if statement not working

[harkly]

I am trying to check a table and determine if it is empty, for some reason I cannot get it to work. I have really dumbed down the code for myself but still can't figure out why it isn't working.

Right now my table is all set to NULL, I have made changes to 0 and to 1 but to no avail.

I have also made so many changes to

if ($results == NULL)

using '' and changing to 0 or 1 and it always prints out one result regardless. Driving me crzy!!

<?php
   include('library/login.php');
   login();
   mysql_select_db('test); 

$sql = mysql_query("SELECT * FROM music WHERE userId = test'");
$results = mysql_fetch_array( $sql );

  if ($results == NULL)
    { 
      echo "Empty  \n";  

    }
  else
    {
      echo "Not Empty \n";
    }

?>

[Paavero]

You're missing a single quote after the word test on the fourth line, at least.

Oh.. and another one before the same word in the query string.

[bradgrafelman]

If the only thing you're doing is trying to see if a table contains any rows that match the WHERE condition, you could do something like this:

$sql = mysql_query("SELECT 1 FROM music WHERE userId = 'test' LIMIT 1"); 
$empty = !((bool)mysql_num_rows($sql));

Thus, $empty will be boolean TRUE if no rows were matched.

If you actually intend to do something with the data in the table as well, you could stick with SELECT'ing the appropriate columns (you should avoid using 'SELECT *') but still use [man]mysql_num_rows/man to determine whether the result set is empty or not.

[harkly]

the missing quote's are not the issue, accidently removed when I took out the actual ones.

My code is much more detailed than what I have but I really dumbed it down to see where my prblem lies.

It is going to check the table and based on the result it will either run one set of code or another, but right now I can't get it to be correct in returning the results.

[bradgrafelman]

Do you actually use the data returned by the query above? If not, it might be simpler to use my example above.

If you do, then it might be easier if you can show us more of your (actual) code.

[harkly]

Here is the code, it works and I am adding the if($results = Null) part. Not sure why this is needed if I can't get the basic if statement to work then ...

$sql = mysql_query("SELECT * FROM music WHERE userId = 'test'");
$results = mysql_fetch_array( $sql );


 if ($results == null)
    { 
      echo " Blank \n";  

    }

  else
    {
      echo "  <br><br>Music I listen to:
              <span class='text'>
      \n";

  $sql = mysql_query("SELECT * FROM music WHERE userId = 'test'");

  while($r = mysql_fetch_array($sql)) {

    $string = "";

      if ($r['band']) $string .= "Band, ";
      if ($r['barbershop']) $string .= "Barbershop, ";
      if ($r['classical']) $string .= "Classical, ";
      if ($r['country']) $string .= "Country, ";
      if ($r['dance']) $string .= "Dance, ";
      if ($r['electric']) $string .= "Electroinical, ";
      if ($r['folk']) $string .= "Folk, ";
      if ($r['jazz']) $string .= "Jazz, ";
      if ($r['metal']) $string .= "Metal, ";
      if ($r['opera']) $string .= "Opera/Musicals, ";
      if ($r['orchestra']) $string .= "Orchestra, ";
      if ($r['punk']) $string .= "Punk Rock, ";
      if ($r['ragtime']) $string .= "Ragtime, ";
      if ($r['rap']) $string .= "Rap/Hip Hop, ";
      if ($r['reggae']) $string .= "Reggae/Ska, ";
      if ($r['religious']) $string .= "Religious, ";
      if ($r['rock_pop']) $string .= "Rock/Pop/Alternative, ";
      if ($r['world']) $string .= "World, ";
      if ($r['othr_music']) $string .= $r['othr_music_txt'];

      $string = rtrim($string, ", ");

      echo " $string \n";
    }
}

[bradgrafelman]

Wait... you're running the same query twice? That doesn't seem very efficient... why not just do something like:

$sql = mysql_query("SELECT * FROM music WHERE userId = 'test'");

if(mysql_num_rows($sql) == 0)
  echo " Blank \n";
else
  while($r = mysql_fetch_array($sql)) {

$string = "";

if ($r['band']) $string .= "Band, ";
if ($r['barbershop']) $string .= "Barbershop, ";
if ($r['classical']) $string .= "Classical, ";
if ($r['country']) $string .= "Country, ";
if ($r['dance']) $string .= "Dance, ";
if ($r['electric']) $string .= "Electroinical, ";
if ($r['folk']) $string .= "Folk, ";
if ($r['jazz']) $string .= "Jazz, ";
if ($r['metal']) $string .= "Metal, ";
if ($r['opera']) $string .= "Opera/Musicals, ";
if ($r['orchestra']) $string .= "Orchestra, ";
if ($r['punk']) $string .= "Punk Rock, ";
if ($r['ragtime']) $string .= "Ragtime, ";
if ($r['rap']) $string .= "Rap/Hip Hop, ";
if ($r['reggae']) $string .= "Reggae/Ska, ";
if ($r['religious']) $string .= "Religious, ";
if ($r['rock_pop']) $string .= "Rock/Pop/Alternative, ";
if ($r['world']) $string .= "World, ";
if ($r['othr_music']) $string .= $r['othr_music_txt'];

$string = rtrim($string, ", ");

echo " $string \n";
  }

EDIT: Oops, didn't read the code very well the first time through... :p

[Weedpacket]

I think some database normalisation would be a benefit here. A second table listing genres (id, text_label); and a third table to link it with music (music_id, genre_id). The long chain of if statements would then disappear into the database. It is, after all, effectively a table already:

$genres = array(
	'band' => 'Band',
	'barbershop' => 'Barbershop',
	//...
	'opera' => 'Opera/Musicals',
	//...
	'rock_pop' => 'Rock/Pop/Alternative',
	'world' => 'World',
);

$selected_genres = array_intersect_key($genres, array_filter($r)));
if($r['othr_music'])
	$selected_genres['othr_music'] = $r['othr_music_txt'];

$genre_string = join(", ", $selected_genres);

[harkly]

I appreciate all the advise but right now I am trying to get my code working for my originally issue. The rest is a work in progress.

I need to verify if the table has info or not and then display an option based on that information.

<?php
   include('library/login.php');
   login();
   mysql_select_db('harkly'); 

$sql = mysql_query("SELECT * FROM music WHERE userId = 'kelly'");
$results = mysql_fetch_array( $sql );

  if ($results == NULL)
    { 
      echo "Empty  \n";  

    }

  else
    {
      echo "Not Empty \n";
    }

?>

[Weedpacket]

Select Count(*) from music where userid='whatever'

This will return one row with one field containing the number of matching records. If it's zero then there aren't any.

But in your original post you were asking for a lot more than just "verify if the table has info or not". You wanted the information itself.

So do the query, and use [man]mysql_num_rows[/man] to see how many rows were returned.

[big0mike]

I'm no expert but would it make a difference if you tried:

if (empty($results))

That's what I would've done...