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();

简单来说:

//$db = new PDO(...);
//$ids = array(...);

$qMarks = str_repeat('?,', count($ids) - 1) . '?';
$sth = $db->prepare("SELECT * FROM myTable WHERE id IN ($qMarks)");
$sth->execute($ids);

稍微编辑一下施纳勒的代码

<?php
$ids     = array(1, 2, 3, 7, 8, 9);
$inQuery = implode(',', array_fill(0, count($ids)-1, '?'));

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

foreach ($ids as $k => $id)
    $stmt->bindValue(($k+1), $id);

$stmt->execute();
?>

//implode(',', array_fill(0, count($ids)-1), '?')); 
//'?' this should be inside the array_fill
//$stmt->bindValue(($k+1), $in); 
// instead of $in, it should be $id

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);

您首先在查询中设置“?”的个数，然后由一个“for”发送参数 像这样:

require 'dbConnect.php';
$db=new dbConnect();
$array=[];
array_push($array,'value1');
array_push($array,'value2');
$query="SELECT * FROM sites WHERE kind IN (";

foreach ($array as $field){
    $query.="?,";
}
$query=substr($query,0,strlen($query)-1);
$query.=")";
$tbl=$db->connection->prepare($query);
for($i=1;$i<=count($array);$i++)
    $tbl->bindParam($i,$array[$i-1],PDO::PARAM_STR);
$tbl->execute();
$row=$tbl->fetchAll(PDO::FETCH_OBJ);
var_dump($row);

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。