I have a query and im getting errors with it...

$query="SELECT b.*, c.* FROM tod_company_information c INNER JOIN tod_bookings_archive b ON c.company_id = b.company_id $filter_string WHERE b.date_fitted < DATE_SUB(now(),INTERVAL 1 YEAR)ORDER BY b.date_in ASC";

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/eeeepcom/public_html/tt/main.php on line 23

any ideas :S

Thanks in advance

crE

what's $filter_string?
you might want to do a echo mysql_error();
it'll help you pinpoint the problem

I have echoed it but it just echos exactly what it says there..

i got rid of the filter string - it was something that was tehre when i cut it from some of my old coding..

still doesnt work :S

crE

this might seem a bit silly but
1) did you make a connection to the db?
2) did you do a mysql_query() or just made the query string?

sample1
$textsql="SELECT * FROM news_details,news_mast WHERE news_mast.item_no = news_details.item_no AND news_details.row_no ='$rwn'";
sample2
SELECT DISTINCT SUBGROUP,ITEM_NO FROM news_mast where NGROUP='".$SubCateSearch."'";

$resultmain=mysql_query($textsql) or die("Invalid ".mysql_error());

$num_of_rows = mysql_num_rows($resultmain)or die ("Data not found from news details");
$row=mysql_fetch_row($resultmain);

This are the parameters
'".$SubCateSearch."'";
'$rwn'

just try...

You have to call mysql_query() and pass the result of that operating to mysql_num_rows(). You don't pass mysql_num_rows() a query.


$query = "SELECT * FROM mytable";
echo mysql_num_rows($query); //wrong, will produce your warning

echo mysql_num_rows(mysql_query($query)); //correct.


This is all assuming that there aren't any errors in your query itself.

i never had a connection to the db :rolleyes:

sorted now!

Originally posted by crEA-tEch
i never had a connection to the db :rolleyes:

sorted now!
don't worry, happens to all of us :)