如果我手中的您的 FLEx 專案副本仍然準確，那麼您有 13 個可以填入屬格槽位的後綴，以及 1 個可以填入連格槽位的後綴。
一種暴力方法是讓您的 and 陳述的第一部分檢查是否存在那 1 個連格後綴，而 and 陳述的第二部分則是一個長的 or 陳述，用來檢查是否與那 13 個後綴中的每一個相等。
第二部分較簡短的寫法是檢查後綴是否以 POSS1 或 POSS2 或 POSS3 結尾。它可能看起來像這樣：
請注意，您不能在 ends with 陳述中使用 literal tag ，您必須使用 literal string 並以 > 結尾，這是標籤的結束標記。
另一種方法是在類別定義中進行所有枚舉。它可能看起來像這樣：
當然，這只是一個部分清單，您可能需要為每個名詞類別（na, ni 等）添加更多項目。
這樣做的好處是，您在類別中完成所有枚舉，並使用該類別建立規則，此時您已知所有條件都已滿足，因此可以進行所需的變更。
我很想知道您最終選擇了哪種做法。
If my copy of your FLEx project is still accurate you have 13 affixes that could go in the genitive slot and 1 that can go in the associative slot.
The brute force method would be to have the first part of your and statement check if the 1 associative affix is present, the second part of your and statement would be a long or statement that would check equality with each of the 13 affixes.
A less long way to do the second part would be to check if the affix ends with POSS1 or POSS2 or POSS3. It might look like this:
Note that you can't use a literal tag with the ends with statement you have to use literal string and end it with a > which is the end marker for a tag.
A different approach would be to do all the enumerating in category definition. It might look like this:
Of course, this is just a partial list and you may have to add to the list for each of your noun categories. na, ni, etc.
The advantage of this, is you do all the enumerating in the category and have a rule using that category where you already know all the conditions have been met and you can do the changes needed.
I'd be curious to know what you end up choosing to do.
機器翻譯自 English 顯示原文
