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返回陣列我提交表單,使用下面的函數(prize.php)
:PHP - 從功能
loadmodule('validate'); //This just loads the validate.php function.
$validate = new Validate;
if($_POST)
{
$validateForm = $validate->validateForm();
switch($validateForm)
{
case 1:
$error = 'You\'re not logged in..';
$stop = true;
break;
//If no error = success.
if($validateForm['code'] == "100"){
$won = $val['prize'];
$type = $val['type'];
$success = 'You have won! The prize was '.$won.' '.$type.'';
die($success);
}
}
die($error);
}
這是爲了驗證表單(validate.php)
功能:
function validate()
{
global $userdata;
if(!is_array($userdata))
return 1; // User not logged in - return error code one.
//If no error, lets show a success message.
$prize = "100";
$text = "dollars";
return array("code"=>"100","prize"=>"$prize","type"=>"$text");
}//If won
}
上面的代碼返回:
Notice: Undefined variable: error in /home/.../public_html/pages/prize.php on line 27
雖然,它不應該在那裏拋出一個錯誤,罪惡ce die($success)
應該由代碼100觸發。
我在做什麼錯?
它很容易被發現。如果案例不是1,那麼您將得到該通知,因爲'$ error'沒有被定義。 – 2014-09-23 17:47:19
@Hanky웃Panky你能舉一個例子嗎?我現在失去了.. – oliverbj 2014-09-23 17:58:51
是'$ validate-> validateForm()'和'validate()'函數你顯示應該是相同的?在那種情況下,這是問題之一... – 2014-09-23 18:22:45