PHP：使用preg_ *函数获取数组

Question

I want to parser and get a two-dimensional matrix using regular expresions of PHP. But I really don't know to parser and get a array(unidimensional). For example, I have this string:

    $str = {a}&{b}&{c}&{d&};

and I want to convert to array like this:

    array( 'a', 'b', 'c', 'd&' )

The pattern should be this:

    $pattern = '#(({[^}]+})(?:&(?1))*#';

The pattern matches correctly but I can't get all elements with $matches parameter:

    preg_match_all( $pattern, $str, $matches );
    print_r( $matches );

Output:

    Array
    (
        [0] => Array
            (
                [0] => {a}&{b}&{c}&{d&}
            )

        [1] => Array
            (
                [0] => a
            )

    )

I try to solve this problem involving the recursive pattern with parentheses.

    $pattern = '#(({[^}]+})(?:&((?1)))*#';

Output:

    Array
    (
        [0] => Array
            (
                [0] => {a}&{b}&{c}&{d&}
            )

        [1] => Array
            (
                [0] => a
            )

        [2] => Array
            (
                [0] => d&
            )

    )

Only is gotten the last element of array. The option flags PREG_PATTERN_ORDER or con PREG_SET_ORDER haven't solved my problem.

Is there any way to get the array of elements?

Greetings and thanks!

Answer 1

You can use this pattern which captures the content inside curly brackets and checks the syntax (i.e. {..}&{..}&{..}&{..}) :

preg_match_all('~\G{([^}]+)}(?:&|$)~', '{a}&{b}&{c}&{d&}', $result);

\G means "contiguous to a precedent match or at the start of the string". Thus, gaps are not allowed and the syntax is checked by the pattern.

The array you are looking for is $result[1]