PHP显示基于昨天和今天的数据(PHP show data based on Yesterday and Today)

 这行代码有一些很容易发生的潜在问题 
 
scanf("%s",&pattern);
 
 你可以让它像这样安全一点 
 
scanf("%6s",&pattern);
 
 这是一个6因为你需要在字符串的末尾添加一个额外的字节'\0' ，这将我们带到下一个问题 
 
for(i=0;i<7;i++)
 
 在这里你假设所有字节都是非nul ，除了你正在用scanf()读取一个字符串，除非你创建的数组比你想要存储在其中的字符数大一个字节， scanf()会溢出数组，即写一个字节超出它的界限。 
 
 将"%6s"长度说明符添加到格式字符串可解决此问题，但您只能在数组中存储6非nul字节，而对于另一个问题 
 
for (i = 0 ; pattern[i] != '\0' ; i++)
 
 会更好，因为您不需要事先知道字符串的长度，也不会冒险读取数组的末尾。 
 
 尝试这个： 
 
#include <stdio.h>

int main(void)
{
    char  letter;
    char  c;
    int   i; 
    char  pattern[7];
    printf("Enter a letter: ");
    if (scanf(" %c",&letter) != 1)
     {
        printf("error: invalid input.\n");
        return -1;
     }
    printf("Enter a pattern: ");
    if (fgets(pattern, sizeof(pattern), stdin) == NULL)
     {
        printf("error: end of file.\n");
        return -1;
     }
    for (i = 0 ; pattern[i] != 0 ; ++i)
    {
        if (letter == pattern[i])
        {
            c++;
            printf("Letter < %c > is found in pattern %s\n", letter, pattern);
            printf("in adress: %p\n", pattern + i);
            printf("index    :%d\n", i);
        }

    }
    if (c == 0)
        printf("The pattern does not include any letter");
    return 0;
}
 
 你也错误地打印了地址，因为在pointer[i]下标操作符取消引用指针，它相当于*(poitner + i) 。 

This line of code has some potential problem that can easily happen
 
scanf("%s",&pattern);
 
you can make it a little bit safer like this
 
scanf("%6s",&pattern);
 
it's a 6 because you need one extra byte '\0' at the end of the string, which takes us to the next problem
 
for(i=0;i<7;i++)
 
here you are assuming that all bytes are non-nul which would be ok except that you are reading a string with scanf() and unless you create the array one byte bigger than the number of characters you want to store in it, scanf() will overflow the array i.e. write a byte beyond it's bounds.
 
Adding the "%6s" length specifier to the format string solves this, but you can only store 6 non-nul bytes in the array, and for the other problem
 
for (i = 0 ; pattern[i] != '\0' ; i++)
 
would be better, because you don't need to know the length of the string in advance and you don't risk reading past the end of the array.
 
Try this:
 
#include <stdio.h>

int main(void)
{
    char  letter;
    char  c;
    int   i; 
    char  pattern[7];
    printf("Enter a letter: ");
    if (scanf(" %c",&letter) != 1)
     {
        printf("error: invalid input.\n");
        return -1;
     }
    printf("Enter a pattern: ");
    if (fgets(pattern, sizeof(pattern), stdin) == NULL)
     {
        printf("error: end of file.\n");
        return -1;
     }
    for (i = 0 ; pattern[i] != 0 ; ++i)
    {
        if (letter == pattern[i])
        {
            c++;
            printf("Letter < %c > is found in pattern %s\n", letter, pattern);
            printf("in adress: %p\n", pattern + i);
            printf("index    :%d\n", i);
        }

    }
    if (c == 0)
        printf("The pattern does not include any letter");
    return 0;
}
 
You also was printg the address wrong, because in pointer[i] the subscript operator dereferences the pointer, it's equivalent to *(poitner + i).