添加列表框值与多值选择SQL数据库在php

Question

我试图从表单帖子添加多个值到我的SQL表(program_celebrity)，我在我的表单帖子如下，读取另一个表:

<label for="ProgramListing"></label>
    <select name="ProgramListing" style="width: 800px;" size="10" multiple id="ProgramListing">
      <?php
            $sql_program_listing = "SELECT * FROM program order by ProgramName ASC"; 
            foreach($conn->query($sql_program_listing) as $row_program_listing){ ?>
      <option value="<?=$row_program_listing["ProgramCode"];?>" <?php if ($row_programlisting["id_program"] == $registrant['id_program']) echo 'selected="selected"';?>><?=$row_program_listing["ProgramName"];?> (<?=$row_program_listing["ReleaseDate"] .")";?>
        </option>
      <?php } ?>
      </select>

什么工作很好现在，我想添加以下信息:

$id_celebrity = $_POST['id_celebrity'];
$id_program_celebrity = $_POST['id_celebrity']+$_POST['ProgramListing'];

And i use:

foreach ($_POST['ProgramListing'] as $id_program)
{
$sql_connect_celebrity = "INSERT INTO 
        program_celebrity 
        (
        id_program,
        id_celebrity,
        id_program_celebrity
        )
        VALUES 
        (?,?,?)";
$stmt = $conn->prepare($sql_connect_celebrity);
$stmt->bindValue(1, $id_program);
$stmt->bindValue(2, $id_celebrity);
$stmt->bindValue(3, $id_program_celebrity);
$stmt->execute();
}

但它的工作很好，当我发布只是一个选择，但我希望表单发布多个选择，并在表中注册它们。似乎不能得到它的权利，你知道我做错了什么?

Answer 1

修改表单

<select name="ProgramListing" style="width: 800px;" size="10" multiple id="ProgramListing">

<select name="ProgramListing[]" style="width: 800px;" size="10" multiple id="ProgramListing">