c ++ strtok跳过第二个令牌或连续的分隔符 [英] c++ strtok skips second token or consecutive delimiter

问题描述

我想读取CSV逗号分隔的文件，文件内容是

I am trying to read CSV comma delimited file, content of file are

      one,,three

读取文件的代码是...

And code to read file is this…

      inFile.getline(line, 500);                            
      token1 = strtok(line, ",");
      token2 = strtok(NULL, ",");
      token3 = strtok(NULL, ",");

      if(token1 != NULL){
             cout << "token1 = " << token1 << "\n";      
      }else{ 
             cout << "token1 = null\n" ;
      }
      if(token2 != NULL){ 
             cout << "token2 = " << token2 << "\n";      
      }else{ 
             cout << "token2 = null\n" ;
      } 
      if(token3 != NULL){ 
             cout << "token3 = " << token3 << "\n";      
      }else{ 
             cout << "token3 = null\n";
      }

输出为

token1 = one
token2 = three
token3 = null

我的期望是输出应该是这样的...

Whereas my expectation are that output should be like this…

token1 = one
token2 = null
token3 = three

如果

I did change if statements from

     if(token1 != NULL) 

$ b b

To

To

     if(token1)

但它也不起作用。

查看此示例 http：/ /www.cplusplus.com/reference/cstring/strtok/ ，我已更新

   token2 = strtok(NULL, ",");

至

   token2 = strtok(NULL, ",,");

同样不起作用

推荐答案

一旦我在读取CSV逗号分隔文件时遇到了这个问题。但是我们可以使用 strtok（）作为我们的解决方案，在这样的问题中分隔字符连续出现。因为根据标准

Once I did face this problem while reading CSV comma delimited file. But we can use strtok() as our solution in such problem were delimited character appears consecutively. Because according to the standard

序列中的第一个调用搜索 s1指向的字符串
s2 指向的当前分隔符字符串中包含的第一个字符

如果没有找到这样的字符，则在
中没有令牌由 s1 指向的字符串和 strtok 函数返回
a空指针。如果找到这样的字符，它是第一个令牌的
开始。 C11§7.24.5.83

The first call in the sequence searches the string pointed to by s1 for the first character that is not contained in the current separator string pointed to by s2. If no such character is found, then there are no tokens in the string pointed to by s1 and the strtok function returns a null pointer. If such a character is found, it is the start of the first token. C11 §7.24.5.8 3

所以，对于我的case，我使用strpbrk（）函数定义了另一个解决方案， 。

So, for my case I did define another solution using strpbrk() function which will also useful for you.

#include<iostream.h>

char *strtok_new(char * string, char const * delimiter){
   static char *source = NULL;
   char *p, *riturn = 0;
   if(string != NULL)         source = string;
   if(source == NULL)         return NULL;

   if((p = strpbrk (source, delimiter)) != NULL) {
      *p  = 0;
      riturn = source;
      source = ++p;
   }
return riturn;
}

int main(){
   char string[] = "one,,three,";
   char delimiter[] = ",";
   char * p    = strtok_new(string, delimiter);

   while(p){
            if(*p)  cout << p << endl;
            else    cout << "No data" << endl;                
            p = strtok_new(NULL, delimiter);
   }
   system("pause");
   return 0;
}

输出

one
No data
three

希望这是您想要的输出。

Hope so this is your desired output.

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