Question

我有2个与此问题相关的表，它们被调用 satellite和satellite_multi。我创建表的思维过程是卫星表包含channel_id（主键），频道名称和频道广播的国家，

satellite_multi 表有一个名为channelid_multi的列，即   作为foreign key链接到卫星表的channel_id，   和其他列是
SatName，频率，极化，符号，   FEC，EncorFTA。

我创建了satellite_multi表，因为有多个频道位于多个卫星上，例如： Thor 0.8w和Hotbird 13.0e可能具有相同的频道广播，因此如果频道在多个卫星上广播，我需要一种能够显示多行数据的方法。

下面是satellite_multi表的表结构：

+-----------------+---------------+-----------+--------------+------------+-----+----------+
| ChannelID_Multi(FK) |    SatName    | Frequency | Polarisation |   Symbrate | FEC | EncorFta |
+-----------------+---------------+-----------+--------------+------------+-----+----------+
|               1 | Thor 0.8w     |     10932 |   H          |     275000 | 5/6 |   ENC    |
|               1 | Hotbird 13.0e |     10654 |   V          |      25000 | 3/4 |   FTA    |
+-----------------+---------------+-----------+--------------+------------+-----+----------+

This is the table structure for the table named satellite:
+-----------+----------------+----------+
| ChannelID (PK) |       Name     | Country  |
+-----------+----------------+----------+
|         1 |   Polsat Sport | Poland   |
|         2 |   Sky Sports   |  England |
+-----------+----------------+----------+

我有网站设置，用户点击主网站上频道名称的超链接，然后将它们带到名为view_channels.php的页面，其中显示的频道详细信息基于来自卫星表。例如 view_channel.php？channelid = 19

当频道在一个卫星上时，这可以正常工作，因为我能够运行SELECT *查询并显示所有数据。

我尝试在每个单独的频道ID下显示多个频道数据，但遗憾的是它不起作用。

我在view_channels.php页面

中使用了以下代码

$sql = "SELECT s.name, s.country, f.satname, f.frequency, f.polarisation, f.symbrate, f.fec, f.encorfta
FROM satellitemulti f
LEFT JOIN satellite s
ON f.channelid_multi=s.channelid
GROUP BY s.name, s.country, f.satname, f.frequency, f.polarisation, f.symbrate, f.fec, f.encorfta";
$stmt = $DB->prepare($sql);
$stmt->execute();

输出是来自satellite_multi表和卫星表的所有信息都显示在每个频道ID中，在本例中，因为Polsat是ID 1，只应显示polsat，但包含不同ID的AFN Sports是也显示出来。（见下图）

All Data showing

我的问题是，我是否需要添加到查询中以检查浏览器链接中的ID，并将其与从表中收到的ID相匹配，因此只会显示特定ID的通道数据？

我尝试添加一个WHERE子句来显示基于channelid_multi的数据

WHERE channelid_multi = $channelid_multi

但是我收到了一个错误：

Fatal error: Uncaught exception 'PDOException' with message 'SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'JOIN satellite s ON f.channelid_multi=s.channelid GROUP BY s.name, s.country, f.' at line 4' in E:\home\students\2132\B00614408\public_html\sportsschedule\view_channels.php:19 Stack trace: #0 E:\home\students\2132\B00614408\public_html\sportsschedule\view_channels.php(19): PDOStatement->execute() #1 {main} thrown in E:\home\students\2132\B00614408\public_html\sportsschedule\view_channels.php on line 19

感谢任何人提供的任何指导

我已经包含了我的整个`

view_channels.php

`以下代码，以防任何人需要查看

<?php

require_once './config.php';
include './header.php';

$sql = "SELECT s.name, s.country, f.satname, f.frequency, f.polarisation, f.symbrate, f.fec, f.encorfta
FROM satellitemulti f
WHERE channelid_multi = $channelid_multi
LEFT JOIN satellite s
ON f.channelid_multi=s.channelid
GROUP BY s.name, s.country, f.satname, f.frequency, f.polarisation, f.symbrate, f.fec, f.encorfta";
$stmt = $DB->prepare($sql);
$stmt->execute();


