php - 在 <select> 中显示 MySQL 数据失败

Question

我只是想显示一个 html <select>显示来自 MySQL 数据库行的选项的下拉列表。据我所知，这应该有效。但是，它没有在下拉列表中显示任何内容。 campaigns表肯定在 campaignid 中有数据和 name列。不知道为什么它不显示任何内容。请帮忙。

应该显示的形式 name

<?php
//include db connect
  include ("db_con.php");



   //campaign change form function
  function changeCampaign() {
    //set variables
    $query = mysqli_query($con,"SELECT * FROM campaigns");
    //echo form
    echo '<table border="1">
          <form action="functions/campaign_change.php" name="campaignChange" method="post">
            <tr>
              <td><select name="campaignList">';
                    //loop campaigns
                    while ($row = mysqli_fetch_array($query)) {
                      echo "<option value='" . $row['campaignid'] . "'>" . $row['name'] . "</option>";
                    }
        echo '</select></td>
            </tr>
            <tr>
              <td><input type="submit" value="Load" /></td>
            </tr>
          </form>
        </table>';
}
?>

然后我只需调用 changeCampaign();函数，它正在显示带有 <select> 的表单只是里面没有内容。

添加了以下代码以显示成功添加到同一个表的脚本

这是创建 campaign 的表单

//campaign creation form function
  function createCampaign() {
    echo '<table border="1">
          <form action="functions/campaign_create.php" name="campaignCreate" method="post">
            <tr>
              <td><input type="text" name="name" placeholder="Enter Campaign Name" required /></td>
            </tr>
            <tr>
              <td><center><input type="submit" name="Create" /></center></td>
            </tr>
          </form>
          </table>';
}

campaign_create.php

<?php
//include db connect
  include ("db_con.php");

//start session
  session_start();

//set variable names
  $username = $_SESSION['username'];
  $campaignname = addslashes($_POST['name']);

//validate campaign name length
  if ((strlen($campaignname) < 6) || (strlen($campaignname) > 55)) {
    echo 'Campaign name must be between 6 and 55 characters - Please go back and try again.';
    exit;
  }

//create new campaign
  $query = mysqli_query($con, "INSERT INTO campaigns (creator, name) VALUES ('".$username."', '".$campaignname."')");
  if ($query) {
    echo 'Campaign created successfully!';
    header( "refresh:2;url=../index.php" );
  } else {
    echo 'There was an unknown error in creating the campaign - Please go back and try again.';
  }

?>

附言- 是的，campaignid是表中的一行，其中确实有数据。

Answer 1

首先，您的 <select> HTML 标签永远不会以 </select> 结束.

二、$con变量在 changeCampaign 中不可用功能范围。添加global $con;到函数的开头来修复它。

此外，我更喜欢用这种东西跳进跳出 PHP block 。考虑这段代码:

<?php
//include db connect
include ("db_con.php");

//campaign change form function
    function changeCampaign() {
    global $con; // we need the connection data in here.
    //set variables
    $query = mysqli_query($con,"SELECT * FROM campaigns");
?>
<table border="1">
    <form action="functions/campaign_change.php" name="campaignChange" method="post">
        <tr>
            <td><select name="campaignList">';
<?php
                    //loop that is NOT displaying the name
                    while ($row = mysqli_fetch_array($query)) {
                        echo "<option value='" . $row['campaignid'] . "'>" . $row['name'] . "</option>";
                    }
?>
            </select></td>
        </tr>
        <tr>
            <td><input type="submit" value="Load" /></td>
        </tr>
        </form>
    </table>';
<?php
}
?>

在这里，您有输出的 HTML 代码，就像它是 echo 一样。 ed，但它更好。