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。