Re: [問題] 也是另一個比對問題?

看板Perl作者 (開心)時間18年前 (2007/04/10 19:56), 編輯推噓0(000)
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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:) -- -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 140.116.96.202
文章代碼(AID): #166tk3Pw (Perl)
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文章代碼(AID): #166tk3Pw (Perl)