如何执行复杂的多个 if-then-else 正则表达式？

Question

I need to do a complex if-then-else with five preferential options. Suppose I first want to match abc but if it's not matched then match ac , then if it's not matched def , then %#@ , then 1z; .

Can I nest the if-thens or how else would it be accomplished? I've never used if-thens before.

For instance, in the string 1z;%#@defarcabcaqcdef%#@1z; I would like the output abc .

In the string 1z;%#@defarcabaqcdef%#@1z; I would like the output arc .

In the string 1z;%#@defacabacdef%#@1z; I would like the output def .

In the string 1z;#@deacabacdf%#@1z; I would like the output %#@ .

In the string foo;%@dfaabaef#@1z;barbbbaarr3 I would like the output 1z; .

Answer 1

You need to force individual matching of each option and not put them together. Doing so as such: .*?(?:x|y|z) will match the first occurrence where any of the options are matched. Using that regex against a string, ie abczx will return z because that's the first match it found. To force prioritization you need to combine the logic of .*? and each option such that you get a regex resembling .*?x|.*?y|.*?z . It will try each option one by one until a match is found. So if x doesn't exist, it'll continue to the next option, etc.

See regex in use here

(?m)^(?:.*?(?=abc)|.*?(?=a.c)|.*?(?=def)|.*?(?=%#@)|.*?(?=1z;))(.{3})

• (?m) Enables multiline mode so that ^ and $ match the start/end of each line
• (?:.*?(?=abc)|.*?(?=ac)|.*?(?=def)|.*?(?=%#@)|.*?(?=1z;)) Match either of the following options
  • .*?(?=abc) Match any character any number of times, but as few as possible, ensuring what follows is abc literally
  • .*?(?=ac) Match any character any number of times, but as few as possible, ensuring what follows is a , any character, then c
  • .*?(?=def) Match any character any number of times, but as few as possible, ensuring what follows is def literally
  • .*?(?=%#@) Match any character any number of times, but as few as possible, ensuring what follows is %#@ literally
  • .*?(?=1z;) Match any character any number of times, but as few as possible, ensuring what follows is 1z; literally
• (.{3}) Capture any character exactly 3 times into capture group 1

If the options vary in length, you'll have to capture in different groups as seen here :

(?m)^(?:.*?(abc)|.*?(a.c)|.*?(def)|.*?(%#@)|.*?(1z;))