三表条件查询-php-mysql

三表条件查询-php-mysql,php,mysql,html-table,Php,Mysql,Html Table,我有三张桌子 使用者 学校 ________________________________________________________________ |School_ID | School name | school address | school contact info| ---------------------------------------------------------------- 图画 _____________________________________

我有三张桌子

使用者

学校

________________________________________________________________
|School_ID | School name | school address | school contact info|
----------------------------------------------------------------
图画

___________________________________________
| Picture ID | User_ID | School_ID|file path|
-------------------------------------------
根据学校ID,我想显示如下信息

学校名称、学校地址、用户名、用户联系信息以及所有具有匹配学校ID的图片

php中的脚本是什么。 我在这个网站上研究了几天,尝试了很多代码,但到目前为止都没有效果

这是我使用的代码

<?php
         $sql="SELECT Schoolname, School_contact_info, email, description, address, website, file, userName FROM school, picture, user WHERE Location_ID=Picture_ID  ORDER BY Location_ID desc";
         $result_set=mysql_query($sql);
         while($row=mysql_fetch_array($result_set))
         {
          ?>
                <div href=<?php $row['Location_ID'] ?>>
                <td><?php echo $row['Location_ID'] ?></td>  
                <td><?php echo $row['name2'] ?></td>
                <td><?php echo $row['email'] ?></td>
                <div><td><?php echo $row['description'] ?></td></div>
                </div>

                <div><img src="uploads/<?php echo $row['file'];?>" alt=" " height="75" width="75"></div>

                <br>
                <?php
         }
     ?>

“alt=”“height=“75”width=“75”>

我想显示学校信息,以及所有以学校ID作为外键的图片。

查询应使用

select school_name,school_address,user_name,user_contact_information
from user,school,picture where user.user_id=picture_id 
and school.school_id=picture.picture_id

    <?php
    $con=mysqli_connect("localhost","my_user","my_password","my_db");
    // Check connection
    if (mysqli_connect_errno())
      {
      echo "Failed to connect to MySQL: " . mysqli_connect_error();
      }

    // Perform queries 
    mysqli_query($con,"select school_name,school_address,user_name,user_contact_information
    from user,school,picture where user.user_id=picture_id 
    and school.school_id=picture.picture_id");
选择学校名称、学校地址、用户名、用户联系信息
来自用户、学校、用户所在的图片。用户\u id=图片\u id
还有学校。学校id=picture.picture\u id

在表结构中,没有提供查询中使用的所有列详细信息。
但是试试这个:

SELECT s.Schoolname, s.School_contact_info, s.email, s.description, s.address, s.website,
 p.file, u.userName 
FROM school as s,
 picture as p on s.School_ID=p.School_ID, 
 user as u on p.User_ID=u.User_ID
WHERE s.Location_ID=p.Picture_ID  ORDER BY s.Location_ID desc

我假设像
Location\u ID,email,website
这样的栏目都是来自学校的表格。

并且尝试了很多代码
…你尝试了什么,你能把它包括进去吗?我想,要求一个完整的PHP脚本和一个查询对于一个问题来说太宽泛了。我同意@TimBiegeleisen,这个问题不会受到欢迎没有演示自己的努力。@TimBiegeleisen我已经用我使用的代码编辑了这个问题。谢谢。
SELECT s.Schoolname, s.School_contact_info, s.email, s.description, s.address, s.website,
 p.file, u.userName 
FROM school as s,
 picture as p on s.School_ID=p.School_ID, 
 user as u on p.User_ID=u.User_ID
WHERE s.Location_ID=p.Picture_ID  ORDER BY s.Location_ID desc