我想創建一個刪除從不斷接觸用戶的功能。這個函數爲常量聯繫調用一個包裝類,並且它可以工作,但是如果給它提供一個在他們的站點上不存在的電子郵件地址,它會返回一個可捕獲的致命錯誤。PHP錯誤信息幫助和try/catch
我是新來的try/catch和我不太得到在哪裏放置,這樣我可以創造一個優美的錯誤信息,而不是我發現了捕致命錯誤。下面是我當前的代碼:
function ccdeleteuser($emailaddress)
{
//this code accesses the constant contact wrapper class to delete a user based on email
$ConstantContact = new ConstantContact("basic", "apikey", "usr", "pwd");
$SearchContact = $ConstantContact->searchContactsByEmail($emailaddress);
$DeleteContact = $ConstantContact->deleteContact($SearchContact[0]);
}
$emailtotry = "[email protected]"; //this is email is NOT in Constant Contact
ccdeleteuser($emailtotry);
現在如果我運行此我得到以下錯誤:
Catchable fatal error: Argument 1 passed to ConstantContact::deleteContact() must be an instance of Contact, null given, called in [path to my page] on line 19 and defined in [path to constant contact php wrapper file] on line 214
任何幫助表示讚賞!
你應該檢查'$ SearchContact'是否是一個數組;那麼你不必擔心異常。 –
DUP:http://stackoverflow.com/questions/2468487/how-can-i-catch-a-catchable-fatal-error-on-php-type-hinting – xdazz