从字符串中删除子字符串？
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。
但是，我想完全忽略一些关键字。
该程序解析VCard文件（.vcf，虚拟名片）。
以下是可能提供的示例行缓冲区：
FN;CHARSET=UTF-8:David Celery FN是我感兴趣的关键词，而David Celery是与FN相关的值。
但是，CHARSET = UTF-8是我根本不关心的事情。
所以我的问题是，有没有办法让我扫描我的缓冲区，只需用“”替换’CHARSET = UTF-8’，这样我就不用担心解析它（以及我想要的其他类似关键字）忽视）。
谢谢，
看看一个简单的ANSI C解决方案，如：
void removeSubstring(char *s,const char *toremove) { while( s=strstr(s,toremove) ) memmove(s,s+strlen(toremove),1+strlen(s+strlen(toremove))); }
其他人有一个C字符串查找和替换function，你可能会发现这里有用。
编辑：根据评论请求，从下面的链接中包含代码段
#include #include #include /* * 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" */
希望这可以帮助。
看一下string.h中的方法。
示例:(和CodePad上 ）
#include #include #include 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; }
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);