mysql_fetch_array error message

dougmcc1

What does this error message mean?
"mysql_fetch_array(): supplied argument is not a valid MySQL result resource"

Pig

it means that whatever is you are feeding the function is no good. Can you show us how you got the $result that you are putting in mysql_fetch_array()?

dougmcc1

Sure, here is my code:

$row['rcnt']=1;

while($row['rcnt']>0)
{
$random_number = rand(00121321000,99867437908);
$result = mysql_query('SELECT COUNT(rnum) as rcnt FROM table

WHERE rnum="'.$random_number.'";');
$row = mysql_fetch_array($result);
}

What I'm trying to do is generate a random number and check to see if it's in the table. I want to do this over and over until a unique random number is generated. Then I want to add that number to the table.

Pig

The end of your query was a little off. You had some weirdness with double semicolons and extra characters. This is how it should look.

$result = mysql_query('SELECT COUNT(rnum) as rcnt FROM table WHERE rnum="'.$random_number.'");

Try that.

dougmcc1

Nope still same error. I included a single quote on the end of yours:
$result = mysql_query('SELECT COUNT(rnum) as rcnt FROM <table> WHERE rnum="'.$random_number.'"');

dougmcc1

The syntax looks ok, is the logic correct?

cgraz

try using mysql_error()

$result = mysql_query("SELECT COUNT(rnum) as rcnt FROM table WHERE rnum='".$random_number."'") or die(mysql_error());

and seeing what that prints

Cgraz

dougmcc1

Ok the code is error free. It only accepts my first entry though. After that, no other entries are entered into the database. Although the random_number never equals the number in the rnum field after that first entry is accepted, so thats good.

So it would be working great if I could get the other entries into the database.

dougmcc1

For example,

I changed the rand() function to this:
$random_number = rand(2,4);

The first entry gave me a random number of 2 and entered everything into the database.

I tried 10 other entries after that and they all produced a random number of 3 or 4, but none got added to the database.

dougmcc1

Haha I forgot to auto_increment the id field for this table...

It works good now. The only thing is if I have this line:
$random_number = rand(2,4);

Then after 3 entries, 2, 3 and 4 get used up and the 4th entry times out after 30 seconds. Instead of timing out, can I display an error message of "Broaden your random number range to enter more data"?

Thanks guys for all your help. I really appreciate it.