c - 如何使用 strstr 但使用用户未指定的关键字搜索字符串

Question

这是代码

int main()
{
    FILE* fptr;
    char c_id[50];
    char str[BUFFER_SIZE];
    int choices;
    char* pos;
    int customer_id;
    printf("Enter customer ID: ");
    scanf("%d", &customer_id);
    sprintf(c_id, "%d", customer_id);
    fptr = fopen("servicestest.txt", "r");
    printf("\n1. Check for unpaid\n2. Check for paid\n");
    scanf("%d", &choices);
    switch (choices)
    {
    case 1:
        while ((fgets(str, BUFFER_SIZE, fptr)) != 0)
        {
            pos = strstr(str, "Not");

            if (pos == 1)
            {
                printf("%s\n", str);
            }
        }
        break;
    case 2:
        while ((fgets(str, BUFFER_SIZE, fptr)) != 0)
        {
            pos = strstr(str, "Not");
            if (pos =! 1)
            {
                printf("%s\n", str);
            }
        }
  }
}

是否计划用 not 检查一行并打印它们如何做到这一点？为什么当我输入“1”作为选项时它不打印任何东西

我想分配

char ch[50];
ch[1] = "Not";

但是当我尝试时strstr (str, ch[1])它导致了一个错误

知道怎么做吗？

标签： c

Answer 1

1. 检查返回值fopen：

if(!fptr) {
   return -1;
}

2. 您尝试在指针和int值之间进行比较。

if (pos == 1)
// and
if (pos =! 1) // even you can compare, it should be if (pos != 1)

如果要计算 substring 的位置"Not"，可以这样做：

pos = strstr(str, "Not");
if(pos) {
   int pos_substring = (int) (str-pos);
   // from here you can compare the position of substring with an `int` value.
   if (pos_substring == 1) {/*do something*/}
   eles {/*do something*/}

}

3. 当您尝试代码时：

char ch[50];
char ch[1] = "Not"; // string "Not" need at least 4 bytes (3 bytes for 'N', 'o' and 't', 1 bytes for null character)

它应该更改为：

char ch[50] = "Not";
// or
char ch[] = "Not";
// or
const char *ch = "Not" // be attention with the string literal

然后当你想申请strstr函数时：

strstr (str, ch); // use ch instead of ch[1]

如果你想使用类似的东西strstr (str, ch[1])，你可以使用 2D 字符数组：

char ch[50][50] = {/*init the string here or later as you want*/}; // this 2D array contents of maximum 50 strings, each string length ups to 49.
strstr (str, ch[0]);
strstr (str, ch[1]);
strstr (str, ch[2]);
// etc.