?>




<div class="panel panel-primary">
<div class="panel-heading">
<h3 class="panel-title"> Whats On</h3>
</div>
<div class="panel-body">
    </div>


<div class="clearfix"></div>
<div class="table-responsive">
<table class="table table-striped table-hover table-bordered ">
<tbody>
<caption> Channel Details</caption>
<tr>
<th>Name</th>
<th>Country</th>
<th>Sat Name</th>
<th>Frequency</th>
<th>Polarisation</th>
<th>Symbol Rate</th>
<th>FEC</th>
<th>Enc or FTA</th>

</tr>
<?php   while ($row = $stmt->fetch(PDO::FETCH_ASSOC))
{

$name = $row['name'];
$country= $row['country'];
$satname = $row['satname'];
$frequency=$row['frequency'];
$polarisation=$row['polarisation'];
$symbrate=$row['symbrate'];
$fec=$row['fec'];
$encorfta=$row['encorfta'];
$channelid_multi=$row['channelid_multi'];



echo "<tr>";
echo "<td>" . $row['name'] . "</td>";
echo "<td>" . $row['country'] . "</td>";
echo "<td>" . $row['satname'] . "</td>";
echo "<td>" . $row['frequency'] . "</td>";
echo "<td>" . $row['polarisation'] . "</td>";
echo "<td>" . $row['symbrate'] . "</td>";
echo "<td>" . $row['fec'] . "</td>";
echo "<td>" . $row['encorfta'] . "</td>";

}
echo "</tr>";
echo "</table>";
?>
</div>


<?php
include './footer.php';
?>

Answer 1

像其他人已经在评论中告诉你的问题查询是您在WHERE语句之前使用JOIN条件是错误的mysql语法。因此，您必须更改此类查询才能工作：

$sql = "SELECT s.name, s.country, f.satname, f.frequency, f.polarisation, f.symbrate, f.fec, f.encorfta
FROM satellitemulti f
LEFT JOIN satellite s
ON f.channelid_multi=s.channelid
WHERE channelid_multi = $channelid_multi
GROUP BY s.name, s.country, f.satname, f.frequency, f.polarisation, f.symbrate, f.fec, f.encorfta";

但是

您的数据库设计不是很好。

例如，当卫星名称发生变化时会发生什么？您必须更新每一行 satellite_multi有这颗卫星。

由于你有多对多的关系，我会使用3个表。

一个名为satellites的卫星。
一个名为channels
一个名为channels2satellites的多对多表。

注意：我假设频率，极化等是卫星的属性。如果它们是频道的属性，只需将它们移到channels表。

satellites表

+-----------------+---------------+-----------+--------------+------------+-----+----------+
|          ID(PK) |    SatName    | Frequency | Polarisation |   Symbrate | FEC | EncorFta |
+-----------------+---------------+-----------+--------------+------------+-----+----------+
|               1 | Thor 0.8w     |     10932 |   H          |     275000 | 5/6 |   ENC    |
|               2 | Hotbird 13.0e |     10654 |   V          |      25000 | 3/4 |   FTA    |
+-----------------+---------------+-----------+--------------+------------+-----+----------+

channels表

+-----------+----------------+----------------+
|        ID (PK) |       Name     |  Country  |
+-----------+----------------+----------+
|         1      |   Polsat Sport |  Poland   |
|         2      |   Sky Sports   |  England  |
+-----------+----------------+----------------+

channels2satellites表

+-----------+----------------+----------------------------+
|        ID (PK) |   channel_id(FK)   | satellite_id(FK)  |
+----------------+--------------------+-------------------+
|         1      |   1                |  1                |
|         2      |   1                |  2                |
|         3      |   2                |  1                |
+-----------+----------------+----------------------------+

当我需要频道的数据时，我会使用此查询。假设你想要频道1的信息

SELECT c.Name,c.Country,s.SatName,s.Frequency,s.Polarization.s.Symbrate,s.FEC,s.EncorfFta FROM channels c INNER JOIN channels2satellites c2s ON c.id=c2s.channel_id INNER JOIN satellites s ON c2s.satellite_id=s.id WHERE c.id=1

`

Answer 2

让它排序，

这对我有用，感谢指导@Anant和@nowhere

UIImageView

mysql / php - 根据id

2 个答案: