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

我扩展了PDO,做一些类似stefs建议的事情,从长远来看,这对我来说更容易:

class Array_Capable_PDO extends PDO {
    /**
     * Both prepare a statement and bind array values to it
     * @param string $statement mysql query with colon-prefixed tokens
     * @param array $arrays associatve array with string tokens as keys and integer-indexed data arrays as values 
     * @param array $driver_options see php documention
     * @return PDOStatement with given array values already bound 
     */
    public function prepare_with_arrays($statement, array $arrays, $driver_options = array()) {

        $replace_strings = array();
        $x = 0;
        foreach($arrays as $token => $data) {
            // just for testing...
            //// tokens should be legit
            //assert('is_string($token)');
            //assert('$token !== ""');
            //// a given token shouldn't appear more than once in the query
            //assert('substr_count($statement, $token) === 1');
            //// there should be an array of values for each token
            //assert('is_array($data)');
            //// empty data arrays aren't okay, they're a SQL syntax error
            //assert('count($data) > 0');

            // replace array tokens with a list of value tokens
            $replace_string_pieces = array();
            foreach($data as $y => $value) {
                //// the data arrays have to be integer-indexed
                //assert('is_int($y)');
                $replace_string_pieces[] = ":{$x}_{$y}";
            }
            $replace_strings[] = '('.implode(', ', $replace_string_pieces).')';
            $x++;
        }
        $statement = str_replace(array_keys($arrays), $replace_strings, $statement);
        $prepared_statement = $this->prepare($statement, $driver_options);

        // bind values to the value tokens
        $x = 0;
        foreach($arrays as $token => $data) {
            foreach($data as $y => $value) {
                $prepared_statement->bindValue(":{$x}_{$y}", $value);
            }
            $x++;
        }

        return $prepared_statement;
    }
}

你可以这样使用它:

$db_link = new Array_Capable_PDO($dsn, $username, $password);

$query = '
    SELECT     *
    FROM       test
    WHERE      field1 IN :array1
     OR        field2 IN :array2
     OR        field3 = :value
';

$pdo_query = $db_link->prepare_with_arrays(
    $query,
    array(
        ':array1' => array(1,2,3),
        ':array2' => array(7,8,9)
    )
);

$pdo_query->bindValue(':value', '10');

$pdo_query->execute();

其他回答

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);

EvilRygy的解决方案对我不起作用。在Postgres中,你可以做另一种解决方法:


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

使用IN语句有这么重要吗?尝试使用FIND_IN_SET操作。

例如,在PDO中有这样一个查询

SELECT * FROM table WHERE FIND_IN_SET(id, :array)

然后你只需要绑定一个数组的值,用逗号括起来,就像这个

$ids_string = implode(',', $array_of_smth); // WITHOUT WHITESPACES BEFORE AND AFTER THE COMMA
$stmt->bindParam('array', $ids_string);

做完了。

UPD:正如一些人在评论中指出的,有一些问题需要明确说明。

FIND_IN_SET没有在表中使用索引,而且它还没有实现——请参阅MYSQL错误跟踪器中的这条记录。感谢@BillKarwin的通知。 不能使用内部带有逗号的字符串作为数组的搜索值。这是不可能在内爆后以正确的方式解析这样的字符串,因为你使用逗号符号作为分隔符。感谢@VaL的提示。

总之,如果您不严重依赖索引,并且不使用带逗号的字符串进行搜索,那么我的解决方案将比上面列出的解决方案更容易、更简单、更快。

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

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]);
...