我想让我的SQL表通过PHP填充到HTML表中。但是,我只是创建了表头。填充有什么问题?这是我的代码:
<?php $con=mysqli_connect("server.com","username","password"); // Check connection if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); } $result = mysqli_query($con,"SELECT * FROM Orders"); echo "<table>"; echo "<table border='1'> <tr> <th>ID</th> <th>orderNumber</th> <th>Price</th> <th>customerName</th> <th>salesRep</th> <th>DatePicker</th> <th>shipMethod</th> <th>trackingNumber</th> <th>Statuscheck</th> <th>Edit</th> </tr>"; while($row = mysqli_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['ID'] . "</td>"; echo "<td>" . $row['orderNumber'] . "</td>"; echo "<td>" . $row['Price'] . "</td>"; echo "<td>" . $row['customerName'] . "</td>"; echo "<td>" . $row['salesRep'] . "</td>"; echo "<td>" . $row['DatePicker'] . "</td>"; echo "<td>" . $row['shipMethod'] . "</td>"; echo "<td>" . $row['trackingNumber'] . "</td>"; echo "<td>" . $row['Statuscheck'] . "</td>"; echo "<td>" . $row['Edit'] . "</td>"; echo "</tr>"; } echo "</table>"; ?> </body> </html>
“编辑”是这样,因此我可以拥有一个编辑超链接以使用SQL UPDATE。它实际上不是数据库中SQL表的一部分。任何帮助是极大的赞赏。谢谢!
好的,这就是问题所在,编辑页面不会预填充已经填写的字段,更新按钮也不会实际更新数据库。这是编辑页面的代码。
$query = "SELECT orderNumber, customerName, salesRep, DatePicker, shipMethod, trackingNumber, StatusCheck FROM Orders WHERE id = '$id'"; $result = @mysql_query($query); mysql_fetch_object($result); ?> <form name="update order" method="post" action="edit.php?a=edit&id=<? echo($ID) ?>&update=1"> <table width="50%" border="0" cellspacing="0" cellpadding="0"> <tr> <td width="50%">Order Number</td> <td><input name="orderNumber" type="text" id="orderNumber" value="<? echo($row->orderNumber) ?>"></td> </tr> <tr> <td>Customer Name</td> <td><input name="customerName" type="text" id="customerName" value="<? echo($row->customerName) ?>"></td> </tr> <tr> <td> </td> <td> </td> </tr> <tr> <td>Sales Rep</td> <td><input name="headline" type="text" id="headline" value="<? echo($row->headline) ?>"></td> <tr> <td>Must Ship By</td> <td><input name="DatePicker" type="text" id="DatePicker" value="<? echo($row->DatePicker) ?>"></td> <tr> <td>Shipping Method</td> <td><input name="shipMethod" type="text" id="shipMethod" value="<? echo($row->shipMethod) ?>"></td> <tr> <td>Tracking Number</td> <td><input name="trackingNumber" type="text" id="trackingNumber" value="<? echo($row->trackingNumber) ?>"></td> </tr> <tr> <td>Status</td> <td><input name="StatusCheck" type="radio" name="status" value="PROCESSING"> PROCESSING<br><input name="StatusCheck" type="radio" name="status" value="PROCESSING"> PICKED<br><input name="StatusCheck" type="radio" name="status" value="PROCESSING" value="<? echo($row->StatusCheck) ?>"> SHIPPED<br> value="<? echo($row->StatusCheck) ?>"></td> </tr> <tr> <td colspan="2"><div align="center"> <input name="hiddenField" type="hidden" value="update"> <input name="add" type="submit" id="add" value="Update"> </div></td> </tr> </table> </form> <?php ?>
您的$result变量为空,对其进行测试。原因是因为您没有设置任何数据库名称,您可以从中获取信息。粘贴此:
$result
mysqli_select_db($con, 'your_database_name');
后
$con=mysqli_connect("server.com","username","password");
并且应该没问题(当然,如果那里有任何记录)。