我有一個代碼。不知何故,它只能夠選擇最後一個隱藏的輸入名稱字段,而不是另一個。我也嘗試使用if和else,但沒有顯示任何內容。請指教。如何區分PHP中的名稱
沒有,如果其他情況下的場景:
HTML:
<div class="tab-label">
<input type="radio" id="ldktech_product_good" name="conditition" value="good" >
<input type="hidden" id="ldktech_product_price_good" name="price" value="7.50">
<label for="ldktech_product_good">Good</label>
注:請在您的iPad以舊換新,或更換費用充電器將從報價 中扣除
<div class="tab-label">
<input type="radio" id="ldktech_product_flawless" name="conditition" value="flawless" >
<input type="hidden" id="ldktech_product_price_flawless" name="price" value="10">
<label for="ldktech_product_flawless">Flawless</label>
PHP:
$condition = $_POST["conditition"];
$price = $_POST["price"];
echo $price;
echo "<br>";
echo $condition;
通過的if else場景:
HTML代碼:
<div class="tab-label">
<input type="radio" id="ldktech_product_good" name="conditition" value="good" >
<input type="hidden" id="ldktech_product_price_good" name="price-good" value="7.50">
<label for="ldktech_product_good">Good</label>
<div class="tab-label">
<input type="radio" id="ldktech_product_flawless" name="conditition" value="flawless" >
<input type="hidden" id="ldktech_product_price_flawless" name="price-flawless" value="10">
<label for="ldktech_product_flawless">Flawless</label>
PHP代碼:
$condition = $_POST["conditition"];
if($condition == "good"){
$price = $_POST["price-good"];}
else if ($condition == "flawless"){
$price = $_POST["price-flawless"];}
echo $price;
echo "<br>";
echo $condition;
沒有工作。請告知
您可能需要通過你的代碼,並檢查拼寫,因爲你拼錯的「狀態」在這裏:$ _ POST [「conditition」] – slevy1