Question

Haven还能找到一个直接的答案..我的阵列$ sitematches：

if (preg_match('/<start>(.*?)<finish>/s', $source, $matches)) {
    if (preg_match_all('/<a\s[^>]*href=\"([^\"]*)\"[^>]*>(.*)<\/a>/siU', $matches[1], $sitematches)); {
    print_r($sitematches);
    }
}

产生

Array
(
    [0] => Array
        (
            [0] => <A HREF="http://site1.com">aaa</A>
            [1] => <A HREF="http://site2.com">bbb</A>
            [2] => <A HREF="http://site3.com">ccc</A>
            [3] => <A HREF="http://site4.com">ddd</A>
        )

    [1] => Array
        (
            [0] => http://site1.com
            [1] => http://site2.com
            [2] => http://site3.com
            [3] => http://site4.com
        )

    [2] => Array
        (
            [0] => aaa
            [1] => bbb
            [2] => ccc
            [3] => ddd
        )

)

我如何输出：

1 - aaa - http://site1.com
2 - bbb - http://site2.com
3 - ccc - http://site3.com
4 - ddd - http://site4.com

Answer 1

for ($i = 0; $i < count($sitematches[0]); $i++)
    print "$i - {$sitematches[2][$i]} - {$sitematches[1][$i]}\n";

只需获取大小（count（$ sitematches [0]））然后遍历数组。

Answer 2

foreach的另一个可能性是：

foreach ($sitematches[1] as $key => $url)
   print "$key - {$sitematches[2][$key]} - $url\n";

但至少需要一次直接阵列访问。

从preg_match_all打印多维数组

2 个答案: