为什么 bakeryid
变量没有发布在我的表单中?我收到的错误是
"Notice:Undefined variable bakeryid"
我有两个页面,一个显示表单,第二个是表单的操作。第二种形式也一直说它是未定义的。 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>';
最佳答案
您使用 $_POST['bakeryid']
设置 $bakeryid
,然后使用 $bakeryid
定义 Bakeryid。
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>";
关于php - 为什么这个变量没有被发布并给出错误?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/28399862/