PHP error with mysql_num_rows - php

Question

I have a simple form which when submitted should count the number of rows in a table which already have the same value as that submitted.
I can't work out why this is returning an error... any ideas?
The error is Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in .../register.php on line 31
$con = mysql_connect("table","user","pass");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("db", $con);
function check_input($value, $quoteIt)
{
// Stripslashes
if (get_magic_quotes_gpc())
{
$value = stripslashes($value);
}
// Quote if not a number
if (is_null($value) || $value=="") {
$value = 'NULL';
} else if (!is_numeric($value) && $quoteIt == 1) {
$value = "'" . mysql_real_escape_string($value) . "'";
}
return $value;
}
$useremail = check_input($_POST['useremail'], 1);
// Check to see if email address already exists in USERS table
$query="SELECT * FROM users WHERE email = $useremail";
$result = mysql_query($query);
echo $query;
echo mysql_num_rows($result); //THIS IS LINE 31
mysql_close();

Answer 1

You can see your query error using this, because of query error only this error showing
$result = mysql_query($query) or die(mysql_error());

Answer 2

<?php
$con = mysql_connect("table","user","pass");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("db", $con);
function check_input($value, $quoteIt)
{
// Stripslashes
if (get_magic_quotes_gpc())
{
$value = stripslashes($value);
}
// Quote if not a number
if (is_null($value) || $value=="") {
$value = 'NULL';
} else if (!is_numeric($value) && $quoteIt == 1) {
$value = "'" . mysql_real_escape_string($value) . "'";
}
return $value;
}
$useremail = check_input($_POST['useremail'], 1);
// Check to see if email address already exists in USERS table
$query="SELECT COUNT(1) FROM users WHERE email = $useremail";
$result = mysql_query($query);
$array = mysql_fetch_array($result);
$whatYouWant = $array[0][0];
mysql_close();
?>
But Leigh is right, prefer PDO instead of mysql_* functions...

Answer 3

Usually this means the query is wrong. Try backquotes for fields:
"SELECT * FROM `users` WHERE `email` = '$useremail'"

Answer 4

Replace
$result = mysql_query($query);
with
$result = mysql_query($query, $con);