Question

为什么bakeryid变量未在我的表单中发布？我收到的错误是

“注意：未定义的变量fabid”

我有两个页面，一个显示表单，第二个是表单的操作。第二种形式一直说它也未定义。 bakeryid是每个蛋糕订单的ID。

$sql = mysqli_query($con,"SELECT `firstname`, `bakeryid`, `order` FROM cakes");

$bakeryid = $_POST['bakeryid'];

?>
  <table border='2'>
    <th>First Name</th>
    <th>Order</th>
<?php
  echo '<form name="display" method="POST" action="cakephp.php">';
  while($row = mysqli_fetch_array($sql))
  {
    echo "<tr>";
    echo "<td>" . $row['firstname'] . "</td>";
    echo "<td>" . $row['order'] . "</td>";
    echo '<td><input type="hidden" name="bakeryid" value="' . $bakeryid . '"/></td>';
    echo '<td><input type="hidden" name="memid" value="' . $memid . '"/><input type="submit" name="takeorder" value="Take Order" ></td>';
    echo "</tr>";
  }
  echo "</form>";
  echo "</table>";

答案：

echo '<td><input type="hidden" name="bakeryid" value="' . $row['bakeryid'] . '"/></td>';

Answer 1

$sql = mysqli_query($con,"SELECT `firstname`, `bakeryid`, `order` FROM cakes c INNER JOIN members m ON c.memid = m.memid");

$bakeryid = $_POST['bakeryid']; ?> 
<table border='2'>
    <th>First Name</th>
    <th>Order</th>
<?php
        echo '<form name="display" method="POST" action="cakephp.php">';
        while($row = mysqli_fetch_array($sql))
        {
                echo '<tr>
                          <td>' . $row['firstname'] . '</td>
                          <td>' . $row['order'] . '</td>
                          <td><input type="hidden" name="bakeryid" value="' . $row['bakeryid'] . '"/></td>
                          <td><input type="hidden" name="memid" value="' . $memid . '"/>
                              <input type="submit" name="takeorder" value="Take Order" >
                          </td>
                     </tr>';
        }
echo '</form></table>';

fabid没有出现在表单中的原因是，在设置字段值时，你使用$ bakeryid设置为尚未发生的帖子。您想将值设置为$ row ['bakeryid']，如上所述。

Answer 2

您使用$bakeryid设置$_POST['bakeryid']，但随后使用$bakeryid定义fabid。

请尝试：

$sql = mysqli_query($con,"SELECT `firstname`, `bakeryid`, `order` FROM cakes c INNER JOIN members m ON c.memid = m.memid");

$bakeryid = $_POST['bakeryid']; // this line is unnecessary

    ?> <table border='2'>
            <th>First Name</th>
            <th>Order</th>
<?php
        echo '<form name="display" method="POST" action="cakephp.php">';
while($row = mysqli_fetch_array($sql))
            {
                    echo "<tr>";
                    echo "<td>" . $row['firstname'] . "</td>";
                    echo "<td>" . $row['order'] . "</td>";
                    echo '<td><input type="hidden" name="bakeryid" value="' . $row['bakeryid'] . '"/></td>'; // this line changed
                    echo  '<td><input type="hidden" name="memid" value="' . $memid . '"/><input type="submit" name="takeorder" value="Take Order" ></td>';
                echo "</tr>";
            }
echo "</form>";

为什么没有发布此变量并给出错误？

2 个答案: