php - Printing a table somewhat complex MYSQL query in PHP

Question

I have a database with 2 tables User, Contact_List.

The User table for example looks like this:

  U_id | U_email | U_password | U_mobileNo | U_name |
  ---------------------------------------------------
  1    | a@b.c   | aaa        | 1234567    | Adam   |

  2    | b@b.c   | bbb        | 1234567    | Ben    |

  3    | c@b.c   | ccc        | 1234567    | Carl   |

The Contact_list table looks like this. This table is table just consisting of foreign keys that relate two users together

  U_id | U_contact_id
  -------------------
  1    |     2

  2    |     3

Now the problem is my SQL/PHP query to display a table that consists of a specific users list of contacts.

This SQL query works fine and gives the results I want:

"SELECT cu.u_name, cu.u_email 
FROM contact_list = c, user = u, user = cu
WHERE c.u_id = 2 
AND c.u_contactId = c.u_id
AND c.u_id = u.u_id"

But this PHP code:

$con = mysqli_connect("dbname","dbuser","pbpass","db");
//Check connection
if (mysqli_connect_errno())
{
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$result = mysqli_query($con,"SELECT cu.u_name, cu.u_email 
  FROM contact_list = c, user = u, user = cu
  WHERE c.u_id = 2 
  AND c.u_contactId = c.u_id
  AND c.u_id = u.u_id" ) or die(mysql_error()); 

echo "<table border='1'>
 <th>Contact List</th>
 <tr>
 <th>Name</th>
 <th>E-mail</th>
 </tr>";

 while($row = mysqli_fetch_array($result))
 {
  echo "<tr>";
  echo "<td>" . $row['u_name'] . "</td>";
  echo "<td>" . $row['u_email'] . "</td>";
  echo "</tr>";


 }
  echo "</table>";  
  mysqli_free_result($result);
  mysqli_close($con);

?>

That code just prints a blank table on the page.

What am I doing wrong here?

I am a complete noob using PHP any help or suggestions would be appreciated.

Answer 1

Try to do a var_dump on $row, you may find out more on what went wrong, if no var_dump appears, no result return was returned, use mysqli_error to find out what went wrong

 mysqli_error($con);

 while($row = mysqli_fetch_array($result))
 {
     var_dump($row);
 }

It could be a case issue

$row['u_name']

may be is

$row['U_name']

Answer 2

SELECT cu.u_name, cu.u_email 
FROM contact_list c, user u, user cu
WHERE c.u_id = 2 
AND c.u_contactId = c.u_id
AND c.u_id = u.u_id

Maybe it's just that I suffer insomnia, but that should work

Answer 3

You want to retrieve all the contacts of user with id 2 so basically your query would look like this:

SELECT u.u_name, u.u_email 
  FROM contact_list cl
  JOIN user u
    ON cl.u_contact_id = u.u_id
 WHERE cl.u_id = 2

Live DEMO.

What the above query does is, it will join the users information based on the u_contact_id and will list all the names and emails registered to it.

And your PHP would look like this:

<?php
// your database info here
$db_host = '';
$db_user = '';
$db_pass = '';
$db_name = '';

$con = mysqli_connect($db_host,$db_user,$db_pass,$db_name);
if($con->connect_error)
    die('Connect Error (' . mysqli_connect_errno() . ') '. mysqli_connect_error());

if (!$stmt = $con->prepare("SELECT u.u_name, u.u_email 
                             FROM contact_list cl
                             JOIN user u
                               ON cl.u_contact_id = u.u_id
                            WHERE cl.u_id = ?"))
    die('Prepare Error: ' . $con->error);

$id = 2;
if (!$stmt->bind_param('i', $id))
    die('Bind Parameters Error ' . $stmt->error);

if (!$stmt->execute())
    die('Select Query Error ' . $stmt->error);
?>
<table border="1">
 <th>Contact List</th>
 <tr>
  <th>Name</th>
  <th>E-mail</th>
 </tr>
<?php
$stmt->bind_result($name,$email);
while ($stmt->fetch())
{
?>
 <tr>
  <td><?php echo $name; ?></td>
  <td><?php echo $email; ?></td>
 </tr>
<?php
}
echo "</table>";
$stmt->close();
$con->close();

Answer 4

In your query:

1. Your syntax for specifying table alias is wrong. There is no need for the = sign.
2. You are aliasing user as both u and cu, but seems like you can get by with just one.

Try the following:

SELECT cu.u_name, cu.u_email 
FROM contact_list c, user u     // Or alternatively: FROM contact_list AS c, user AS u
WHERE c.u_id = 2 
AND c.u_contactId = c.u_id
AND c.u_id = u.u_id

By the way, when printing the <table>, you need to place <th> inside <tr> for valid HTML.

Also, in your $result = mysql_query (....) you ended it with or die (mysql_error()). It should be or die (mysqli_error()).