Question

我有一个单词列表，在标记后，我在列表中得到一些我实际上不需要的空格。建议请...

示例字符串

String str = "3) type an \"l\" in the search field (\"bl\")"+
"4) startHistorySearch()'s previous result contains [\"blah\", \"baaa\", \"bloop\", \"bzzz\"], Satchel filters this down to [\"blah\", \"bloop\"] to match the new \"bl\" search string"+
and so on.....

这是代码和输出代码段

String[] splitString = (EXAMPLE_TEST.split("[\\[\\],\\'\"  \\(\\)\\{\\}\\*\\.]"));

输出

nsIAutoCompletResult, , no, , Simple, , , , so, the, QI, fails, , historyResult

在某些地方，我看到这样......

finds, 1, entry, , , blah, , , , search-suggestions, finds, , baaa, , , , bloop, , , , bzzz, , , the, autocompete, menu, shows, these, in, order, with, a, divider, between, , blah, , and, , baaa, , 3, , type, , l, , in, the, search, field, , , bl, , 4, , startHistorySearch, , , s, previous, result, contains, , , blah, , , , baaa, , , , bloop, , , , bzzz, , , , Satchel, filters, this, down, to, , , blah, , , , bloop, , , to, match, the, new, , bl, , search, string5, , nsSearchSuggestions, s, onReadyState, , , change, is, called, with, updated, search,

Answer 1

只需在表达式中添加+即可在两个令牌之间进行拆分。您也可以稍微简化表达式，不必转义字符类中的所有字符：

String[] splitString = (EXAMPLE_TEST.split("[\\[\\],'\" (){}*.]+"));

标记化后列表中的空格

1 个答案: