Question

我试图制作一个下拉列表，允许用户选择符合其需求的培训计划，因此当他们选择它时，该选项将进入MySQL数据库中的表。

我的代码是：

<form action="sessionreq.php" method="POST">
    <div class="form-group">
        <label for="exampleSelect1">Category</label>
        <select class="form-control" id="exampleSelect1">
            <option name='eg1'>Example option</option>
            <option name='eg2'>Example option 2</option>
            <option name='eg3'>Example option 3</option>
        </select>
    </div>
    <button type="submit" class="button button-block" name="delete">Request</button>
    <br>

</form>

我需要的是一个SQL查询，因此当他们选择＆＃39;示例选项3＆＃39;时，它将在名为method的列中输入到数据库中。我应该使用值插入method（取决于它们在下拉列表中的选项）。例如，如果用户选择Example option 2，则会将method插入Example option 2，其值为class CompaniesController < ApplicationController before_action :authenticate_user!, only: [ :new, :create, :edit, :update, :destroy ] def new @company = Company.new @company.quotes.build @company.employees.build end def create @company = current_user.companies.new(company_params) if @company.save redirect_to company_quote_url(@company.id), notice: 'Quote request created' else render :new end end private def company_params params.require(:company).permit(:co_name, :co_number, :postcode, :industry, :quotes_attributes => [:id, :lives_overseas, :payment_frequency], :employees_attributes => [:id, :first_name, :last_name, :email, :gender, :date_of_birth, :salary, :_destroy] ) end end。

TL; DR如何使用下拉列表填充数据库表/列？

固定见Using $_POST to get select option value from HTML

Answer 1

首先为您的select命名。在这种情况下，我给了training

<form action="sessionreq.php" method="POST">
    <div class="form-group">
        <label for="exampleSelect1">Category</label>
        <select class="form-control" id="exampleSelect1" name="training">
            <option value='Example 1'>Example option 1</option>
            <option value='Example 2'>Example option 2</option>
            <option value='Example 3'>Example option 3</option>
        </select>
    </div>
    <button type="submit" class="button button-block" name="delete">Request</button>
    <br>

</form>

现在，在您的sessionreq.php中，您可以通过$_POST获取值

$_POST['training']; //You can manipulate the data anyway you like.

现在您可以运行要插入的查询。鉴于是准备好的声明（首选）

$stmt = $connection->prepare("INSERT INTO table (column) VALUES (?)");
$stmt->bind_param("s",$_POST['training']);
$stmt->execute();
$stmt->close();

从下拉列表中将数据插入MySQL

1 个答案: