Question

我正在为学校做一个项目，需要在数据库（csv）中搜索产品。它通过将经过验证的EAN-8代码与每行的第0列匹配，在for循环中扫描来完成此操作。有一个尝试，除了试图匹配代码，然后显示产品，除非它无法找到它，当它意味着打印＆＃34;产品未找到，再试一次＆＃34 ;.我尝试了不同的迭代if，else，if，elif结合尝试，除了但无济于事。请帮忙。

#code is the barcode
        for row in csv_file:
            try:
                if code == row[0]:
                    print("This product is: ", row[1], "\n", "Quantity left in stock: ", row[2], "\n", "Cost: ", row[3], "\n")
                    quant = int(input("How many do you you want?: "))
                    if quant > int(row[2]):
                        print("We don't have that many, try again")
                        order()
                    else:
                        orderrow = row
                        orderrow[2] = quant
                        orderrow[3] = float(orderrow[3]) * quant
                        totalcost = totalcost + orderrow[3]
                        orderlist.append(orderrow)


            except:
                print("ITEM NOT FOUND, TRY AGAIN")
                order()

Answer 1

您需要一个合适的0dp，而不是if-else。这里没有try-catch被抓住。怎么样：

Exception

Answer 2

执行此操作的正确方法是在找到匹配的行时设置变量。循环完成后，检查变量以查看是否找到了任何内容，如果没有，则打印该消息。

found = false
for row in csv_file:
    if code == row[0]:
        found = true
        print("This product is: ", row[1], "\n", "Quantity left in stock: ", row[2], "\n", "Cost: ", row[3], "\n")
        quant = int(input("How many do you you want?: "))
        if quant > int(row[2]):
            print("We don't have that many, try again")
            order()
        else:
            orderrow = row
            orderrow[2] = quant
            orderrow[3] = float(orderrow[3]) * quant
            totalcost = totalcost + orderrow[3]
            orderlist.append(orderrow)
if !found
    print("ITEM NOT FOUND, TRY AGAIN")
    order()

我无法理解的Python行为

2 个答案: