內部連接ot返回第二個和第三個表的結果

Question

我對我做錯了什麼感到困惑。我已經在許多資源中看到了這一點，包括我正在學習PHP的書，看起來它應該是正確的......但它不起作用。

<?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'; 

這將返回第一個表（公園）中的所有數據就好了。項目評論名稱，評論，事件，日期和描述2來自連接表（事件和評論）

如果我回應'$ row ['state']'我得到一個正確的答案。但是，如果我回應其他表中的任何項目（例如$ row ['comment']），我不會得到任何結果。

我錯過了什麼？

Answer 1

SQL SH由於如何從連接表

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' 
; 

這不是一個完整的解決方案，但豪的演示，從連接表純粹從主表

Answer 2

不是你寫的查詢返回他們...

<?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，具有相同的名稱ES，這是不行的！）：

$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'; 

Answer 3

請試試這個：

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); 
}