将连续的单个字符匹配为整个单词

1.)从 穿绳短裙 和 用户输入 .

.replaceAll("(?<=\\b\\w) +(?=\\w\\b)","")

• (?<=\b\w) look 后面检查前面是否有 \b word boundary , \w word character
• (?=\\w\\b) 向前看,检查是否紧随其后 \w(重量) 单词字符, \(b) 字边界

See demo at regexplanet (click Java)

2.)检查 穿绳短裙 .contains 用户输入 .

听起来你只是想忽略空白。您可以通过在查找匹配之前从目标字符串和用户输入中删除空白来轻松做到这一点。

您基本上希望修改每个搜索词以允许插入空格,因此

"abcd" becomes regex "\ba ?b ?c ?d\b"

要实现这一点,请在匹配之前对每个单词执行以下操作:

word = "\\b" + word.replaceAll("(?<=.)(?=.)", " ?") + "\\b";

单词断了 \b 必须停止匹配 "comma bcd" 或 "abc duck" .

此正则表达式将匹配由一个或多个空格分隔的所有单个字符

(^(\w\s+)+)|(\s+\w)+$|((\s+\w)+\s+)

以下正则表达式(在多行模式下)可以帮助您:

^(?<first>\w+)(?<chars>(?:.(?!(?:\b\w{2,}\b)))*)
# assure that it is the beginning of the line
# capture as many word characters as possible in the first group "first"
# the construction afterwards consumes everything up to (not including)
# a word which has at least two characters...
# ... and saves it to the group called "chars"

您只需要替换第二组中的空白(又名“ 字符 ").
查看演示 on regex101.com

str = str.replaceAll("\\s","");