您可以使用正则表达式。它看起来像这样(未测试):
#include <regex.h>
regex_t *re = malloc(sizeof(regex_t));
const char *restr = "by ([A-Za-z.]+) \(([^\)]*)\)";
check(regcomp(re, restr, REG_EXTENDED | REG_ICASE), "regcomp");
size_t nmatch = 1;
regmatch_t *matches = malloc(sizeof(regmatch_t) * nmatch);
int ret = regexec(re, YOUR_STRING, nmatch, matches, 0);
check(ret != 0, "regexec");
int size;
size = matches[2].rm_eo - matches[2].rm_so;
char *host = malloc(sizeof(char) * size);
strncpy(host, YOUR_STRING + matches[2].rm_so, size);
host[size] = '\0';
size = matches[3].rm_eo - matches[3].rm_so;
char *ip = malloc(sizeof(char) * size);
strncpy(ip, YOUR_STRING + matches[3].rm_so, size);
ip[size] = '\0';
检查是宏来帮助你找出是否有任何问题:
#define check(condition, description) if (condition) { fprintf(stdout, "%s:%i - %s - %s\n", __FILE__, __LINE__, description, strerror(errno)); exit(1); }
你一定要明白,你不能对任何接收头数也许除了最后一个是正确的,对不对? – 2009-02-02 17:57:29