<?php
require 'connect.inc.php';
$food= $_POST['food'];
$cost= $_POST['cost'];
$sqlinsert = "INSERT INTO test (eat, pay) VALUES ('$food','$cost')";
?>
<form method = "POST" action = "index.php">
Food: <input type = "text" name = "food">
<br/>
Cost: <input type = "text" name = "cost">
<br/>
<input type = "submit" value = "order">
</form>
爲什麼我的代碼不工作? index.php是好的,它連接好,但沒有輸出顯示在我的數據庫中時,我運行此代碼。爲什麼我的代碼不能將數據插入數據庫?
你在哪裏運行查詢? – Federkun
由於將查詢文本分配給變量並不意味着該查詢將被執行。 –
執行mysqli_query() –