我正在尝试使用类似于SQL的语法来查询数组,并且我开始了解YaLinqo。
我设法让它对带有上级或下级运算符的where子句起作用,但我不能让它对等于运算符起作用。
我做错了什么?
下面是一个示例:
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);发布于 2015-10-07 19:32:42
我发现使用==而不是=是可行的:
// 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();https://stackoverflow.com/questions/32989067
复制相似问题