如何在同一頁面中顯示輸入和輸出？
[英]how to display input and output in same page?
問題描述
我只有一個PHP文件。 但使用兩個單獨的PHP標簽。 如果單擊第一個PHP標記表中的第一行，則需要在同一頁面的第二個PHP標記表中顯示第一行的值。 兩個表的公共字段是order_id輸出應顯示在第一個表的旁邊。
<head>
<script type="text/javascript" src="http://code.jquery.com/jquery-1.8.2.js"></script>
<script type="text/javascript">
$(function() {
$('.one').click(function() {
$(".slidediv").slideToggle();
});
});
</script>
</head>
<body>
<?php
session_start();
include "db.php";
$query = "select * from orders_list";
$result = mysql_query($query);
$num_rows = mysql_num_rows($result);
if($num_rows >= 1)
{
echo "<div id='showmenu' class='scroll'>";
echo "<table id='table_struct' cellspacing='0' cellpadding='1' border='1' width='400'>
<tr class='tr_class' bgcolor='#0066CC'>
<td align='center' style='font-weight:bold'> Select </td>
<td align='center' style='font-weight:bold'> order_id </td>
<td align='center' style='font-weight:bold'> customer_name </td>
<td align='center' style='font-weight:bold'> no_of_pack </td>
<td align='center' style='font-weight:bold'> price </td>
<td align='center' style='font-weight:bold'> Weight </td>
<td align='center' style='font-weight:bold'> payment mode </td>
</tr>";
while($row = mysql_fetch_array($result))
{
echo "<tr class='one' data-id='".$row['order_id']."'>
<td align='center'><input type='checkbox' class='case' name='case' value='1'></td>
<td align='center'>".$row['order_id']."</td>
<td align='center'>".$row['customer_name']."</td>
<td align='center'>".$row['number_of_pack']."</td>
<td align='center'>".$row['price']."</td>
<td align='center'>".$row['weight']."</td>
<td align='center'>".$row['payment']."</td>";
/*echo "<td align='center'><a href='pin.php?increment_id=".$row['increment_id']."&postcode=".$row['postcode']."&proposed_shipping=".$proposed_shipping." '>SUBMIT</a></td>";*/
echo "</tr>";
}
$order_id = $row['order_id'];
global $order_id;
echo "</table>";
echo "</div>";
}
if(!mysql_close($con))
{
echo "failed to close";
}
?>
<div>
<?php
include "db.php";
global $order_id;
$query = "select * from orders_details where order_id=$order_id";
echo $query;
$result = mysql_query($query);
$num_rows = mysql_num_rows($result);
if($num_rows >= 1)
{
echo "<div class='menu' class='scroll'>";
echo "<table id='table_struct' cellspacing='0' cellpadding='1' border='1' width='400'>
<tr class='slidediv' bgcolor='#0066CC'>
<td align='center' style='font-weight:bold'> Product </td>
<td align='center' style='font-weight:bold'> Quantity </td>
<td align='center' style='font-weight:bold'> Sku </td>
<td align='center' style='font-weight:bold'> Price </td>
<td align='center' style='font-weight:bold'> Total </td>
</tr>";
while($row = mysql_fetch_array($result))
{
echo "<tr class='slidediv'>
<td align='center'><input type='checkbox' class='case' name='case' value='1'></td>
<td align='center'>".$row['product']."</td>
<td align='center'>".$row['quantity']."</td>
<td align='center'>".$row['sku']."</td>
<td align='center'>".$row['price']."</td>
<td align='center'>".$row['total']."</td>";
echo "</tr>";
}
echo "</table>";
echo "</div>";
}
if(!mysql_close($con))
{
echo "failed to close";
}
?>
</body>
1 個解決方案
解決方案1
0 2015-03-11 10:24:01
無需全局聲明$ order_id，只需將您的代碼$ order_id = $ row ['order_id']; 在while循環內，並在您的頁面中的任何地方使用它並享受。
如何在同一頁面中顯示html輸出？
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