我缺乏使用C的经验,我需要使用PCRE来获得匹配 以下是我的源代码示例:
int test2()
{
const char *error;
int erroffset;
pcre *re;
int rc;
int i;
int ovector[OVECCOUNT];
char *regex = "From:([^@]+)@([^\r]+)";
char str[] = "From:regular.expressions@example.com\r\n"\
"From:exddd@43434.com\r\n"\
"From:7853456@exgem.com\r\n";
re = pcre_compile (
regex, /* the pattern */
0, /* default options */
&error, /* for error message */
&erroffset, /* for error offset */
0); /* use default character tables */
if (!re) {
printf("pcre_compile failed (offset: %d), %s\n", erroffset, error);
return -1;
}
rc = pcre_exec (
re, /* the compiled pattern */
0, /* no extra data - pattern was not studied */
str, /* the string to match */
strlen(str), /* the length of the string */
0, /* start at offset 0 in the subject */
0, /* default options */
ovector, /* output vector for substring information */
OVECCOUNT); /* number of elements in the output vector */
if (rc < 0) {
switch (rc) {
case PCRE_ERROR_NOMATCH:
printf("String didn't match");
break;
default:
printf("Error while matching: %d\n", rc);
break;
}
free(re);
return -1;
}
for (i = 0; i < rc; i++) {
printf("%2d: %.*s\n", i, ovector[2*i+1] - ovector[2*i], str + ovector[2*i]);
}
}
在此演示中,输出仅为:
0: From:regular.expressions@example.com
1: regular.expressions
2: example.com
我想输出所有的比赛;我怎么能这样做?
答案 0 :(得分:7)
我使用一个类来包装PCRE以使其更容易,但是在pcre_exec之后,ovector包含了在原始字符串中找到匹配项所需的子字符串索引。
所以它会是这样的:
#include <string>
#include <iostream>
#include "pcre.h"
int main (int argc, char *argv[])
{
const char *error;
int erroffset;
pcre *re;
int rc;
int i;
int ovector[100];
char *regex = "From:([^@]+)@([^\r]+)";
char str[] = "From:regular.expressions@example.com\r\n"\
"From:exddd@43434.com\r\n"\
"From:7853456@exgem.com\r\n";
re = pcre_compile (regex, /* the pattern */
PCRE_MULTILINE,
&error, /* for error message */
&erroffset, /* for error offset */
0); /* use default character tables */
if (!re)
{
printf("pcre_compile failed (offset: %d), %s\n", erroffset, error);
return -1;
}
unsigned int offset = 0;
unsigned int len = strlen(str);
while (offset < len && (rc = pcre_exec(re, 0, str, len, offset, 0, ovector, sizeof(ovector))) >= 0)
{
for(int i = 0; i < rc; ++i)
{
printf("%2d: %.*s\n", i, ovector[2*i+1] - ovector[2*i], str + ovector[2*i]);
}
offset = ovector[1];
}
return 1;
}
答案 1 :(得分:5)
注意:pcre_exec()的最后一个参数必须是element-count,而不是sizeof()! (http://www.pcre.org/readme.txt)