Question

我有以下表格，但我无法在数据库中添加SELECT（具有多项选择的下拉菜单）值。我哪里错了？我正在使用PHP，除了选择部分，我可以正确添加所有其他值。我想有一个带有一些选项的下拉菜单，当我按“提交”时，所选的值进入数据库。

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

                            <div class="form-group">
                    <label for="name">Customer Name:</label>
                    <input type="text" name="name" class="form-control" id="name">
                        </div>

                <div class="form-group">
                    <label for="email">Email address:</label>
                    <input type="email" name="email" class="form-control" id="email">
                </div>

                <div class="form-group">
                    <label for="number">Mobile Number:</label>
                    <input type="text" name="number" class="form-control" id="number">
                </div>

                <div class="form-group">
                    <label for="price">Price:</label>
                    <input type="text" name="price" class="form-control" id="price">
                </div>

                <div class="form-group">
                    <label for="payment">Payment Type</label>
                    <input type="text" name="payment" class="form-control" id="payment">
                </div>

                <div class="form-group">
                    <label for="device">Device Name:</label>
                    <input type="text" name="device" class="form-control" id="device">
                </div>

                 <div class="form-group">
                    <label for="status">Status:</label>
             <select name="status[]" multiple class="form-control" id="status">
<option value="New">New</option>
<option value="Green">Progress</option>
<option value="Blue">Wait</option>
<option value="Pink">Done</option>
<option value="Yellow">Close</option>
</select>

      </div>             
                 <button type="submit"  name="submit" class="btn btn-default">Submit</button>
</form>

和查询：

$sql = "INSERT INTO ... (name, mail, number, device, price, paymenttype,status,date) VALUES ('$name', '$email', '$number', '$device', '$price', '$payment','$status',NOW())";
if(mysqli_query($link, $sql)){
    // echo "Records added successfully.";
    header("location:....php?message=The customer has been added to the database");

} else{
    echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}

Answer 1

您必须更改以下HTML代码：

name="status[]"

要：

name="status"

另外，根据this，您应该在重定向页面后添加exit()：

header("location:....php?message=The customer has been added to the database");
exit();

如何使用php将dropdownmenu的值传递给数据库

1 个答案: