有時在搜索第二個按鈕時，站點無法正確加載，因此我想檢查第一個按鈕是否可用，以便我知道站點加載正確。目前它只檢查一次 firstbutton 是否可用，如果這是真的，它會停止回圈。那么我如何回圈這段代碼，以便它每次檢查 firstbutton 是否可用，即使在 else 函式中運行 while not 代碼之后也是如此。
firstbutton 和 secondbutton 之后的代碼不能相同，所以像 Prophet's answer 這樣的東西不起作用。
while not (driver.find_elements(By.XPATH, "firstbutton")):
driver.get('different url')
driver.refresh()
else:
secondButton = driver.find_elements(By.XPATH, "secondbutton")
while not (driver.find_elements(By.XPATH, "secondbutton")):
time.sleep(1)
driver.refresh()
secondbutton[0].click()
uj5u.com熱心網友回復：
# wait while firstbutton is present
while not (driver.find_elements(By.XPATH, "firstbutton")):
driver.get('different url')
driver.refresh()
# firstbutton is found
secondButton = driver.find_elements(By.XPATH, "secondbutton")
# keep waiting and refreshing till we have both first and second button
while (not (driver.find_elements(By.XPATH, "secondbutton")) or
not (driver.find_elements(By.XPATH, "firstbutton"))):
time.sleep(1)
driver.refresh()
secondbutton[0].click()
應該適用于您的情況。
uj5u.com熱心網友回復：
看起來您正在嘗試重繪 頁面，直到 first_button和second_button 都出現在那里？
如果是這樣，您只需要重繪 ，直到找到兩個元素。
像這樣的東西：
first_button = driver.find_elements(By.XPATH, "firstbutton")
second_button = driver.find_elements(By.XPATH, "secondbutton")
while not (first_button and second_button):
first_button = driver.find_elements(By.XPATH, "firstbutton")
second_button = driver.find_elements(By.XPATH, "secondbutton")
time.sleep(1)
secondbutton[0].click()
轉載請註明出處，本文鏈接：https://www.uj5u.com/gongcheng/398283.html
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