warning: mysql_fetch_array () EXPECTS PARAMETER 1 TO BE RESOURCE, BOOLEAN GIVEN IN N: \ HOME \ Localhost \ www \ phpsite \ TestReg.php on line 36

How to make not issued an error?

Answer 1, Authority 100%

so as not to give out – @ in front of the function.
To avoid Errors, write like this:

$ result = @mysql_query ("Select` col` from `oshibka` Where ip = '$ ip'", $ db );
$ MYROW = Array ();
if (mysql_num_rows ($ result) & gt; 0) {
 $ myrow = mysql_fetch_array ($ result);
}

even better

try {
$ result = mysql_query ("Select` col` from `oshibka` Where ip = '$ ip'", $ db);
} Catch (Exception E) {
//Error processing
}

Answer 2, Authority 100%

It is also easier. Error is most likely in the request and we simply do not see it, but better to bring it and see!

$ result = mysql_query ($ sql) || DIE (mysql_error ());

Answer 3, Authority 50%

$ result = mysql_query ("Select col from oshibka where ip = '$ ip'", $ db);
if ($ result) {
  $ myrow = mysql_fetch_array ($ result);
}
ELSE {
  echo mysql_error ();
}

and everything ..

Answer 4

If you rinsed it right, then the request came empty and you throw it to another function.
It is necessary to make a plug, so that this does not happen. I do this:

$ result = mysql_query ("Select col from oshibka where ip = '$ ip'", $ db) or Die (' CANN`t Query '.Mysql_error);

Error occurs due to not matching types: The function accepts the type of Resource data, your request came empty, which is equivalent to False (Boolean).