[英]copy the tracking id from Table 2's column to Table 1's column
在下圖中,一旦單擊第三列中的“ submit ”按鈕,我想將跟蹤ID從表2的列復制到表1的列 ...。
表訂單
在第4列中顯示的跟蹤ID”
我將訂單信息保存在表訂單中 [表1]
我將跟蹤ID保存在表awbno [表2]和列tracking_two中
track.php
<?php
$con = mysqli_connect("localhost","root","","do_management4");
$result = mysqli_query($con,"SELECT * FROM orders");
echo "<table border='1'>
<tr>
<th>order</th>
<th>payment</th>
<th>generate</th>
<th>tracking id</th>
</tr>";
while($row = mysqli_fetch_array($result))
{
$id = $row['id'];
echo "<tr>";
echo "<td>" . $row['order_id'] . "</td>";
echo "<td>" . $row['payment_type'] . "</td>";
echo "<td>";
if (empty($row['tracking_one'])) {
echo "<form method='post' action='call.php'>";
echo "<input type ='hidden' name='id' value='$id'>
<input type='submit'>
</form>";
}
echo "</td>";
echo "<td>" . $row['tracking_one'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
call.php
<?php
$con = mysqli_connect("localhost","root","","do_management4");
$result = mysqli_query($con,"SELECT * FROM orders");
$id = $_POST['id'];
$r = "";
$sql = $con->query("update orders set tracking_one = '$r' WHERE id ='$id'");
mysqli_close($con);
?>
我在google中找不到任何特定的查詢,什么查詢對我有幫助?
您的查詢應該是這樣的
$sql = $con->query("update orders set tracking_one = (select tracking_two from awbno WHERE id =$id)");
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