regex - 正規表現 - ブラケットに一致しますが、結果から除外します

Question

私はこの例の文字列を持っています

[can be anything here %+^-_][can be anything here %+^-_][can be anything here %+^-_][can be anything here %+^-_][can be anything here %+^-_][can be anything here %+^-_][can be anything here %+^-_][can be anything here %+^-_]

私のパターンは(\[[^\]]+\])

結果としてこれを取得します

(
    [0] => Array
        (
            [0] => [can be anything here %+^-_]
            [1] => [can be anything here %+^-_]
            [2] => [can be anything here %+^-_]
            [3] => [can be anything here %+^-_]
            [4] => [can be anything here %+^-_]
            [5] => [can be anything here %+^-_]
            [6] => [can be anything here %+^-_]
            [7] => [can be anything here %+^-_]
        )

    [1] => Array
        (
            [0] => [can be anything here %+^-_]
            [1] => [can be anything here %+^-_]
            [2] => [can be anything here %+^-_]
            [3] => [can be anything here %+^-_]
            [4] => [can be anything here %+^-_]
            [5] => [can be anything here %+^-_]
            [6] => [can be anything here %+^-_]
            [7] => [can be anything here %+^-_]
        )

)

問題1

結果に2つの配列があるのはなぜですか? とにかく大したことではないのですが不思議です。

問題 2

正規表現のみを使用して、各配列値の最初と最後にある括弧を取り除くにはどうすればよいですか。このような。

[0] => Array
            (
                [0] => can be anything here %+^-_
                [1] => can be anything here %+^-_
                [2] => can be anything here %+^-_
                [3] => can be anything here %+^-_
                [4] => can be anything here %+^-_
                [5] => can be anything here %+^-_
                [6] => can be anything here %+^-_
                [7] => can be anything here %+^-_
            )