mySQL warning problems

adavis

I'm getting the following two warnings when I hit my database.

Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in /data/web/html/folder/buildoutput.php on line 183

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /data/web/html/folder/buildoutput.php on line 187

I have echoed my sql statement because it's being build dynamically. My first sql statement looks like this:

SELECT count(money) AS tot, round(sum(money),0) AS value
FROM mainTable
WHERE xt = '2'
AND ADate BETWEEN '2005-10-01' AND '2005-12-29'
AND code not in ('x','y','z')

When I cut/paste it into mysqlyog, the correct results are returned. Which are tot = 0 and value = NULL. 'ob' is a bigint, but the default is set to NULL. Is this where my problems is? Should the default be set to 0?

Or is this an entirely different problem. What I want to do is:
if tot = 0, then I push a 0 into my variable,
else, I push value in.

Do I need to do this in 2 seperate sql statements? The first one to check the count and if it's > 0, submit the sql statement to sum the field?

devinemke

that first error suggests that you are not connected to the database. are you? as always you should use [man]mysql_error[/man] to debug.

adavis

I think I'm connected. This is all inside a function and I call my connect just before I enter the switch that calls this function about 8 or so times. I had this, but it wasn't giving me anything special, except killing my script:

$result = mysql_query($sql,$dblink) or die(mysql_errno().":".mysql_error()."<br />ERROR: Query failed:<br />$sql\n");

I guess I could connect and close inside the function and see if that changes anything.

adavis

okay, well, I thought I was calling the database, but apparently I wasn't. I move the call and close to inside the function and viola, it worked. I can try to blame my momentary lapse on my head cold, but... I won't. I don't know if it's a PHP5 issue or what, but PHP5 is definately challenging. 😉

Thanks,
Alisa

Roger_Ramjet

It may be that the Variable Scope has been changed in php 5 - certainly that seems to be your problem.

Sxooter

Or is this an entirely different problem. What I want to do is:
if tot = 0, then I push a 0 into my variable,
else, I push value in.

OK, here you want a case statement:

select case where tot=0 then 0 else value end from sometable;

adavis

I generally do opt for a case statement unless I KNOW I will only have 1 of 2 values, then for ease of coding, I go with an if or if/else. I probably went with the if/else statement here since the value was either 0 or not 0. I'd have to go back and look at my code because
1. I've slept since then and
2. I've coded different stuff since then

🙂