内部连接ot返回第二个和第三个表的结果

-1

我对我做错了什么感到困惑。我已经在许多资源中看到了这一点，包括我正在学习PHP的书，看起来它应该是正确的......但它不起作用。内部连接ot返回第二个和第三个表的结果

<?php 

try 
{ 
    $sql = 'SELECT parks.id, parks.state, parks.name, parks.description, parks.site, parks.sname, parks.street, parks.city, parks.zip, parks.phone FROM parks 
INNER JOIN comments ON parks.parkid = comments.parkid 
INNER JOIN photos ON parks.parkid = photos.parkid 
INNER JOIN events ON parks.parkid = events.parkid'; 
$result = $pdo->query($sql); 
} 
catch (PDOException $e) 
{ 
    $error = 'Error fetching data: ' . $e->getMessage(); 
    include 'output.html.php'; 
    exit(); 
} 

foreach ($result as $row) 
{ 
    $datas[] = array ('id' =>$row['id'], 
    'parkid' =>$row['parkid'], 
    'state' =>$row['state'], 
    'name' =>$row['name'], 
    'description' =>$row['description'], 
    'site' =>$row['site'], 
    'sname' =>$row['sname'], 
    'street' =>$row['street'], 
    'city' =>$row['city'], 
    'phone' =>$row['phone'], 
    'zip' =>$row['zip'], 
    'commentname' =>$row['commentname'], 
    'comment' =>$row['comment'], 
    'event' =>$row['event'], 
    'date' =>$row['date'], 
    'description2' =>$row['description2']); 
} 

include 'writing.html.php';

这将返回第一个表（公园）中的所有数据就好了。项目评论名称，评论，事件，日期和描述2来自连接表（事件和评论）

如果我回应'$ row ['state']'我得到一个正确的答案。但是，如果我回应其他表中的任何项目（例如$ row ['comment']），我不会得到任何结果。

我错过了什么？

来源

2012-10-20 user1483042

你不上市返回数据，而不是返回列来自任何连接表的列返回 –

马克，我不明白你的答案。我应该在哪里列出它们，而不是在数组中？他们在阵列中。 – user1483042

该数组是从$ row中由SQL返回的$ result生成的...但如果您的$ sql没有列出列，那么它们将不在$ row中...查看下面的列表示例列，而不是一个完整的解决方案 –

SQL SH由于如何从连接表

SELECT parks.id, 
     parks.state, 
     parks.name AS park_name, -- use an alias when column names exist in more than one table 
     parks.description, 
     parks.site, 
     parks.sname, 
     parks.street, 
     parks.city, 
     parks.zip, 
     parks.phone, 
     comments.comment, -- to return the comment from the comments table 
     events.name AS event_name, -- return name from the event table 
     event_date -- return date from the event table 
    FROM parks 
    INNER JOIN comments 
      ON parks.parkid = comments.parkid 
    INNER JOIN photos 
      ON parks.parkid = photos.parkid 
    INNER JOIN events 
      ON parks.parkid = events.parkid' 
;

这不是一个完整的解决方案，但豪的演示，从连接表纯粹从主表

来源

2012-10-20 22:51:29

WOW这正是我正在寻找的。有效。非常感谢。我已经尝试过好几天了，你是一个救生员。 – user1483042

不是你写的查询返回他们...

<?php 

try 
{ 
    $sql = 'SELECT parks.id, parks.state, parks.name, parks.description, parks.site, parks.sname, parks.street, parks.city, parks.zip, parks.phone 
    --insert here any other column names you want to have in PHP 
    FROM parks 
    INNER JOIN comments ON parks.parkid = comments.parkid 
    INNER JOIN photos ON parks.parkid = photos.parkid 
    INNER JOIN events ON parks.parkid = events.parkid'; 
    $result = $pdo->query($sql); 
}

或者你可以得到所有（提防，如果有在TABL列，具有相同的名称ES，这是不行的！）：

$sql = 'SELECT * 
    FROM parks 
    INNER JOIN comments ON parks.parkid = comments.parkid 
    INNER JOIN photos ON parks.parkid = photos.parkid 
    INNER JOIN events ON parks.parkid = events.parkid';

来源

2012-10-20 21:52:21 ppeterka

我想获取所有的数据。我正在排序并在include中显示它。每个表都有一个名为id的列。这可能是问题吗？ – user1483042

请试试这个：

try 
{ 
    $sql = 'SELECT parks.id, parks.state, parks.name, parks.description, parks.site, parks.sname, parks.street, parks.city, parks.zip, parks.phone FROM parks 
INNER JOIN comments INNER JOIN photos INNER JOIN events ON parks.parkid = comments.parkid and parks.parkid = photos.parkid and parks.parkid = events.parkid'; 
$result = $pdo->query($sql); 
}

来源

2012-10-20 21:55:11

这给我一个语法错误，意外） – user1483042

请尝试我的更新 –

我做到了。我收到一个语法错误：“意外）” – user1483042

内部连接ot返回第二个和第三个表的结果

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