Question

我对自己做错了什么感到困惑。我在很多资源中都看过这个，包括我正在学习PHP的那本书，看起来它应该是正确的......但它只是不起作用。

<?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';

这将返回第一个表（公园）中的所有数据。条目注释名称，注释，事件，日期和描述2来自连接表（事件和注释）

如果我回应'$ row ['state']'我会得到一个正确答案。但是，如果我回显其他表中的任何项目（例如$ row ['comment']），我就没有结果。

我错过了什么？

Answer 1

您没有编写查询以返回它们......

<?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tabl中有列; es具有相同的名称，这将无效！）：

    $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 2

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);
}

Answer 3

SQL显示如何从连接表中返回列

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，而是演示ho从连接表中返回数据而不是纯粹从主表中返回

内连接不返回第二和第三个表的结果

3 个答案: