从字符串中删除子字符串?

12
我有一个C函数,它接受一个名为“buffer”的字符串,并解析它,它将匹配关键字并使用它来分配结构中的值。
然而,我想要完全忽略一些关键字。
该程序解析VCard文件(.vcf,虚拟商务卡)。
这是一个示例行缓冲区可能提供的内容:
FN;CHARSET=UTF-8:David Celery

我对FN这个关键字很感兴趣,而David Celery是与FN相关的值。

然而,CHARSET=UTF-8这个东西我根本不关心。

所以我的问题是,有没有一种方式可以扫描我的缓冲区,并简单地用""替换'CHARSET=UTF-8',这样我就不必担心解析它(以及其他类似的我只想忽略的关键字)。

谢谢。


1
你目前尝试了什么?如果你正在使用C语言处理字符串,但是无法实现这么基本的操作,那我建议你考虑使用内置字符串支持的编程语言(比如C ++、Java、C#、Python、Delphi等)。 - David Heffernan
4个回答

16

可以看一下一个简单的 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)));
}

2
我在尝试使用这段代码时收到了警告和段错误。是否有什么地方出错了?如果有影响的话,我正在使用c99。 - Blackbinary
8
如果您存储strlen(toremove),将会更有效率——循环中就不需要再计算两次了。 - ThiefMaster
性能不是问题;这段代码对于更多的出现情况也可以正常工作,我认为这是你或者你的编译器的问题。 - user411313
5
我同意ThiefMaster的看法,手册中充斥着糟糕的例子。为什么不在这里发布更有效率和简洁的代码呢?这样或许能够使你的回答成为更好的得到赞同的目标。 - Alexander Pogrebnyak
我不明白这是如何工作的,可以有人解释一下吗?我也使用了memmove来解决类似的问题,但我给出了字符数组源和目标的确切位置以及要移动的字节数超过了源的大小。例如- memmove(s+pos, s+pos+strlen(toremove), (strlen(s)-pos)); 这里的pos是通过strstr获取toremove的第一个出现位置。 - Ghos3t
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3

这里有一个 C 语言字符串查找和替换的函数,你可能会觉得有用。可以在这里找到。

编辑:根据评论请求,下面包含链接中的代码片段。

#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"
*/

希望这能帮到你。

0

请查看 string.h 中的方法。

例如:(并在 CodePad 上查看)

#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;
}

0

你可以选择忽略它们,而不是将它们删除,例如像这样:

#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);

我认为我们并不是要解决所有世界的问题,只是提供一些方法建议,但结果可能因人而异。 - jlehr

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