.$v] = $v;
}

最后的代码:

$ids = array(1,2,3,7,8,9);
$db = new PDO(...);

$in = array();
$consultaParam = array();
foreach($ids as $k => $v){
    $in[] = ':id'.$v;
    $consultaParam[':id'.$v] = $v;
}

$stmt = $db->prepare(
    'SELECT *
     FROM table
     WHERE id IN('.$in.')'
);
$stmt->execute($consultaParam);

在MySQL和PDO中，我们可以使用JSON数组和JSON_CONTAINS() (https://dev.mysql.com/doc/refman/8.0/en/json-search-functions.html#function_json-contains)进行搜索。

$ids = [123, 234, 345, 456]; // Array of users I search
$ids = json_encode($ids); // JSON conversion

$sql = <<<SQL
    SELECT ALL user_id, user_login
    FROM users
    -- Cast is mandatory beaucause JSON_CONTAINS() waits JSON doc candidate
    WHERE JSON_CONTAINS(:ids, CAST(user_id AS JSON))
    SQL;

$search = $pdo->prepare($sql);
$search->execute([':ids' => $ids]);
$users = $search->fetchAll();

我们也可以使用JSON_TABLE() (https://dev.mysql.com/doc/refman/8.0/en/json-table-functions.html#function_json-table)进行更复杂的情况和JSON数据探索:

$users = [
    ['id' => 123, 'bday' => ..., 'address' => ...],
    ['id' => 234, 'bday' => ..., 'address' => ...],
    ['id' => 345, 'bday' => ..., 'address' => ...],
]; // I'd like to know their login

$users = json_encode($users);

$sql = <<<SQL
    SELECT ALL user_id, user_login
    FROM users
    WHERE user_id IN (
        SELECT ALL user_id
        FROM JSON_TABLE(:users, '$[*]' COLUMNS (
            -- Data exploration...
            -- (if needed I can explore really deeply with NESTED kword)
            user_id INT PATH '$.id',
            -- I could skip these :
            user_bday DATE PATH '$.bday',
            user_address TINYTEXT PATH '$.address'
        )) AS _
    )
    SQL;

$search = $pdo->prepare($sql);
$search->execute([':users' => $users]);
...

以下是我的解决方案:

$total_items = count($array_of_items);
$question_marks = array_fill(0, $total_items, '?');
$sql = 'SELECT * FROM foo WHERE bar IN (' . implode(',', $question_marks ). ')';

$stmt = $dbh->prepare($sql);
$stmt->execute(array_values($array_of_items));

注意array_values的使用。这可以修复键排序问题。

我正在合并id数组，然后删除重复的项。我是这样写的:

$ids = array(0 => 23, 1 => 47, 3 => 17);

那就是失败。