的位置。假設我有以下字符串:分割字符串,請記住分裂
I have | been very busy lately and need to go | to bed early
通過分割上「|」,您將獲得:
$arr = array(
[0] => I have
[1] => been very busy lately and need to go
[2] => to bed early
)
第一分割是後2個字,然後是第二個分裂的8個單詞。存儲多少個單詞之後的位置將被存儲:array(2,8,3)。然後,該字符串崩盤傳遞到自定義字符串惡搞:
tag_string('I have been very busy lately and need to go to bed early');
我不知道爲tag_string的輸出將是究竟是什麼,除了總的話將保持不變。輸出示例如下:
I have-nn been-vb very-vb busy lately and-rr need to-r go to bed early-p
I-ee have been-vb very busy-df lately-nn and need-f to go to bed-uu early-yy
這會延長字符串中未知數量的字符。我無法控制tag_string。我所知道的是(1)單詞的數量將與之前相同,(2)數組在2之後分開,然後在8個單詞之後分開。我現在需要一個解決方案引爆標記串入同一陣列爲前:
$string = "I have-nn been-vb very-vb busy lately and-rr need to-r go to bed early-p"
function split_string_again() {
// split after 2nd, and thereafter after 8th word
}
隨着輸出:
$arr = array(
[0] => I have-nn
[1] => been-vb very-vb busy lately and-rr need to-r go
[2] => to bed early-p
)
所以要清楚(我是不是之前):我不能記住拆分strpos,因爲strtring在字符串前後經過tagger,是不一樣的。我需要計算單詞的數量。我希望我已經讓自己更清楚了:)
你可以爆炸它,然後'strlen'數組的每個部分 – seanbreeden 2012-02-13 17:04:32
我會對你想要做的事情感興趣,如果它更高級的話你可能會對[Rope數據結構]感興趣http://en.wikipedia.org/wiki/Rope_%28computer_science%29) - 雖然在PHP中實現它可能會「慢」。 – tplaner 2012-02-13 17:16:07
@evolve我想我現在更好地解釋它 - 請參閱OP的編輯。 – Pr0no 2012-02-14 00:05:09