這個是pdo的bug嗎？

$user='test';$pass='123456';$dbh = new PDO('mysql:host=127.0.0.1;dbname=test', $user, $pass);$sth = $dbh->prepare("SELECT * FROM user WHERE id IN (?) ORDER BY id ASC");$sth->execute(['456089015, 456089016, 456089017, 456089018, 456089019, 456089020, 456089021, 456089022, 456089023, 456089024, 456089025, 678689287']);$result = $sth->fetchAll();print_r($result);

其中只有678689287是存在的，結果返回為空白；如果把678689287放在第一位，就有返回。

是不是個bug？

回複內容：

$user='test';$pass='123456';$dbh = new PDO('mysql:host=127.0.0.1;dbname=test', $user, $pass);$sth = $dbh->prepare("SELECT * FROM user WHERE id IN (?) ORDER BY id ASC");$sth->execute(['456089015, 456089016, 456089017, 456089018, 456089019, 456089020, 456089021, 456089022, 456089023, 456089024, 456089025, 678689287']);$result = $sth->fetchAll();print_r($result);

其中只有678689287是存在的，結果返回為空白；如果把678689287放在第一位，就有返回。

是不是個bug？

用預先處理的的select...in語句，你得產生與查詢數量相同的預留位置

$queryParams = [456089015, 456089016, 456089017, 456089018, 456089019, 456089020, 456089021, 456089022, 456089023, 456089024, 456089025, 678689287];$markers = str_repeat('?,', count($queryParams) - 1) . '?';$st = $dbh->prepare("SELECT * FROM user WHERE id IN (${markers}) ORDER BY id ASC");$st->execute($queryParams);$result = $st->fetchAll();

pdo不支援這種數組綁定

和我以前犯的錯誤一樣，in後面的數值範圍，有多少個數值，就來多少個問號，一個問號不行。

