Question

使用微处理器，我的内存不足，我必须以非常有效和安全的方式执行此操作。

所以我有一些来自服务器的数据，我必须找到它的标题。例如：

char *meida="+IPD,0,206:GET /setWifi:home:0545881255 HTTP/1.1  Host: 192.168.4.1" ;

我只需要setWifi:home:0545881255。

strstr在句子中找到字符串的出现。我如何用它来找到两个单词之间的句子？

Answer 1

编辑现在提取两个关键字之间的短语。

假设您读入的数据不是字符串文字，则可以对其进行修改。此代码查找第一个关键字的位置，跳过它，然后查找下一个关键字，并截断字符串。

#include <stdio.h>
#include <string.h>

int main (void){
    char meida[] ="+IPD,0,206:GET /setWifi:home:0545881255 HTTP/1.1  Host: 192.168.4.1";
    char *keyworda = "GET /";
    char *keywordb = " HTTP/";
    char *aptr, *bptr;

    aptr = strstr(meida, keyworda);
    if (aptr != NULL) {
        aptr += strlen(keyworda);       // skip past first keyword
        bptr = strstr(aptr, keywordb);
        if (bptr != NULL) {
            *bptr = '\0';               // truncate
            printf("%s\n", aptr);
        }
    }
    return 0;
}

节目输出：

setWifi:home:0545881255

Answer 2

这取决于您是否可以修改输入字符串，如果您的解决方案更有效率。

这是两个版本，一个复制输入字符串，另一个修改它，返回子字符串。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <ctype.h>

char *
substring_enclosed(char *const input, const char *const left, const char *const right)
{
    char *result;
    char *tail;
    result = strstr(input, left);
    if (result == NULL)
        return NULL;
    /* At this point `result' points to the occurrance of `left', skip
     * the same number of characters of `left' and you are there 
     */
    result += strlen(left);
    tail = strstr(result, right);
    if (tail != NULL)
        tail[0] = '\0';
    return result;
}

char *
substring_enclosed_const(const char *const input, const char *const left, const char *const right)
{
    char *result;
    char *tail;
    char *copy;

    copy = strdup(input);
    if (copy == NULL)
        return NULL;

    result = strstr(copy, left);
    if (result == NULL)
        return NULL;
    /* At this point `result' points to the occurrance of `left', skip
     * the same number of characters of `left' and you are there 
     */
    result += strlen(left);

    /* You have to call `free' later and you cannot do it on the 
     * pointer that is a result of arithmetic above, you need the
     * one returned by `strdup()' 
     */
    memmove(copy, result, strlen(result) + 1);

    tail = strstr(result, right);
    if (tail != NULL)
        tail[0] = '\0';

    return copy;
}

int
main(void)
{
    const char *meida = "+IPD,0,206:GET /setWifi:home:0545881255 HTTP/1.1  Host: 192.168.4.1";
    char writeable[] = "+IPD,0,206:GET /setWifi:home:0545881255 HTTP/1.1  Host: 192.168.4.1";
    char *substr;

    substr = substring_enclosed_const(meida, "GET /", " HTTP");
    if (substr != NULL)
    {
        puts(substr);
        free(substr);
    }

    substr = substring_enclosed(writeable, "GET /", " HTTP");
    if (substr != NULL)
        puts(substr);

    return 0;
}

以有效的方式在字符串之间查找字符串

2 个答案: