c - 如何在数组中使用字符串查找所需的特定字母

Question

如何获取包含 P 或 p 的项目并返回此数组中出现的匹配项数？

            {  SKU:  275, "Royal Gala Apples"   },
            {  SKU:  386, "Honeydew Melon"      },
            {  SKU:  240, "Blueberries"         },
            {  SKU:  916, "Seedless Grapes"     },
            {  SKU:  385, "Pomegranate"         },
            {  SKU:  495, "Banana"              },
            {  SKU:  316, "Kiwifruit"           },
            {  SKU:  355, "Chicken Alfredo"     },
            {  SKU:  846, "Veal Parmigiana"     },
            {  SKU:  359, "Beefsteak Pie"       },
            {  SKU:  127, "Curry Chicken"       },
            {  SKU:  238, "Tide Detergent"      },
            {  SKU:  324, "Tide Liq. Pods"      },
            {  SKU:  491, "Tide Powder Det."    },
            {  SKU:  538, "Lays Chips S&V"      },
            {  SKU:  649, "Joe Org Chips"       },
            {  SKU:  731, "Allen's Apple Juice" },
            {  SKU:  984, "Coke 12 Pack"        },
            {  SKU:  350, "Nestea 12 Pack"      },
            {  SKU:  835, "7up 12 Pack"         }
    };

Answer 1

这可能是一个简单的解决方案:

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

#define NPROD 6
struct product {
    int sku;
    char *name;
};

struct product products[NPROD] = {
{  275, "Royal Gala Apples"   },
{  386, "Honeydew Melon"      },
{  240, "Blueberries"         },
{  916, "Seedless Grapes"     },
{  385, "Pomegranate"         },
{  127, "Curry Chickens"      }
};

int main()
{
    int i, j, nfound;
    nfound = 0;
    for (i = 0; i < NPROD; i++) {
        for (j = 0; j < (int)strlen(products[i].name); j++) {
            if (products[i].name[j] == 'p' ||
                products[i].name[j] == 'P') {
                /* Do whathever you want with this product */
                printf("%d %s\n", products[i].sku,
                       products[i].name);
                nfound++;
                break;
            }
        }
    }

    printf("found %d matching products\n", nfound);
    return 0;
}

您也可以使用 strpbrk()，因为 TonyB 建议将内部 for 循环替换为:

if (strpbrk("p", products[i].name) ||
    strpbrk("P", products[i].name)) {
    /* Do whathever you want with this product */
    printf("%d %s\n", products[i].sku, products[i].name);
    nfound++;
}

但是，我不确定 strbpbrk 是如何实现的，但如果它只是对字符串的大小执行 for 循环，我不会感到惊讶。在这种情况下，如果两次调用 strbpbrk，则该解决方案的效率甚至会比第一个解决方案低。可能有更好的解决方案，但我希望你明白。这是输出:

275 Royal Gala Apples
916 Seedless Grapes
385 Pomegranate
found 3 matching products