在<select>失败</select>中显示MySQL数据

Question

我只是想显示一个显示MySQL数据库行选项的html <select>下拉列表。据我所知，这应该是有效的。但是，它没有在下拉列表中显示任何内容。 campaigns表肯定包含campaignid和name列中的数据。不知道为什么它没有显示任何东西。请帮忙。

应显示name

的表单

<?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>';
}
?>

然后我只是调用changeCampaign();函数，该函数显示带有<select>的表单，其中没有任何内容。

添加了以下代码以显示添加到同一表的成功脚本

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.';
  }

?>

P.S。 - 是的，campaignid是表格中的一行，并且其中包含数据。

Answer 1

您需要回显一个关闭的选择标记。只需在循环后添加此行，就可以了。

echo '</select>';

还可以尝试将循环的第一行更改为：

while ($row = mysqli_fetch_array($con, $query)) {

Answer 2

首先，您的<select> HTML代码永远不会被</select>关闭。

其次，$con函数范围内的changeCampaign变量不可用。将global $con;添加到函数的开头以进行修复。

另外，我喜欢用这种东西跳进和跳出PHP块。请考虑以下代码：

<?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
}
?>

在此，您输入的HTML代码就像echo编辑一样，但它更好。