从一个表中返回与另一个表中的条件匹配的数据

Question

I want to display some data from a MYSQL database in a table format on my webpage. So everyone dosent get confused between the website table and mysql table, i'll call the website table 'web table' from now on.

Most of the data for this web table is being pulled from a MYSQL table called 'image', however for one of the web table's columns I want to display data from a different MYSQL table (within the same database) called 'users'.

Both 'image' and 'users' have a 'user_id' field, so I was looking to find the entry in 'users' where 'user_id' matched the 'user_id' in image.

Here is the code I have so far. I know it is depreciated, but his project requires it.

For your reference, the user_id field from 'image' is echo'd as $data[1]

mysql_connect('XX','XX','XX');
mysql_select_db('XX')

$order = "SELECT * FROM image ORDER BY id";

$result = mysql_query($order) or die ("Bad query: " . mysql_error() );  

while($data = mysql_fetch_row($result)){
  echo("<tr><td>$data[0]</td><td>'SELECT users.full_name FROM users WHERE users.user_id = $data[1]'</td><td>$data[2]</td><td>$data[3]</td></tr>");
}

Cheers in advanced. I know this is probably basic, but I'm not great with PHP.

Answer 1

您可以通过一个INNER JOIN查询获得所需的所有数据

$order = "SELECT a.*, b.full_name FROM image a INNER JOIN user b ON a.user_id = b.user_id ORDER BY a.id";

Answer 2

Another way to do that.

<?php
mysql_connect('XX','XX','XX');
mysql_select_db('XX')

$order = "SELECT * FROM image as i, users as u WHERE  i.user_id =u.user_id ORDER BY i.id";

$result = mysql_query($order) or die ("Bad query: " . mysql_error() );  

while($data = mysql_fetch_array($result)){
    ?>

  <tr><td><?php echo $data[1]?></td><td><?php echo $data[2]?></td><td><?php echo $data[3]?></td><td><?php echo $data[4]?></td></tr>;

<?php
}

?>

Answer 3

Error is here.

    <?php 
 while($data = mysql_fetch_row($result)){
    ?>
  <tr><td><?php echo $data[0]; ?></td><td><?php $sql_res = mysql_fetch_assoc(mysql_query('SELECT users.full_name FROM users WHERE users.user_id ="'.$data[1].'"')); ?></td><td><?php echo $data[2] ?></td><td><?php echo $sql_res['full_name'] ?></td></tr>

  <?php
}

?>