Error: Resource id #6

svenski

Help! I've got an inner join that works when i try to use it in an SQL interface, but returns 'Resource id #6' when used in a n php enviroment. Can't figure out why! any help appreciated, heres the code:

$query ="
SELECT phonetable.phone_number
FROM phonetable
INNER JOIN camptable
ON camptable.inbound_phone_no_id = phonetable.sms_phone_no_id
WHERE camptable.campaign_id = $campaign_id";

$sender = mysql_query($query);

given any value of $campaign_id the result of $sender = Resource ID #6

Anybody got any ideas? TIA

OhLordy

Performing the query (mysql_query($sql)) returns a resource handle which you then use to rerieve the matching lines as follows.

<?
$query =" 
SELECT phonetable.phone_number 
FROM phonetable 
INNER JOIN camptable 
ON camptable.inbound_phone_no_id = phonetable.sms_phone_no_id 
WHERE camptable.campaign_id = $campaign_id"; 
$sender = mysql_query($query); 
while($row=mysql_fetch_array($sender))
{
  echo($row[0]." - ".$row[1]."<br />\n");
}
?>

HTH
Rob

svenski

Thanks, having some trouble with the code, i get the message:

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

What do you mean exactly that it returns a resource handle?
There should always be only one result rather than an array... im confused

😕

OhLordy

With your query maybe but many queries multiple elements and multiple rows take for example

select user_id, f_name,s_name,email from users where f_name LIKE '%fr%';

This would return four elements and two rows (in this case the entire table)
so to access the first name you would do

<?
while($row=mysql_fetch_array($result))
{
  echo("the first name of user ".$row['user_id']." is ".$row['f_name']."<br />\n";
}
?>

With regards to your error. try putting an or die statement after the mysql_query like this

<?
$result=mysql_query($sql) or die(mysql_error());
?>

HTH
Rob

svenski

Well the following still gives me an error! argggghhhh!

<?
$query ="

SELECT phonetable.phone_number

FROM phonetable

INNER JOIN camptable

ON camptable.inbound_phone_no_id = phonetable.sms_phone_no_id

WHERE camptable.campaign_id = $campaign_id";

$sender = mysql_query($query) or die(mysql_error());

while($row=mysql_fetch_array($sender))
{
echo($row[0]." - ".$row[1]."<br>");
} ?>

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

WTH is going wrong? am i doing what you mean?

laserlight

I'm guessing that you need to use single quotes here:

<?
$query = "SELECT phonetable.phone_number
	FROM phonetable
	INNER JOIN camptable
	ON camptable.inbound_phone_no_id = phonetable.sms_phone_no_id
	WHERE camptable.campaign_id = '$campaign_id'";
$sender = mysql_query($query) or die(mysql_error());
while ($row = mysql_fetch_array($sender))
{
	echo $row[0] . " - " . $row[1] . "<br>";
}
?>