是否有可能在异常捕获时打印超出范围的变量

Question

我想打印出我的程序崩溃的网址。

我的主像是这样：

if __name__ == '__main__':
    if sys.argv[1] == "allkeyshop":
        try:
            scrape_allkeyshop()
        except AttributeError as e:
            print("page structure probably changed or loaded improperly:", e)
            sys.exit(1)
        except RequestException as e:
            print("Error during requests:", e)
            sys.exit(1)
        except TypeError as e:
            print(e)
            print(LOG_VAR)
            sys.exit(1)

我的功能：

def scrape_allkeyshop():
    mydb = connection_parameters()
    mycursor = mydb.cursor()
    sql = "SELECT link FROM links WHERE domain='AKS' ORDER BY priority"
    mycursor.execute(sql)
    allkeyshop_links = mycursor.fetchall()

    priority = 1
    for i, url in enumerate(allkeyshop_links):
        time.sleep(SLEEP_TIME)
        page_html = simple_get(url[0].strip())
        page_html = BeautifulSoup(page_html, 'html.parser')

是否可以打印出它在main中崩溃的特定URL。该程序通常在simple_get（）上崩溃并抛出TypeError。

我不确定是否可以打印出不同范围的变量，但是我的同事坚持要保持这种结构。