我怎样才能轻松搞定正则表达式选择用C？ [英] How can I easily get regex selections in C?

问题描述

我使用的 regex.h （ POSIX 的），定期EX pressions。是否有任何C选择方法的正则表达式匹配？

I'm using regex.h (POSIX) for regular expressions. Are there any selection methods for regex matches in C?

我可以很容易地检查定期EX pressions但如果我需要检索匹配的价值，我必须手动查找和存储。

I can quite easily check for regular expressions but if I need to retrieve the matched value, I have to manually find and store it.

{{( )*(([[:alnum:]]+\\.)*)?[[:alnum:]]+( )*}}

这正则表达式查找在双大括号的变量匹配。不过，我只需要在字符串中的最核心的项目。 如何可以在C检索与常规的前pressions价值？

This regex looks for any variable matches in double curly braces. But I only need the most central item in the string. How can I retrieve the value with regular expressions in C?

推荐答案

您需要的是一组 regmatch_t S中的正则表达式可以用的指数填写传递火柴。试试下面的程序使用一个命令行参数（字符串进行测试）。

You need to pass in a set of regmatch_ts that the regex can fill with the indices of the matches. Try the below program with a single command line argument (the string to test).

一旦有匹配的指数，它应该是很容易拔出你以后。 （注：比赛[0] 将是整个前pression的比赛，所以SUBEX pressions在开始匹配[ 1] ）

Once you have the indices of the matches, it should be fairly easy to pull out what you're after. (Note: matches[0] will be the match of the entire expression, so the subexpressions start at matches[1].)

#include <stdlib.h>
#include <stdio.h>
#include <regex.h>

int main(int argc, char* argv[])
{
    const char* pattern = "{{( )*(([[:alnum:]]+\\.)*)?[[:alnum:]]+( )*}}";
    regex_t rex;
    int rc;

    if ((rc = regcomp(&rex, pattern, REG_EXTENDED))) {
        fprintf(stderr, "error %d compiling regex\n", rc);
        /* retrieve error here with regerror */
        return -1;
    }

    regmatch_t* matches = malloc(sizeof(regex_t) * (rex.re_nsub + 1));

    if ((rc = regexec(&rex, argv[1], rex.re_nsub + 1, matches, 0))){
        printf("no match\n");
        /* error or no match */
    } else {
        for(int i = 0; i < rex.re_nsub; ++i) {
            printf("match %d from index %d to %d: ", i, matches[i].rm_so,
                   matches[i].rm_eo);
            for(int j = matches[i].rm_so; j < matches[i].rm_eo; ++j) {
                printf("%c", argv[1][j]);
            }
            printf("\n");
        }
    }

    free(matches);
    regfree(&rex);

    return 0;
}

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