Question

这只是我的程序中的一个函数，它应该搜索用户在struct数组中输入的字符串或整数。我尝试做的是添加一条错误消息，并在结构数组中找不到输入的字符串或整数时再次尝试。如果输入正确的字符串或整数，它会搜索得很好，但如果你没有，则没有任何反应。我试图改变它，但无法找到解决方案。

我已经尝试了一段时间，其中一个案例在我的switch语句中，但我需要为这三个案例执行此操作。但到目前为止，我只尝试过案例3.

search(struct items aItems[], int *num_items)
{
    int choice_of_search, b=0, found_word=0, search_num, i=0, j=0, k=0;
    char search_phrase[20]; struct items search[MAX];

    printf("Choose what to search for? (1) Item number, (2) Name and (3) Balance. ");
    scanf("%d", &choice_of_search);

    while(choice_of_search < 1 || choice_of_search > 3)
    {
        printf("Wrong choice!\n");
        printf("Choose what to search for? (1) Item number, (2) Name and (3) Balance. ");
        scanf("%d", &choice_of_search);
    }


    switch(choice_of_search)
    {
        case 1:
            printf("Item number?\n");
            scanf("%d", &search_num);
            for(i = 0; i < *num_items; i++)
            {
                if(search_num == aItems[i].itemnumber)
                {
                    printf("Item number found!\n");
                    search[found_word]=aItems[i];
                    found_word+=1;
                }
            }
            break;

        case 2:
            printf("Name?\n");
            scanf("%s", search_phrase);
            for(i = 0; i < *num_items; i++)
            {
                if(strstr(aItems[i].name, search_phrase))
                {
                    printf("Name found!\n");
                    search[found_word]=aItems[i];
                    found_word+=1;
                }
            }
            break;

        case 3:
            printf("Balance?\n");
            scanf("%d", &search_num);
            for(i = 0; i < *num_items; i++)
            {
                if(search_num == aItems[i].balance)
                {
                    printf("Balance found!\n");
                    search[found_word]=aItems[i];
                    found_word+=1;
                }
                else
                {
                    printf("Balance not found! Try again.\n");
                    printf("Balance?\n");
                    scanf("%d", &search_num);
                }
            }
            break;
    }


    while(b < found_word)
    {
        printf("Item number: %d Name: %s  Balance: %d\n", search[b].itemnumber, search[b].name, search[b].balance);
        b++;
    }
}

Answer 1

也许这可以帮助

int done;
...
...

    case 3:
        done = 0;
        while(1);
        {
            printf("Balance?\n");
            scanf("%d", &search_num);
            for(i = 0; i < *num_items; i++)
            {
                if(search_num == aItems[i].balance)
                {
                    printf("Balance found!\n");
                    search[found_word]=aItems[i];
                    found_word+=1;
                    done = 1;
                }
            }
            if (done) break;

            printf("Balance not found! Try again.\n");
        } 
        break;

但请注意，代码不是用户友好的，因为它不允许用户在没有匹配的情况下停止搜索。所以也许你应该考虑添加一个“你想再试一次”选项。

一种简单的方法可能是

            printf("Balance not found! Would like to try again?.\n");
            scanf(" %c", &some_char);
            if (some_char != 'y') break;

在struct array中搜索用户输入的字符串

1 个答案: