Question

我迷路了。我想要做的是，我希望我的网站允许用户在我的数据库中选择某一行（记录），然后将它们重定向到另一个网页，向他们显示有关该特定记录的完整信息..我不知道知道如何使用动态表格/文本将这两个网页连接在一起.. 以下是我的代码的一部分:(这是第一个网页：）

mysql_select_db($database_rfq_portal, $rfq_portal);
$query_rfqrecord = "SELECT tblrfq.`RFQ_ID`, tblrfq.`Company_Name`, tblrfq.Service,
tblrfq.`Kind_of_Request`, tblrfq.Status, tblrfq.`Date` FROM tblrfq";
$rfqrecord = mysql_query($query_rfqrecord, $rfq_portal) or die(mysql_error());
$row_rfqrecord = mysql_fetch_assoc($rfqrecord);
$totalRows_rfqrecord = mysql_num_rows($rfqrecord);


<form id="viewform" name="viewform" method="get" action="ViewSpecificRFQ.php">
<table width="716" border="1" align="center" cellpadding="5">
<tr>
<td>RFQ ID</td>
<td>Company Name</td>
<td>Service</td>
<td>Kind of Request</td>
<td>Status</td>
<td>Date</td>
</tr>
<?php do { ?>
  <tr>
    <td><a href="ViewSpecificRFQ.phpRFQID=<?php echo $row_rfqrecord['RFQ_ID'];?>"><?php echo $row_rfqrecord['RFQ_ID']; ?></a></td>
    <td><a href="ViewSpecificRFQ.phpRFQID=<?php echo $row_rfqrecord['RFQ_ID'];?>"><?php echo $row_rfqrecord['Company_Name']; ?></a></td>
    <td><a href="ViewSpecificRFQ.phpRFQID=<?php echo $row_rfqrecord['RFQ_ID'];?>"><?php echo $row_rfqrecord['Service']; ?></a></td>
    <td><a href="ViewSpecificRFQ.phpRFQID=<?php echo $row_rfqrecord['RFQ_ID'];?>"><?php echo $row_rfqrecord['Kind_of_Request']; ?></a></td>
    <td><a href="ViewSpecificRFQ.phpRFQID=<?php echo $row_rfqrecord['RFQ_ID'];?>"><?php echo $row_rfqrecord['Status']; ?></a></td>
    <td><a href="ViewSpecificRFQ.phpRFQID=<?php echo $row_rfqrecord['RFQ_ID'];?>"><?php echo $row_rfqrecord['Date']; ?></a></td>
  </tr>
<?php } while ($row_rfqrecord = mysql_fetch_assoc($rfqrecord)); ?>
</table>
}
</form>

这是将获得表单的网页..（我的代码的一部分）

$RFQID = $_GET['RFQ_ID'];
mysql_select_db($database_rfq_portal, $rfq_portal);
$query_rfqrecord = "SELECT * FROM tblrfq WHERE $RFQID";

$rfqrecord = mysql_query($query_rfqrecord, $rfq_portal) or die(mysql_error());
$row_rfqrecord = mysql_fetch_assoc($rfqrecord);
$totalRows_rfqrecord = mysql_num_rows($rfqrecord);

mysql_select_db($database_rfq_portal, $rfq_portal);
$query_user = "SELECT tbluser.Username, tbluser.Password FROM tbluser";
$user = mysql_query($query_user, $rfq_portal) or die(mysql_error());
$row_user = mysql_fetch_assoc($user);
$totalRows_user = mysql_num_rows($user);



<table width="716" border="0" align="center">
<tr>
  <th colspan="2" scope="row">RFQ ID:</th>
  <td><?php echo $row_rfqrecord['RFQ_ID']; ?></td>
</tr>
<tr>
  <th colspan="2" scope="row">Company Name:</th>
  <td width="511"><?php echo $row_rfqrecord['Company_Name']; ?></td>
</tr>
<tr>
  <th width="101" rowspan="2" scope="row">Address:</th>
  <th width="90" scope="row">Site A:</th>
  <td><?php echo $row_rfqrecord['Address_A']; ?></td>
</tr>
<tr>
  <th scope="row">Site B:</th>
  <td><?php echo $row_rfqrecord['Address_B']; ?></td>
</tr>
<tr>
  <th colspan="2" scope="row">Contact Number:</th>
  <td><?php echo $row_rfqrecord['Contact_Number']; ?></td>
</tr>
<tr>
  <th colspan="2" scope="row">Contact Person:</th>
  <td><?php echo $row_rfqrecord['Contact_Person']; ?></td>
</tr>
<tr>
  <th colspan="2" scope="row">Service:</th>
  <td><?php echo $row_rfqrecord['Service']; ?></td>
</tr>
<tr>
  <th colspan="2" scope="row">Bandwidth:</th>
  <td><?php echo $row_rfqrecord['Bandwidth']; ?></td>
</tr>
<tr>
  <th colspan="2" scope="row">Telco:</th>
  <td><?php echo $row_rfqrecord['Telco']; ?></td>
</tr>
<tr>
  <th colspan="2" scope="row">Account Manager:</th>
  <td><?php echo $row_rfqrecord['Account_Manager']; ?></td>
</tr>
<tr>
  <th colspan="2" scope="row">Status:</th>
  <td><?php echo $row_rfqrecord['Status']; ?></td>
</tr>
<tr>
  <th colspan="2" scope="row">Kind of Request:</th>
  <td><?php echo $row_rfqrecord['Kind_of_Request']; ?></td>
</tr>
<tr>
  <th colspan="2" scope="row">Date:</th>
  <td><?php echo $row_rfqrecord['Date']; ?></td>
</tr>
<tr>
  <th colspan="2" scope="row">Remarks:</th>
  <td><?php echo $row_rfqrecord['Remarks']; ?></td>
</tr>

</table>
</form>

这将用户重定向到下一页但我的问题是它继续显示相同的记录，这是我数据库中的第一条记录。

Answer 1

1-考虑page1.php是用户选择记录的第一页，然后在page2.php中显示完整的详细信息。

for page1.php

您需要发送一些参数，例如记录ID或任何其他密钥，以便能够识别记录并选择数据库中的详细信息。

例如：记录1一旦用户点击此链接，你必须在page2.php中做这样的事情，记住我们的参数是record_id

<?php
$id = $_GET['record_id'];
//we have to check it for validity
//if id is an integer (numbers) then simply we can check it
//if(!is_numeric($id)) { die("invalid id") }

// now the id is there
//lets build query
$query = "SELECT * from tablename where id= '$id' ";
?>

对于你的代码，它必须是这样的： //添加where子句 $ query_rfqrecord =“SELECT * FROM tblrfq where RFQ_ID ='$ id'”;

$rfqrecord = mysql_query($query_rfqrecord, $rfq_portal) or die(mysql_error());
$row_rfqrecord = mysql_fetch_assoc($rfqrecord);
$totalRows_rfqrecord = mysql_num_rows($rfqrecord);

Answer 2

如果您希望用户在点击特定行时加载特定记录，则必须确保将其重定向到特定网址。在您的示例中，您的工作时间为<a href="ViewSpecificRFQ.php">，因此表格中的每一行都会有相同的链接，然后是相同的目标网页。

您可能需要<a href="ViewSpecificRFQ.php?recordId=<?php echo $row_rfqrecord['RFQ_ID'] ?>">之类的内容，然后在您的目标网页中，您将根据$_GET['recordId']中的密钥执行查询。

编辑：执行问题

你使用do-while，所以在第一次迭代中你没有加载记录，因为fetch指令位于代码块的底部，所以你应该把它改为：

<?php
while ($row_rfqrecord = mysql_fetch_assoc($rfqrecord)) {
    ?>
    <tr>
        <td><a href="ViewSpecificRFQ.php?recordId=<?php echo $row_rfqrecord['RFQ_ID'] ?>"><?php echo $row_rfqrecord['RFQ_ID']; ?></a></td>
        <td><a href="ViewSpecificRFQ.php?recordId=<?php echo $row_rfqrecord['RFQ_ID'] ?>"><?php echo $row_rfqrecord['Company_Name']; ?></a></td>
        <td><a href="ViewSpecificRFQ.php?recordId=<?php echo $row_rfqrecord['RFQ_ID'] ?>"><?php echo $row_rfqrecord['Service']; ?></a></td>
        <td><a href="ViewSpecificRFQ.php?recordId=<?php echo $row_rfqrecord['RFQ_ID'] ?>"><?php echo $row_rfqrecord['Kind_of_Request']; ?></a></td>
        <td><a href="ViewSpecificRFQ.php?recordId=<?php echo $row_rfqrecord['RFQ_ID'] ?>"><?php echo $row_rfqrecord['Status']; ?></a></td>
        <td><a href="ViewSpecificRFQ.php?recordId=<?php echo $row_rfqrecord['RFQ_ID'] ?>"><?php echo $row_rfqrecord['Date']; ?></a></td>
    </tr>
    <?php
}
?>

使用动态文本链接提交表单

2 个答案: