我想在 if-else 中定義 $roomprice 但結果什么也沒顯示。我做錯什么了嗎？
$answer = $_POST['ans'];
$roomprice = roomprice($answer, $roomprice);
function roomprice($answer, $roomprice)
{
if($answer == "singleroom")
{
$roomprice = 150;
return $roomprice;
}
else if($answer == "deluxeroom")
{
$roomprice = 180;
return $roomprice;
}
}
roomprice($answer, $roomprice);
echo "$roomprice";
uj5u.com熱心網友回復：
您必須將函式呼叫分配給一個變數，否則它回傳的值將丟失。例如：
$result = roomprice($answer, $roomprice);
uj5u.com熱心網友回復：
您應該將函式中回傳的資料分配給一個變數。
$roomprice = roomprice($answer, $roomprice);
uj5u.com熱心網友回復：
你的功能沒有問題。只能獲取$answer 的值。
$answer = 'deluxeroom'; //$_POST['ans'] Cannot be retrieved here.
$roomprice = roomprice($answer, $roomprice);
function roomprice($answer, $roomprice)
{
if($answer == "singleroom")
{
$roomprice = 150;
return $roomprice;
}
else if($answer == "deluxeroom")
{
$roomprice = 180;
return $roomprice;
}
}
roomprice($answer, $roomprice);
echo "$roomprice"; // 180
uj5u.com熱心網友回復：
我看到了一些問題，盡管代碼仍應為您提供答案（以及警告）。您不會得到答案的一個原因是 $_POST['ans'] 設定不正確（它需要通過表單發布設定）。
您在這里兩次呼叫 roomprice 一次在第二行，一次在倒數第二行
您還試圖將 $roomprice 變數傳遞給 roomprice 函式，但該變數未設定（并且不需要）
//$answer = $_POST['ans'];
//set answer to known value for testing
$answer = "singleroom";
//remove $roomprice variable
//$roomprice = roomprice($answer, $roomprice);
$roomprice = roomprice($answer);
//echo the $roomprice
echo "$roomprice";
//change the function to not need $roomprice
//function roomprice($answer, $roomprice)
function roomprice($answer)
{
if($answer == "singleroom")
{
$roomprice = 150;
return $roomprice;
}
else if($answer == "deluxeroom")
{
$roomprice = 180;
return $roomprice;
}
}
//not needed
//roomprice($answer, $roomprice);
//echo "$roomprice";
轉載請註明出處，本文鏈接：https://www.uj5u.com/qukuanlian/384803.html
上一篇：使用參考號的多個if陳述句