如何使用preg_split捕获不同数组或指定键中的分隔符

Question

更新:第一个问题解决了，是HTML/浏览器的问题

第二个问题:是否有办法将分隔符输出到单独的数组中?如果我使用PREG_SPLIT_DELIM_CAPTURE，它会将分隔符混合到同一个数组中，并使其混淆，而不是PREG_SPLIT_OFFSET_CAPTURE，它为偏移量指定自己的键。

代码:

$str = 'world=earth;world!=mars;world>venus';
$arr = preg_split('/([;|])/', $str, -1, PREG_SPLIT_DELIM_CAPTURE);
echo "<pre>";
print_r($arr);
echo "</pre>";

DELIM CAPTURE示例:

Array
(
    [0] => world=earth
    [1] => ;
    [2] => world!=mars
    [3] => ;
    [4] => world>venus
)

OFFSET CAPTURE示例:

Array
(
    [0] => Array
        (
            [0] => world=earth
            [1] => 0
        )
    [1] => Array
        (
            [0] => world!=mars
            [1] => 12
        )
    [2] => Array
        (
            [0] => world>venus
            [1] => 34
        )
)

Answer 1

从技术上讲，您不能这样做，但在这种特殊情况下，您可以在事后很容易地做到这一点:

print_r(array_chunk($arr, 2));

输出:

Array
(
    [0] => Array
        (
            [0] => world=earth
            [1] => ;
        )
    [1] => Array
        (
            [0] => world!=mars
            [1] => ;
        )
    [2] => Array
        (
            [0] => world>venus
        )
)

参见:array_chunk()

Answer 2

你不能!

一种方法是用preg_match_all代替preg_split来获得你想要的:

$pattern = '~[^;|]+(?=([;|])?)~';
preg_match_all($pattern, $str, $arr, PREG_SET_ORDER);

的想法是在前瞻性中放置一个可选的捕获组，用于捕获分隔符。

如果您想要确保不同的结果是连续的，您必须将这个检查添加到模式中:

$pattern = '~G[;|]?K[^;|]+(?=([;|])?)~';

模式细节:

G           # the end of the last match position ( A at the begining )
[;|]?        # optional delimiter (the first item doesn't have a delimiter)
K           # reset all that has been matched before from match result
[^;|]+       # all that is not the delimiter
(?=          # lookead: means "followed by"
    ([;|])?  # capturing group 1 (optional): to capture the delimiter 
)            # close the lookahead

Answer 3

当我运行PHP代码时，这是我从Chrome浏览器得到的结果:

<pre>Array
(
    [0] =&gt; world=earth
    [1] =&gt; world!=mars
    [2] =&gt; world<sun [3]=""> world&gt;venus
)
</sun></pre>

因此preg_split函数运行良好。这只是HTML中数组转储的问题