如何用数组中的字符串查找特定字母
问题描述:
如何获得包含P或p的项目并返回此数组中出现的匹配数量?如何用数组中的字符串查找特定字母
{ 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" }
};
答
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;
}
你也可以使用strpbrk()作为TonyB建议以取代内部的循环:
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++;
}
不过,我不是确定strbpbrk是如何实现的,但如果仅仅为字符串的大小执行for循环,我不会感到惊讶。在这种情况下,通过两次调用strbpbrk,这个解决方案的效率会比第一个解决方案低。可能有更好的解决方案,但我希望你明白了。 这是输出:
275 Royal Gala Apples
916 Seedless Grapes
385 Pomegranate
found 3 matching products
看手册页strpbrk()...这应该让你开始。如果你仍然有问题,发布您的代码,并要求您的具体问题。 – TonyB