我可以在PDO查询中绑定一个数组到IN()条件吗?

我很好奇是否有可能使用PDO将值数组绑定到占位符。这里的用例试图传递一个值数组，以便与IN()条件一起使用。

我希望能够这样做:

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

并让PDO绑定并引用数组中的所有值。

目前我正在做:

<?php
$ids = array(1,2,3,7,8,9);
$db = new PDO(...);
foreach($ids as &$val)
    $val=$db->quote($val); //iterate through array and quote
$in = implode(',',$ids); //create comma separated list
$stmt = $db->prepare(
    'SELECT *
     FROM table
     WHERE id IN('.$in.')'
);
$stmt->execute();
?>

这当然是工作，但只是想知道是否有一个内置的解决方案，我错过了?

当前回答

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]);
...

2021-04-08 22:16:25

其他回答

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]);
...

2021-04-08 22:16:25

使用postgres数组("{}")是一种非常简洁的方法:

$ids = array(1,4,7,9,45);
$param = "{".implode(', ',$ids)."}";
$cmd = $db->prepare("SELECT * FROM table WHERE id = ANY (?)");
$result = $cmd->execute(array($param));

2014-04-15 13:07:04

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);

2022-10-03 10:55:10

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。

2009-05-28 12:08:51

您必须构造查询字符串。

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

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

// bindvalue is 1-indexed, so $k+1
foreach ($ids as $k => $id)
    $stmt->bindValue(($k+1), $id);

$stmt->execute();
?>

chris(评论)和一些人都有麻烦了，建议foreach-loop…

(...)
// bindvalue is 1-indexed, so $k+1
foreach ($ids as $k => $id)
    $stmt->bindValue(($k+1), $id);

$stmt->execute();

．.．可能是多余的，因此foreach循环和$stmt->执行可以被替换为…

<?php 
  (...)
  $stmt->execute($ids);

2009-05-28 12:02:04

我可以在PDO查询中绑定一个数组到IN()条件吗?

推荐文章

最新文章

标签