YaLinqo - 获取包含等于运算符的 where 子句的正确结果

Question

我正在尝试使用类似 SQL 的语法查询一个数组，并且我正在了解 YaLinqo。

我设法让它适用于具有优越或劣等运算符的 where 子句，但我不能让它与等于运算符一起使用。

我做错了什么？

这里是一个例子：

require_once __DIR__ . '/vendor/autoload.php';

use \YaLinqo\Enumerable;

$users = [
    [
        'UserId' => '1',
        'username' => 'joe',
        'password' => 'joepw',
        'mail' => 'joe@mail.com'
    ],
    [
        'UserId' => '2',
        'username' => 'nancy',
        'password' => 'nancypw',
        'mail' => 'nancy@mail.com'
    ],
    [
        'UserId' => '3',
        'username' => 'alice',
        'password' => 'alicepw',
        'mail' => 'alice@mail.com'
    ]
];

$working = \YaLinqo\Enumerable::from($users)
    ->where('$users ==> $users["UserId"] > 2')
    ->toArray();

$notWorking = \YaLinqo\Enumerable::from($users)
    ->where('$users ==> $users["UserId"] = 2')
    ->toArray();

$workingAndUgly = \YaLinqo\Enumerable::from($users)
    ->where('$users ==> $users["UserId"] > 1')
    ->where('$users ==> $users["UserId"] < 3')
    ->toArray();

$notWorkingEither = \YaLinqo\Enumerable::from($users)
    ->where('$users ==> $users["username"] = "nancy"')
    ->toArray();

var_dump($working, $notWorking, $workingAndUgly, $notWorkingEither);

Answer 1

我发现使用== 而不是= 可以：

// previously $notWorking
$nowWorking = \YaLinqo\Enumerable::from($users)
    ->where('$users ==> $users["UserId"] == 2')
    ->toArray();

// previously $notWorkingEither
$nowWorkingAlso = \YaLinqo\Enumerable::from($users)
->where('$users ==> $users["username"] == "nancy"')
->toArray();