当我的用户进入像“密码”这样的术语时,我想把它当作一个野心查询“密码*”,这样就找到了开始“密码”的所有术语。 难道有办法告诉QueryParser,以便自动制造野心问询,还是我不得不把问话说成 par? 这只是简单询问的一个问题,但可能会成为更复杂问题的trick。
除非我缺少一些东西——对每一个问题来说,一个野心 que常常是不可思议的——它非常昂贵,可能造成许多问题。 如果你试图找到结果,包括 stem的变体(例如,赢家、赢家等) 您应考虑采用无克办法。
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