: mysql_fetch_array(): supplied argument is not a valid MySQL

[leatherback]

Hi,

This is the error message I get:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource

The querying done is:

        $plotquery = "select * from plots where pl_name = '$plot[$i]'";
           $result= mysql_query($plotquery);
           $rows = mysql_num_rows($result);
           echo $rows;
          // echo $plotquery;
           if(! $rows>0 )
             {
             $insertplot = mysql_query("insert into plots (pl_name, pl_country) values 
             ('$plot[$i]', '$countries[$i]')");
             }

         $result = mysql_query($plotquery);
         echo $plotquery;
         while ( $row = mysql_fetch_array($result))
          {
           }

When I take the echod query from the screen and parse it in phpmyadmin I get nicely two rows from the database. What else could I be doing wrong?
J.

[lanjoky]

When using mysql_fetch_array you must have a Select Statement as your base SQL argument ... you have an instert statement and there for you are getting the error... Not sure what your trying to do.... but that is your error

HTH

[leatherback]

??

As far as I can tell I am using a select query??
Why do you think I am using an insert query?

😕

J.

[kevinsequeira]

$insertplot = mysql_query("insert into plots (pl_name, pl_country) values
('$plot[$i]', '$countries[$i]')");

thats why ...

EDIT : actually im not sure if ure code is written to do what u want, u have blanlk while loop ??

another thing is that the result may be null cause $plot[$i] changes

what exactly do u want to do

reg
kevin

[ahundiak]

What does mysql_error() say? My guess is that $plot[$i] does not have a correct value.

$result = mysql_query($plotquery);
if (mysql_errno()) die(mysql_error());

[leatherback]

Yes, but I am not refering to that query:

$plotquery = "select * from plots where pl_name = '$plot[$i]'"; 

// blablabla

$result = mysql_query($plotquery); 
while ( $row = mysql_fetch_array($result)) 

[leatherback]

Ahundiak:
mysql_error() does NOT give anything. I tried the query in phpmyadmin, and now I tried with your snippet.. Nothing..

J.

[ahundiak]

Well that seems strange. I have never had mysql_query return a bad value without getting an error code set. I'll be interested in seeing how the problem get's resolved. Of course, there has been a few times where I thought I was checking the error code but wasn't.

Are you sure you don't have a simple typo somewhere? Maybe misspelled the $result variable name?

[leatherback]

Hm..

Script seems to have been mixed up with an other mistake further down: I found a missing ) 10 lines down, I replaced that, and th the error dissappeared.. Is my PC infected by the GATE$ virus??

J.

[leatherback]

Aaah..

The script does it again. Don't know why. Did not change anything. Just reloaded the page!?

J.

The script now is;

        $plotquery = "select * from plots where pl_name = '$plot[$i]'";
           $result= mysql_query($plotquery);
           $rows = mysql_num_rows($result);

       if(! $rows>0 )
         {
         $insertplot = mysql_query("insert into plots (pl_name, pl_country) values ('$plot[$i]', '$countries[$i]')");
         }

       $plotquery = "select * from plots where pl_name = '$plot[$i]'";
       $result= mysql_query($plotquery);


         while ( $row = mysql_fetch_array($result))
          {
          $date = mktime($datehour, $dateminute, $datesecond, $datemonth, $dateday, $dateyear);

Please help!?

[leatherback]

OK. Now I have it.

It is due to looping through a var:
The query:

$plotquery = "select * from plots where pl_name = '$plot[$i]'";

returns an error in the first instance. Then, after inserting the new variable into the database, the query should return the correct results. BUT in some weird way the script maintains that error contents, so that when I call the query again, an eror is displayed.

I now run it like:

        $plotquery = "select * from plots where pl_name = '$plot[$i]'";
           $result= mysql_query($plotquery);
           $rows = mysql_num_rows($result);

       if(! $rows>0 )
         {
         $insertplot = mysql_query("insert into plots (pl_name, pl_country) values ('$plot[$i]', '$countries[$i]')");
         }

       $plotqueryn = "select * from plots where pl_name = '$plot[$i]'";
       $resultn= mysql_query($plotqueryn);

         while ( $row = mysql_fetch_array($resultn))
          {

which doesn't give errors (nope, not even after several reloads ;-) )