Question

我正在尝试创建一个下拉菜单，供用户将他们的订单插入我的数据库系统。

我在创建下拉菜单时遇到了问题，但我对我应该做的事情有了一般的了解。我是新手，所以请耐心等待。如果它有帮助，这是我的uml：

<?php

$databaseName = 'pizza_db';
$databaseUser = 'root';
$databasePassword = 'root';
$databaseHost = '127.0.0.1';
$conn = new mysqli($databaseHost, $databaseUser, $databasePassword, $databaseName);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}
echo "Connected sucessfully";
if(isset($_POST['submit'})){

$sql = "SELECT size FROM size";
$result = mysqli_query($sql); 
$opt = "<select name='size'>";
while($row = mysqli_fetch_assoc($result)) {
    $opt .= "option value='{$row['size']}' >{$row['size']}></option>";
}

$opt .="</select>";

if ($conn->query($sql)){
    $msg = "Data inserted successfully";
} else{
    $msg = "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
}
?>


<!DOCTYPE html>
<html>
<head>
    <title>Pizza</title>
</head>
<body>
<form action="size.php" method="POST">
    <select name="size">

    </select>
    <button type ="submit" name="submit"> Submit</button
</form>
</body>
</html>

我知道我的HTML代码错误或不完整。如何创建多个下拉菜单以及如何在一个脚本中使用多个$ sql insert语句以将数据提交到我的数据库中？任何帮助将不胜感激。

Answer 1

你几乎没有错别字。在第13行中，if(isset($_POST['submit'})){应为if(isset($_POST['submit'])){。在while循环中，你应该有$opt .= "<option value='{$row['size']}' >{$row['size']}></option>";（缺少“＆lt;”）
此外，还有一些冗余代码。 mysqli_query()和$conn->query()执行相同的操作（see this SO tread）。简单的OOP与程序语法。

您可以删除

if ($conn->query($sql)){
    $msg = "Data inserted successfully";
}
else{
     $msg = "Error: " . $sql . "<br>" . $conn->error;
}

之后，这样的事情就足够了

<?php

$databaseName = 'pizza_db';
$databaseUser = 'root';
$databasePassword = 'root';
$databaseHost = '127.0.0.1';
$conn = new mysqli($databaseHost, $databaseUser, $databasePassword, $databaseName);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}
echo "Connected sucessfully";
if(isset($_POST['submit'])){

    $sql = "SELECT size FROM size";
    $result = mysqli_query($sql); 
    $opt = "<select name='size'>";
    while($row = mysqli_fetch_assoc($result)) {
        $opt .= "<option value='{$row['size']}' >{$row['size']}></option>";
    }

    $opt .="</select>";
    $conn->close();
}
?>

但是，正如我在此理解的那样，您有一个空表单，在提交时会获得所有大小。如果你想让它像 php get sizes ＆gt; 生成表单＆gt; 提交表单并重定向到下一步你应该像这样或类似的那样（并从php部分删除if $_POST['submit']条件）

<!DOCTYPE html>
<html>
<head>
    <title>Pizza</title>
</head>
<body>
<form action="submit-chosen-size.php" method="POST">
    <label for="size">Select your size:</label>
    <?php echo($obj); ?>
    <button type ="submit" name="submit"> Submit</button
</form>
</body>
</html>

编辑：请尝试使用此代码。它与上面的代码非常相似，但有助于找到问题

<?php
$databaseName = 'pizza_db';
$databaseUser = 'root';
$databasePassword = 'root';
$databaseHost = '127.0.0.1';
$conn = new mysqli($databaseHost, $databaseUser, $databasePassword, $databaseName);
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}
echo "Connected sucessfully";

$sql = "SELECT size FROM size";
$result = mysqli_query($sql); 
$opt = "<select name='size'>";
while($row = mysqli_fetch_assoc($result)) {
    $opt .= "<option value='{$row['size']}' >{$row['size']}></option>";
    echo $row['size']; // you can remove this later
}
$opt .="</select>";
echo $obj; //you can remove this later
$conn->close();
?>

<!DOCTYPE html>
<html>
    <head>
        <title>Pizza</title>
    </head>
    <body>
        <form action="submit-chosen-size.php" method="POST">
            <label for="size">Select your size:</label>
            <?php echo(isset($obj) ? $obj : "Empty"); ?>
            <button type ="submit" name="submit"> Submit</button
        </form>
    </body>
</html>

PHP创建下拉菜单，插入数据库

1 个答案: