Cannot retrieve name from Users database (mysql) using PHP

Question

I'm trying to display a "Welcome back, <name>." to my blog when I log in back. I'm using php to access the database, get the name and last name of the username currently in $_SESSION['username'], and then print it back in the index.

So the function to query the database is:

 function get_full_name($username){

    $real = array();

    $query = mysql_query("SELECT `name`, `last` FROM `users` WHERE `user` = `{$username}`");

    $row = mysql_num_rows($query);

    foreach($row as $k => $v)
    {
        $real[$k] = $v;
    }

       return $real;
  }

Then the part of the html where it calls the function above:

<div id="menu">
    <?php 
            $temp = $_SESSION['username'];
            $real[] = get_full_name($temp);
            if(isset($_SESSION['username']))
            {
                echo '<br />'.'Welcome back, '. $real['name'] . '.';
            }
    ?>
    </div>

The output of the above codes is:

 Welcome back, .

var_dump($real) gives:

array(1) { [0]=> array(0) { } }

var_dump($real) after changing to mysql_fectch_assoc:

array(0) { }

Fixed: The error was not using the single quotes ' {$username} '

By changing them, it worked like a charm, cheers to all!

try var_dump($real) to see what indexes it has. I seem to remember mysql_num_rows being the wrong function if you want an associative array back. — Brian Warshaw
– Brian Warshaw, Commented Jun 12, 2012 at 10:55
Have you done var_dump after using mysql_fetch_assoc? Good way to confirm it's working and to see what you're actually getting back from your indexes. — Brian Warshaw
– Brian Warshaw, Commented Jun 12, 2012 at 11:13

John C · Accepted Answer · 2012-06-12 11:21:42Z

2

Curiously you are using mysql_num_rows to get the results of your query. This will only return the number of rows. You'll be better off using mysql_fetch_assoc to get the associative array of results.

You also need to change your backticks to single quotes around {$username} in the query.

Something like:

function get_full_name($username){

    $real = array();

    $query = mysql_query("SELECT `name`, `last` FROM `users` WHERE `user` = '{$username}'");

    $row = mysql_fetch_assoc($query);

    foreach($row as $k => $v)
    {
        $real[$k] = $v;
    }

    return $real;
}

edited Jun 12, 2012 at 11:21

answered Jun 12, 2012 at 10:59

John C

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3 Comments

nullwriter Over a year ago

Somehow this is not working, still showing me no printed name, but i didnt understand why before it wouldnt work at all, this seems logical but still its not working for me =/, must be doing something else wrng?

John C Over a year ago

See also Nick's solution, you need to remove the [] from $real[] = get_full_name($temp); bsdnoobz has an even more comprehensive writeup!

nullwriter Over a year ago

Yes, it worked!! So the error was just the ` {$username} ` quotes instead of '{$username}' ! Thanks a lot!

flowfree · Accepted Answer · 2012-06-12 11:16:17Z

Your get_full_name() function should be like this:

function get_full_name($username) {
   $real  = array();
   $query = mysql_query("SELECT `name`, `last` 
                         FROM `users` 
                         WHERE `user` = '{$username}'");
   if (mysql_num_rows($query)) {
     $real = mysql_fetch_assoc($query);
   }

   return $real;
}

Notes:

Use single quotes for the variable: '{$username}' instead of `{$username}`
Use mysql_num_rows() for checking if the result is not empty.
Use mysql_fetch_assoc() for retrieving the row.

Calling the function:

$temp = $_SESSION['username'];
$real = get_full_name($temp);
if (isset($temp) && count($real)) {
   echo '<br />Welcome back, '. $real['name'] . '.';
}

Notes:

Use $real instead of $real[].
Check if $real empty with count().

Miqdad Ali · Accepted Answer · 2012-06-12 10:59:03Z

0

Change this line $row = mysql_fetch_array($query);

answered Jun 12, 2012 at 10:59

Miqdad Ali

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Comments

Nick · Accepted Answer · 2012-06-12 11:01:52Z

Two issues I see.

1) You are using mysql_num_rows, not fetching the array of data from the database.

function get_full_name($username){

    $real = array();

    $query = mysql_query("SELECT `name`, `last` FROM `users` WHERE `user` = `{$username}`");

    $row = mysql_fetch_assoc($query);

    foreach($row as $k => $v)
    {
        $real[$k] = $v;
    }

    return $real;
}

2) Looks like $real is a multi-dimention array in your code. Try this instead:

<div id="menu">
    <?php 
            $temp = $_SESSION['username'];
            $real = get_full_name($temp);
            if(isset($_SESSION['username']))
            {
                echo '<br />'.'Welcome back, '. $real['name'] . '.';
            }
    ?>
    </div>

Notice I've removed the [] at the end of $real.

Tom · Accepted Answer · 2012-06-12 11:02:36Z

0

Use mysql_query() to get the result of running the query, then use mysql_fetch_assoc() to get individual rows.

$result = mysql_query($query);

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

Note that mysql_num_rows($result) will return the number of rows in a given result.

answered Jun 12, 2012 at 11:02

Tom

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