Medoo中有IN()函数吗？

Question

假设我有一段代码：

$this->database->debug()->select(
            'client',
            [
                'id',
                'name',
                'phone',
                'email',
                'address'
            ],
            Medoo::raw(
                'WHERE `id` IN(:clientIds)',
                [
                    ':clientIds' => $clientIDs
                ]
            )
        );

它给出了这个错误：

Notice: Undefined index: array in vendor\catfan\medoo\src\Medoo.php on line 519
Call Stack
#   Time    Memory  Function    Location
1   0.4014  404880  {main}( )   ...\test.php:0
2   0.4269  1080712 Writers\InvoicesWriter->Write( )    ...\test.php:9
3   0.4430  1136296 Writers\InvoicesWriter->getClientInfo( )    ...\InvoicesWriter.php:18
4   0.4430  1136752 Medoo\Medoo->select( )  ...\InvoicesWriter.php:83
5   0.4431  1136816 Medoo\Medoo->selectContext( )   ...\Medoo.php:1365
6   0.4433  1137496 Medoo\Medoo->whereClause( ) ...\Medoo.php:1075
7   0.4433  1137496 Medoo\Medoo->buildRaw( )    ...\Medoo.php:983
8   0.4434  1137496 Medoo\Medoo->typeMap( ) ...\Medoo.php:471

那么，我猜它不支持它了？如果没有，您认为我应该如何安全地使用IN()？

Answer 1

看起来像是is

$database->select("account", "user_name", [
    "OR" => [
        "user_id" => [2, 123, 234, 54],
        "email" => ["foo@bar.com", "cat@dog.com", "admin@medoo.in"]
    ]
]);
// WHERE
// user_id IN (2,123,234,54) OR
// email IN ('foo@bar.com','cat@dog.com','admin@medoo.in')

但我不认为您可以将它与原始对象一起使用。

Answer 2

是的。*这就是它是如何做到的*

$database->select("table_name", "column_name", 
    [
        "Value" => [option1, option2, option3, e.t.c]
  ]
);

• example

$database->select("users", "user_id", 
    [
      "age" => [12, 15, 18, 14]
    ]
);

上面的示例向>> "SELECT `user_id` FROM `users` WHERE `age` IN(12, 15, 18, 14)"解释

你可以找到更多的例子here