Re: [問題] 也是另一個比對問題?
真的沒人有什麼好方法可以建議= =
總覺得應該有點什麼規律或好的方法才對
若我先將 ACT with PTN PTN 用hash存成key
class2:ACT PTN PTN 存成value
再進行更進一步的處理呢???
以取得4個相對應的字呢???
※ 引述《Yaowei (開心)》之銘言:
: 若我知道 以下這些條件
: ACT with PTN PTN 與 class2:ACT PTN PTN
: ACT of PTN with PTN and PTN 與 class2:ACT PTN PTN
: ACT PTN and PTN 與 class2:ACT PTN PTN
: with PTN to ACT PTN . 與 class1:PTN ACT PTN
: PTN and PTN ACT with and PTN 與 class1:PTN ACT PTN
: . PTN ACT with PTN in 與 class1:PTN ACT PTN
: ACT of PTN PTN ACT 與 class3:PTN PTN ACT
: PTN in PTN ACT of PTN ACT by 與 class3:PTN PTN ACT
: ACT with PTN PTN ACT 與 class3:PTN PTN ACT
: 例子一:
: ACT with PTN PTN 想取到下面()內的字
: ==> () ACT (with) PTN () PTN ()
: 即我想取得 class2:ACT PTN PTN 的前後與中間兩個範圍內的字,若沒有字則以空白代替
: ==> ($left, $middle1, $middle2, $right) = ("","with","","");
: 例子二:
: with PTN to ACT PTN . 與 class1:PTN ACT PTN
: ==> (with) PTN (to) ACT () PTN (.)
: ==> ($left, $middle1, $middle2, $right) = ("with","to","",".");
: 例子三:
: ACT of PTN PTN ACT class3:PTN PTN ACT
: ==>(ACT of) PTN () PTN () ACT ()
: ==> ($left, $middle1, $middle2, $right) = ("ACT of","","","");
: 請問有人可以解決或回答嗎^^?感謝:)
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