正则表达式,用于计算字符串中逗号的数量

Question

如何构建一个正则表达式,它将匹配包含任何字符但必须包含21个逗号的任何长度的字符串？

Answer 1

/^([^,]*,){21}[^,]*$/

那是:

^     Start of string
(     Start of group
[^,]* Any character except comma, zero or more times
,     A comma
){21} End and repeat the group 21 times
[^,]* Any character except comma, zero or more times again
$     End of string

Answer 2

如果您正在使用支持占有量词的正则表达式(例如Java),您可以:

^(?:[^,]*+,){21}[^,]*+$

占有量词可以比贪心量词更好.

说明:

(?x)    # enables comments, so this whole block can be used in a regex.
^       # start of string
(?:     # start non-capturing group
[^,]*+  # as many non-commas as possible, but none required
,       # a comma
)       # end non-capturing group
{21}    # 21 of previous entity (i.e. the group)
[^,]*+  # as many non-commas as possible, but none required
$       # end of string

Answer 3

正好21个逗号:

^([^,]*,){21}[^,]$

至少21个逗号:

^([^,]?,){21}.*$

Answer 4

迭代字符串可能更快,更容易理解,计算找到的逗号数量,然后将其与21进行比较.