我有一个 C 函数，它接受一个名为 'buffer' 的字符串并对其进行解析，它将匹配关键字并使用它在结构中分配值。
但是，我想完全忽略一些关键字。
此程序解析 VCard 文件（.vcf，虚拟名片）。
这是一个可能提供的示例行缓冲区：
FN;CHARSET=UTF-8:David Celery
FN是我感兴趣的关键字，David Celery是与FN关联的值。
但是，我根本不关心 CHARSET=UTF-8。
所以我的问题是，有没有办法让我扫描我的缓冲区并简单地将“CHARSET=UTF-8”替换为“”，这样我就不必担心解析它（以及其他类似的关键字我只是想忽略）。
谢谢，
【问题讨论】：
看看一个简单的 ANSI C 解决方案，例如：
void removeSubstring(char *s,const char *toremove)
{
while( s=strstr(s,toremove) )
memmove(s,s+strlen(toremove),1+strlen(s+strlen(toremove)));
}
【讨论】：
strlen(toremove)，效率会更高——每个循环不需要确定两次。
别人有一个 C 字符串查找和替换函数，您可能会发现它很有用 here。
编辑：根据评论请求，包含来自下面链接的代码 sn-p
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
/*
* Description:
* Find and replace text within a string.
*
* Parameters:
* src (in) - pointer to source string
* from (in) - pointer to search text
* to (in) - pointer to replacement text
*
* Returns:
* Returns a pointer to dynamically-allocated memory containing string
* with occurences of the text pointed to by 'from' replaced by with the
* text pointed to by 'to'.
*/
char *replace(const char *src, const char *from, const char *to)
{
/*
* Find out the lengths of the source string, text to replace, and
* the replacement text.
*/
size_t size = strlen(src) + 1;
size_t fromlen = strlen(from);
size_t tolen = strlen(to);
/*
* Allocate the first chunk with enough for the original string.
*/
char *value = malloc(size);
/*
* We need to return 'value', so let's make a copy to mess around with.
*/
char *dst = value;
/*
* Before we begin, let's see if malloc was successful.
*/
if ( value != NULL )
{
/*
* Loop until no matches are found.
*/
for ( ;; )
{
/*
* Try to find the search text.
*/
const char *match = strstr(src, from);
if ( match != NULL )
{
/*
* Found search text at location 'match'. :)
* Find out how many characters to copy up to the 'match'.
*/
size_t count = match - src;
/*
* We are going to realloc, and for that we will need a
* temporary pointer for safe usage.
*/
char *temp;
/*
* Calculate the total size the string will be after the
* replacement is performed.
*/
size += tolen - fromlen;
/*
* Attempt to realloc memory for the new size.
*/
temp = realloc(value, size);
if ( temp == NULL )
{
/*
* Attempt to realloc failed. Free the previously malloc'd
* memory and return with our tail between our legs. :(
*/
free(value);
return NULL;
}
/*
* The call to realloc was successful. :) But we'll want to
* return 'value' eventually, so let's point it to the memory
* that we are now working with. And let's not forget to point
* to the right location in the destination as well.
*/
dst = temp + (dst - value);
value = temp;
/*
* Copy from the source to the point where we matched. Then
* move the source pointer ahead by the amount we copied. And
* move the destination pointer ahead by the same amount.
*/
memmove(dst, src, count);
src += count;
dst += count;
/*
* Now copy in the replacement text 'to' at the position of
* the match. Adjust the source pointer by the text we replaced.
* Adjust the destination pointer by the amount of replacement
* text.
*/
memmove(dst, to, tolen);
src += fromlen;
dst += tolen;
}
else /* No match found. */
{
/*
* Copy any remaining part of the string. This includes the null
* termination character.
*/
strcpy(dst, src);
break;
}
}
}
return value;
}
void test(const char *source, const char *search, const char *repl)
{
char *after;
after = replace(source, search, repl);
printf("\nsearch = \"%s\", repl = \"%s\"\n", search, repl);
if ( after != NULL )
{
printf("after = \"%s\"\n", after);
free(after);
}
}
int main(void)
{
const char before[] = "the rain in Spain falls mainly on the plain";
printf("before = \"%s\"\n", before);
test(before, "the", "THEE");
test(before, "the", "A");
test(before, "cat", "DOG");
test(before, "plain", "PLANE");
test(before, "ain", "AINLY");
return 0;
}
/* my output
before = "the rain in Spain falls mainly on the plain"
search = "the", repl = "THEE"
after = "THEE rain in Spain falls mainly on THEE plain"
search = "the", repl = "A"
after = "A rain in Spain falls mainly on A plain"
search = "cat", repl = "DOG"
after = "the rain in Spain falls mainly on the plain"
search = "plain", repl = "PLANE"
after = "the rain in Spain falls mainly on the PLANE"
search = "ain", repl = "AINLY"
after = "the rAINLY in SpAINLY falls mAINLYly on the plAINLY"
*/
希望这会有所帮助。
【讨论】：
查看 string.h 中的方法。
示例：（在 CodePad 上）
#include <string.h>
#include <stdio.h>
#include <stddef.h>
int main()
{
const char * source = "FN;CHARSET=UTF-8:David Celery";
const char * newBegin = strrchr(source, ':');
if (!newBegin)
{
puts("Error!");
return -1;
}
newBegin++;
puts(newBegin);
return 0;
}
【讨论】：
您可以忽略它们，而不是删除它们，例如：
#define KEY_TO_IGNORE "CHARSET=UTF-8"
char key[80];
char value[80];
char *text = "FN;CHARSET=UTF-8:David Celery";
sscanf(text, "%2s;" KEY_TO_IGNORE ":%s", key, value);
printf("key: %s, value: %s\n", key, value);
【讨论】：