Question

我正在尝试将表单数据插入到数据库中，但它无法正常工作，我已添加下面的代码。请检查并告诉我问题是什么。由于删除和搜索查询正在运行，但插入是唯一无效的查询。感谢

<form method="post" action= "assign5.php">

Name : <input type ="text" name="name" >

<br>

Flavor : <select name="flavor">

     <option value="Chocolate">Chocolate</option>
     <option value="Vanilla">Vanilla</option>
     <option value="Strawberry">Strawberry</option>
     <option value="MahngiVanilla">MahngiVanilla</option>
     <option value="SastiStrawberry">SastiStrawberry</option>

</select>

<br>

Scoops:
<input type ="radio" name="scoops" value="1">1</input>
<input type ="radio" name="scoops" value="2">2</input>
<input type ="radio" name="scoops" value="3">3</input>
<input type ="radio" name="scoops" value="4">4</input>
<input type ="radio" name="scoops" value="5">5</input>

<input type ="submit" name="button" value="Place Order"/>

<br>
<br>

  </form>
 <?php
require_once 'login.php';

$connection = new mysqli($db_hostname,$db_username,$db_password,$db_database);

if($connection ->connect_error) die($connection ->connect_error);

if(isset($_POST['name']) && isset($_POST['Flavour']) && isset($_POST['Scoops']) ){

    $CName=$_POST['name'];
    $Flavor=$_POST['flavor'];
    $Scoops=$_POST['scoops'];

    $sql ="INSERT INTO orders VALUES (CName, Flavour, Scoops) VALUES ('$CName', '$FLAVOUR', '$Scoops')";
    $result=$connection->query($sql);
    if(!$result) die($connection->error);
    header("Location: assign5.php");

}

$connection->close();

?>

Answer 1

你必须提供名字，味道和来自形式的scopps。即使你提供，但仍然你的查询不起作用。因为您在表单上的输入键是小写字母，并且您使用第一个字符的大写字母在条件下进行检查。所以试试吧

if(isset($_POST['name']) && isset($_POST['flavour']) && isset($_POST['scoops']) ){
.....
.....
}

您曾两次使用过值。请从

更正此问题

$sql ="INSERT INTO orders VALUES (CName, Flavour, Scoops) VALUES ('$CName', '$FLAVOUR', '$Scoops')";

要：

 $sql ="INSERT INTO orders (CName, Flavour, Scoops) VALUES ('$CName', '$FLAVOUR', '$Scoops')";

数据未插入数据库

1 个答案: