查找字符串之間的字符串中有效的方式

Question

使用微處理器，我跑出來的內存，我不得不這樣做在一個非常有效和安全的方式。

所以我有一些數據從服務器來了，我一定要找到它的頭。例如：

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

我只需要得到setWifi:home:0545881255。

strstr發現在一個句子串的發生。我怎樣才能用它找到2個單詞之間的句子？

Answer 1

編輯現在提取兩個關鍵詞之間的短語。

假設您讀入的數據是而不是是一個字符串文字，它可以被修改。該代碼找到第一個關鍵字的位置，跳過它，然後找到下一個關鍵字，並截斷該字符串。

#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; 
} 

程序輸出：

setWifi:home:0545881255 

Answer 2

這取決於你是否可以修改輸入的字符串或沒有，如果你能解決的辦法是更有效的。

這裏有兩個版本，一個是將複製修改到位返回字符串輸入字符串和一個。

#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; 
}