mysql_fetch_array() error? please help me. :)

I have a code that needs to count for the total value and each value that has matched in the column. Please the code:

//Parts of Speech, contains in ss_type for echo 
$part_speech = array("n"=>"n.","v"=>"verb", "r"=>"adv./adj.", "a"=>"adj.", "s"=>"adj."); 

$counting = mysql_query('SELECT count(ss_type) FROM wn_synset WHERE word="'.str_replace(' ', '_', $word).'"'); 
$countnoun = mysql_query('SELECT count(WHERE ss_type="n") FROM wn_synset WHERE word="'.str_replace(' ', '_', $word).'"');

$total = mysql_fetch_array($counting);
$totalnoun = mysql_fetch_array($countnoun);

$count = 0;
$countnoun = 0;

//this is in a loop
$count = $count + 1;
$countnoun = $countnoun + 1;

echo "Total: $total[0]<br />";
if ($totalnoun[0] > 0) {
echo "-$totalnoun[0] nouns</h2>";
}

PHP:

I thought my code was correct now but I had this error:

The line with error is this line:

$totalnoun = mysql_fetch_array($countnoun);

PHP:

What's wrong with code? The echo for the total cound value works but the count for the nound doesn't.Thank you in advance for trying to help me.

 

try this

$totalnoun = @mysql_fetch_array($countnoun);

If the code is ok it will remove warning but be sure that code is ok

Regards

Alex

 

The error was gone but it doesn't echo the $totalnoun[0].. What is "@" for? Thanks for the help.

 

The @ is simply error suppression, so even though it wants to tell you that it doesn't have a valid result resource, it won't show up at all. Usually I see this error being caused because the query has an error in it. Add an or die('Error: '.mysql_error()) to the end of the query with the problem and see if it's actually valid or if there is a problem with your query. I've honestly never seen "count(WHERE ss_type="n")" before, generally you do count(*) or count(field) (if you have NULL values count(field) will ignore them so it's faster than * in that scenario). I know there are generally 500 ways to do the same thing in SQL, I'm not sure if what you wrote is proper but I just don't see it like that. That's where i'd start checking =)

 

Supressing errors wont help solve the problem only hide them.

you should change

$countnoun = mysql_query('SELECT count(WHERE ss_type="n") FROM wn_synset WHERE word="'.str_replace(' ', '_', $word).'"');

Code (markup):

to

$countnoun = mysql_query('SELECT count(ss_type) FROM wn_synset WHERE ss_type="n" AND word="'.str_replace(' ', '_', $word).'"');

Code (markup):

you can also use mysql_num_rows() to see if there are valid records pulled back like

if(mysql_num_rows($countnoun) > 0)
{
   //proceed

}
else
{
   echo "No results found";
}

Code (markup):

 

supplied argument is not a valid MySQL result resource

normally means it is trying to break results up into an array, when there are no results

 

Here's a little code modification

//Parts of Speech, contains in ss_type for echo 
$part_speech = array("n"=>"n.","v"=>"verb", "r"=>"adv./adj.", "a"=>"adj.", "s"=>"adj."); 

$counting = mysql_query('SELECT count(ss_type) FROM wn_synset WHERE word="'.str_replace(' ', '_', $word).'"'); 
$countnoun = mysql_query('SELECT count(WHERE ss_type="n") FROM wn_synset WHERE word="'.str_replace(' ', '_', $word).'"');

// mod here
$total = ($counting) ? mysql_fetch_array($counting) : 0;
$totalnoun = ($countnoun) ? mysql_fetch_array($countnoun) : 0;

$count = 0;
$countnoun = 0;

//this is in a loop
$count = $count + 1;
$countnoun = $countnoun + 1;

echo "Total: $total[0]<br />";
if ($totalnoun[0] > 0) {
echo "-$totalnoun[0] nouns</h2>";
}[

PHP:

 

Thanks for the help guys! Thanks for the modification of the script xrvel! lfhost is correct, the cause of the error is in the mysql_query in the variable $countnoun. the count() function was incorrect. It should be like this:

$countnoun = mysql_query('SELECT [B]count(ss_type)[/B] FROM wn_synset WHERE word="'.str_replace(' ', '_', $word).'" [B]AND ss_type="n"[/B]');

PHP:

By the way, I will try the code you modified also xrvel. Thank you.