我试图解析输入字符串,看起来像数组[位或前pression或阵列,数字或前pression或阵列] 所以,我需要在来获取值[,] 。我试图让他们使用这个正则表达式:
(数组1)\\ [(。*)\\(。*)\\]获得值捕获组,但它doen't的工作,因为它是贪婪的量词,所以在的情况下,(*):
数组1 [数组2 [4,3]数组2 [1,6]我将获得数组2 [4,3]数组2 [1,作为第一个捕获组和 6 作为第二个这是不正确的。
我如何获得数组2 [4,3] 作为第一和数组2 [1,6] 作为第二捕获组?或数组2 [ARRAY3 [1,1],3] 和 5 + 3 如果输入字符串为数组1 [数组2 [ARRAY3 [1,1],3,5 + 3] ?
解决方案您可以均衡组的使用:
array\\d*\\[\\s*((?:[^\\[\\]]|(?<o>\\[)|(?<-o>\\]))+(?(o)(?!))),\\s*((?:[^\\[\\]]|(?<o>\\[)|(?<-o>\\]))+(?(o)(?!)))\\]ideone演示你的最后一个字符串。
一个细分:
阵列\\ D * \\ [\\ S *#匹配阵列,它的编号(如果有的话),第一个'['和空格( (?: [^ \\ [\\]#匹配所有非支架 | (小于2 O&GT; \\ [)#匹配'[',并捕捉到O(代表开) | (?&LT; -O - GT; \\])#匹配']',并删除O捕获 )+ (?(O)(?!))#如果'O'不存在失败),\\ S *#匹配逗号和空格(#重复高于... (?: [^ \\ [\\]#匹配所有非支架 | (小于2 O&GT; \\ [)#匹配'[',并捕捉到O(代表开) | (?&LT; -O - GT; \\])#匹配']',并删除O捕获 )+ (?(O)(?!))#如果'O'不存在失败)\\]#最后大括号I'm trying parse input string which looks like array[digit or expression or array, digit or expression or array] So I need to get values in [ , ]. I was trying to get them using this regex:
(array1)\[(.*)\,(.*)\]to get values of (.*) capturing groups, but it doen't work, because it's greedy quantifier, so in the case of:
array1[ array2[4,3] , array2[1,6] ]I will get array2[4,3] , array2[1, as first capturing group and 6 as a second which is not right.
How can I get array2[4,3] as first and array2[1,6] as second capturing group? Or array2[array3[1,1],3] and 5+3 if the input string is array1[ array2[array3[1,1],3] , 5+3 ]?
解决方案You can make use of balancing groups:
array\d*\[\s*((?:[^\[\]]|(?<o>\[)|(?<-o>\]))+(?(o)(?!))),\s*((?:[^\[\]]|(?<o>\[)|(?<-o>\]))+(?(o)(?!)))\]ideone demo on your last string.
A breakdown:
array\d*\[\s* # Match array with its number (if any), first '[' and any spaces ( (?: [^\[\]] # Match all non-brackets | (?<o>\[) # Match '[', and capture into 'o' (stands for open) | (?<-o>\]) # Match ']', and delete the 'o' capture )+ (?(o)(?!)) # Fails if 'o' doesn't exist ) ,\s* # Match comma and any spaces ( # Repeat what was above... (?: [^\[\]] # Match all non-brackets | (?<o>\[) # Match '[', and capture into 'o' (stands for open) | (?<-o>\]) # Match ']', and delete the 'o' capture )+ (?(o)(?!)) # Fails if 'o' doesn't exist ) \] # Last closing brace
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