cannot get mysql_num_rows to work

[mattastic]

Hi,

I cant get the following piece of code to work, i keep getting Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource

$query="SELECT *
      FROM tblevent e
INNER JOIN tblroombooking r
        ON e.eventID = r.eventID
     WHERE e.startdate <= '$startdate'
       AND e.enddate >= '$enddate'
       AND r.roomID = $room";
	   echo $query;


$result = mysql_query(query);

$num = mysql_num_rows(result);

if ($num>0) #if $num=0 the table is empty
{

error_message('Room is booked for that date');

}

Can anyone help?

THankyou

[Piranha]

$result = mysql_query(query); 
$num = mysql_num_rows(result);

Don't forget the dollar signs before query and result.

$result = mysql_query($query); 
$num = mysql_num_rows($result);

[NogDog]

Also, it's best not to assume your query worked. You should check that $result is not FALSE before continuing. For debugging, if it is false you can die() with some debug info, such as:

$result = mysql_query($query) or die("Query failed: $query - " . mysql_error());

[coldfused]

I think by having multiple lines, it messes up the query so either put the query into all one line or try using a "heredoc string" so by that:

$query= <<<QUERY
SELECT *
FROM tblevent e
INNER JOIN tblroombooking r
ON e.eventID = r.eventID
WHERE e.startdate <= '{$startdate}'
AND e.enddate >= '{$enddate}'
AND r.roomID = {$room}
QUERY;

or as i just realised you might need to escape the variables in the query strings by using braces {$variable}.

[Pehape]

or as i just realised you might need to escape the variables in the query strings by using braces {$variable}.

what are these braces for ?

[NogDog]

Pehape wrote:

what are these braces for ?

In your example, since all the variables are simple variables with word boundaries on either side, the braces (called "complex variable notation") are not necessary, though they won't hurt anything. But when using a variable name within a string literal where the parser might get confused, such as if it is an array or object reference or the variable name would butt up against another word, then the braces delineate the variable and allow the parser to know exactly where it begins and ends without having to close the quotes and use "." for concatenation:

$test = "This {$variable}runs into the next word.";
$test2 = "This string {$includes['an']['array']} in the middle."
$test3 = "It could {$also->be} an object attribute.";

[angeljyt]

mattastic wrote:

Hi,

I cant get the following piece of code to work, i keep getting Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource

$query="SELECT *
      FROM tblevent e
INNER JOIN tblroombooking r
        ON e.eventID = r.eventID
     WHERE e.startdate <= '$startdate'
       AND e.enddate >= '$enddate'
       AND r.roomID = $room";
	   echo $query;


$result = mysql_query(query);

$num = mysql_num_rows(result);

if ($num>0) #if $num=0 the table is empty
{

error_message('Room is booked for that date');

}

Can anyone help?

THankyou

Notice this:
$result = mysql_query(query);

$num = mysql_num_rows(result);

you forget to write the $ signs before query and result , they are treated as constants.and then your SQL statements is not correctly executed.

It should like this one:
$result = mysql_query($query);
if (!result) {
die(mysql_error());
}
$num = mysql_num_rows($result);

[bradgrafelman]

angeljyt wrote:

It should like this one:

Except you also forgot a dollar sign in the if() statement. At any rate, your code can be simplified to this:

$result = mysql_query($query) or die(mysql_error());
$num = mysql_num_rows($result);

[angeljyt]

I am very sorry.I am so carefulless that I make the same error. :mad:
May i make friends with you,bradgrafelman .My gmail is [e-mail] a million.

[bradgrafelman]

Don't forget to mark this thread resolved (if it is).

[Pehape]

hmmm...so using {$var} instead ".$var." is better. it has no difference right.?
so i can replace all my ".$var." to {$var}, simpler... 🙂