我需要的strtok和strtok_single的混合 [英] I need a mix of strtok and strtok_single

问题描述

我有以下字符串，我试图分析变量。

I have the following string that I am trying to parse for variables.

char data[]="to=[e-mail]&cc=[e-mail]&title=&content=how are you?&signature=best regards."

我开始用的strtok及以下code

I started with strtok and the following code

char *to=parsePostData("to",data);

char* parsePostData(char s[],char t[])
{
  char *postVal;
  char *pch;
  char tCpy[512];//Make a copy. Otherwise, strtok works on the char pointer, and original char array gets modified/ corrupted.
  strcpy(tCpy,t);
  pch = strtok (tCpy,"=&");
  while (pch != NULL)
  {
      if(strcmp(pch,s)==0) {
            pch= strtok (NULL, "&");
                return pch;          
      }else{
        pch = strtok (NULL, "=&");  
      }
  }      
}

这工作得很好，当谈到连续的分隔符，如称号后的人，除了。所以，我发现这个习俗strtok_single实现。
<一href=\"http://stackoverflow.com/questions/8705844/need-to-know-when-no-data-appears-between-two-token-separators-using-strtok\">Need在使用strtok的双令牌分离器之间似乎没有数据就知道（）

This works fine, except when it comes to consecutive delimiters such as the one after "title". So I found this custom strtok_single implementation. Need to know when no data appears between two token separators using strtok()

char * strtok_single (char * str, char const * delims)
{
  static char  * src = NULL;
  char  *  p,  * ret = 0;

  if (str != NULL)
    src = str;

  if (src == NULL)
    return NULL;

  if ((p = strpbrk (src, delims)) != NULL) {
    *p  = 0;
    ret = src;
    src = ++p;
  }

  return ret;
}

但有了这个，问题是我不能让签名，因为没有和放大器;之后分隔符。

But with this, the problem is I cannot get "signature", as there is no & delimiter after that.

怎样才能得到这个二者的混合，所以我不错过了最后一个变量，我可以应付连续的分隔符？

How can I get a mix of this two, so I dont miss out the last variable, and I can handle consecutive delimiters?

推荐答案

您还没有确切地告诉我们，你说的这工作正常，但它似乎足以说，要分析一个<$ C $是什么意思C>应用程序/ x-WWW的形式urlen codeD 字符串。你为什么不摆在首位这么说？

You haven't exactly told us what you mean by "this works fine", though it seems sufficient to say that you want to parse an application/x-www-form-urlencoded string. Why didn't you say so in the first place?

考虑到第一个字段，键，可由第一任=或终止'和; 。这将是适当的搜索，在任一这些字符的结束令牌，提取键。

Consider that the first field, key, may be terminated by the first of either '=' or '&'. It would be appropriate to search for a token that ends in either of those characters, to extract key.

第二个字段，值，但不是由=字符终止，所以它的不适当的要寻找该字符提取值。你想要搜索'和;仅

The second field, value, however, isn't terminated by an '=' character, so it's inappropriate to be searching for that character to extract value. You'd want to search for '&' only.

当然。您的可能的使用 strtok的来解析这一点，但是我敢肯定有很多更合适的工具。 strcspn ，例如，不会对数据，这意味着你不需要做任何改变数据副本，你是...

Sure. You could use strtok to parse this, however I'm sure there are many more suitable tools. strcspn, for example, won't make any changes to data, which means you won't need to make a copy of data as you are...

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

int main(void) {
    char data[]="to=[e-mail]&cc=[e-mail]&title=&content=how are you?&signature=best regards.";

    char *key = data;
    do {
        int key_length = strcspn(key, "&=");

        char *value = key + key_length + (key[key_length] == '=');
        int value_length = strcspn(value, "&");

        printf("Key:   %.*s\n"
               "Value: %.*s\n\n",
               key_length,   key,
               value_length, value);

        key = value + value_length + (value[value_length] == '&');
    } while (*key);
    return 0;
}

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