嘿,我已经试了好几个小时了。我对PHP还不太熟悉,所以请同情我吧!
我正在尝试调用$_POST[$id],并使用foreach将其放入表中。它还告诉我,id是一个身份不明的索引。知道为什么吗?如果你对初学者有任何其他的建议,请随意分享
提前谢谢你!
索引格式:
foreach ($rows as $row) {
$food = $row["food"];
$price = $row["price"];
$picture = $row["picture"];
$description = $row["description"];
$id = $row['id'];
echo "<tr>
<td><img src='$picture' width='120px' /></td>
<td>$food</td>
<td>$$price</td>
<td><input type='number' min='0' max='10' enctype='multipart/form-data' placeholder='#' name='<?php echo $id; ?>' maxlength='1'></td>
</tr>";
}
submitorder.php表:
while($row = $result->fetch_array()){
$rows[] = $row;
}
foreach ($rows as $row) {
$food = $row["food"];
$price = $row["price"];
$id = $_POST[$id];
if(!empty($_POST[$id])){
if(isset($_POST[$id])){
$qty = $_POST[$id];
}else{
echo "Is NOT SET";
}
}else{
echo "Is EMPTY";
}
echo "<tr>
<td>$food</td>
<td></td>
<td>$$price</td>
</tr>";
}
最佳答案
foreach ($rows as $row) {
$food = $row["food"];
$price = $row["price"];
$picture = $row["picture"];
$description = $row["description"];
$id = $row['id'];
echo '<tr>
<td><img src='.$picture.' width=\'120px\' /></td>
<td>'.$food.'</td>
<td>'.$price.'</td>
<td><input type=\'number\' min=\'0\' max=\'10\' enctype=\'multipart/form-data\' placeholder=\'#\' name='.$id.' maxlength=\'1\'></td>
</tr>';
}
while($row = $result->fetch_array()){
$rows[] = $row;
}
foreach ($rows as $row) {
$food = $row["food"];
$price = $row["price"];
if(isset($_POST['id'])){
$qty = $_POST['id'];
}else{
echo "Is NOT SET";
}
echo '<tr>
<td>' . $food . '</td>
<td></td>
<td>' .$price. '</td>
</tr>';
}
修正了错误,我唯一能想到的是你的表单实际上并没有使用POST作为方法。另外,为了您和其他人的利益,如果$_POST['id']值实际上是quantity,请将其重命名为quantity,以减轻您的一些头痛。
关于php - 为什么$ _POST为空?使用foreach将变量$ row用作输入名称,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/21356519/