我需要一个可以与类似东西相匹配的管理机构。
"4f0f30500be4443126002034"
以及
“4f0f30500be4443126002034>4f0f31310be4443126005578”
but not like
“4f0f30500be4443126002034>4f0f31310be4443126005578>4f0f31310be4443126005579”
Try:
^[da-f]{24}(>[da-f]{24})?$
[da-f]{24} is exactly 24 characters consisting only of 0-9, a-f. The whole pattern is one such number optionally followed by a > and a second such number.
I think you want something like:
/^[0-9a-f]{24}(>[0-9a-f]{24})?$/
1 D-1, 1 D-1, 1 D-1, 1 D-1, 1 D-1, 1 D-1, 1 D-1, 1 D-1, 1 D-1, 1 D-1, 1 D-1, 1 D-1, 1 D-1, 1 D-1, 1 D-1, 1 D-1, 1 D-1, 1 D-1, 1 D-1, 1 D-1, 1 D-1, 1 D-1, 1 D-1, 1 D-1, 1 D-1, 1 D-1, 1 D-1 页: 1
不需要reg。
str = “4f0f30500be4443126002034>4f0f31310be4443126005578>
match = str.count( >) < 2
在有1个或0个或1个“体”的对应体内,将确定对应点。 否则,会伪造。
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