C - 如何从没有数字,特殊字符(,。?&)的文件中提取单词?

时间:2015-05-14 16:14:50

标签: c file word scanf

我想从文件中取出单词。我发送给数组。我用fscanf做到了这一点。但是,它需要数字,字符(,。%和#!?)和其他东西。我该如何控制这个陈述?

int main(void) 
{       
  char path[100];
  printf("Please, enter path of file: "); scanf("%s",&path);
  FILE *dosya;
  char kelime[1000][50];
  int i=0;

  if((dosya = fopen(path,"r")) != NULL)
  {
    while(!feof (dosya))
    {
        fscanf(dosya,"%s",&kelime[i]);
        i++;
    }
  }
  else{printf("Not Found File !");}
  fclose(dosya);
}

2 个答案:

答案 0 :(得分:2)

使用"%[]"区分字母和非字母。

#define NWORDS (1000)
char kelime[NWORDS][50];

size_t i;
for (i=0; i<NWORDS; i++) {
  // Toss ("*" indicates do not save) characters that are not ("^" indicates `not`) letters.
  // Ignore return value
  fscanf(dosya, "%*[^A-Za-z]");

  // Read letters
  // Expect only a return value of 1:Success or EOF:no more file to read
  // Be sure to limit number of characters read: 49
  if (fscanf(dosya, "%49[A-Za-z]", kelime[i]) != 1) break;
}

// do something with the `i` words

答案 1 :(得分:-1)

您可以使用fgetc()

char str[1000];
while((ch=fgetc(fname))!=EOF)
{
        // Write your code here;
        if((ch>='0' && ch<='9') || (ch<'a' && ch>'z') || (ch<'A' && ch>'Z'))
              continue;
        str[i++]=ch;
}
str[i]='\0';

使用fscanf()时,无法解析语句。所以你必须

  1. 通过char

  2. 检查它
  3. 如果不是特殊字符或数字,则添加到字符串