小心!此答案包含一个严重的SQL注入漏洞。不要使用这里给出的代码示例，除非确保任何外部输入都被清除。

$ids = join("','",$galleries);   
$sql = "SELECT * FROM galleries WHERE id IN ('$ids')";

正如Flavius Stef的回答，你可以使用intval()来确保所有id都是int值:

$ids = join(',', array_map('intval', $galleries));  
$sql = "SELECT * FROM galleries WHERE id IN ($ids)";

假设你事先正确地清理了你的输入……

$matches = implode(',', $galleries);

然后调整你的查询:

SELECT *
FROM galleries
WHERE id IN ( $matches ) 

根据数据集适当地引用值。

小心!此答案包含一个严重的SQL注入漏洞。不要使用这里给出的代码示例，除非确保任何外部输入都被清除。

$ids = join("','",$galleries);   
$sql = "SELECT * FROM galleries WHERE id IN ('$ids')";

更安全。

$galleries = array(1,2,5);
array_walk($galleries , 'intval');
$ids = implode(',', $galleries);
$sql = "SELECT * FROM galleries WHERE id IN ($ids)";

对于带有转义函数的MySQLi:

$ids = array_map(function($a) use($mysqli) { 
    return is_string($a) ? "'".$mysqli->real_escape_string($a)."'" : $a;
  }, $ids);
$ids = join(',', $ids);  
$result = $mysqli->query("SELECT * FROM galleries WHERE id IN ($ids)");

对于带准备语句的PDO:

$qmarks = implode(',', array_fill(0, count($ids), '?'));
$sth = $dbh->prepare("SELECT * FROM galleries WHERE id IN ($qmarks)");
$sth->execute($ids);