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这个是pdo的bug吗?

程序员文章站 2022-04-26 09:57:42
...
$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,结果返回为空;如果把678689287放在第一位,就有返回。

是不是个bug?

回复内容:

$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,结果返回为空;如果把678689287放在第一位,就有返回。

是不是个bug?

用预处理的的select...in语句,你得生成与查询数量相同的占位符

$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不支持这种数组绑定

和我以前犯的错误一样,in后面的数值范围,有多少个数值,就来多少个问号,一个问号不行。

相关标签: php pdo