如何在输入中找到随机数?

时间:2018-11-13 05:07:41

标签: python python-3.x

我希望print语句打印上下限之间的数字。

我不断收到错误代码:

Traceback (most recent call last):File "python", line 5, in <module> ValueError: non-integer arg 1 for randrange()

从程序中:

from random import*
lowRange = input('What is the lower range number?')
hiRange = input('What is the higher range nunmber?')

ran = randrange (lowRange,hiRange)
print (ran)

2 个答案:

答案 0 :(得分:2)

nextClikcedHandle(continuationToken: string): void { this.customerService.searchCustomersPaged(this.customerService.searchTerm, this.currentContinuationToken) .subscribe(resp => { //add current continuation token, to previous now, as this will be used for 'previous' searching this.previousContinuationTokens.push(this.currentContinuationToken); //set next continuation token received by server this.currentContinuationToken = resp.headers.get('continuationToken'); //return search results this.customerService.searchResults.next(resp.body); }); 函数始终返回一个字符串。如果要使用整数(在这种情况下),则必须使用input()将这些字符串转换为整数。但是,如果用户输入的内容无法转换为整数(例如“ hi”,或仅按回车键),则尝试转换时会收到错误消息。为了解决这个问题,您将需要研究try和except语句。希望有帮助!

答案 1 :(得分:0)

尝试一下:

在这里,直到您输入数字,它都不会停止。 int以外的其他输入将被视为无效输入。

代码中的问题是从输入中读取的所有内容都被当作字符串。

from random import*
while True:
        try:
            lowRange = int(input('What is the lower range number?'))
            break
        except:
            print("That's not a valid input!")
while True:
        try:
            hiRange = int(input('What is the higher range nunmber?'))
            break
        except:
            print("That's not a valid input!")

ran = randrange (lowRange,hiRange)
print (ran)