field name is a number and treated like a number

nephish

hey there,

cant seem to find a work around for this,
i have a table where each hour of the day is a field , 01, 02, 03 ......

and when i pass a variable to the table it treats it like the variable result,

<?php
$Hour = strftime("%H");
print $Hour;

function OneDay($Unit){
	$result = mysql_query("SELECT `VOffset`, '{$Hour}' FROM Rain_Gauges WHERE `Sensor_ID` = {$Unit} $connection");
	$Row = mysql_fetch_row($result);
	//subtract the Hour Value from the VOffset value
	$Unit =  ($Row[0] - $Row[1]);
	return $Unit;
	}

can you maybe tell me what i am messing up here?
thanks

leatherback

nephish wrote:
hey there,

and when i pass a variable to the table it treats it like the variable result,

Sorry. But this doesn't make sense. What are you trying to say?

nephish

instead of getting the value in the row for field 11 it is returning the number 11
11 would represent the value at 11:00 am,
it is freaking out about passing a number as a variable to select a field . The actual field name is 11. there is one for each hour of the day.
thanks

leatherback

I doubt it is freaking out about the variable. You usually do not need quotes, but maybe for numbers you do.. But in principle, I think you shouldn't use fieldnames which consist of only a number.

Have you tried like this:

function OneDay($Unit){
	$result = mysql_query("SELECT VOffset, $Hour FROM Rain_Gauges WHERE sensor_ID = $Unit", $connection);
	$Row = mysql_fetch_row($result);
	//subtract the Hour Value from the VOffset value
	$Unit =  ($Row[0] - $Row[1]);
	return $Unit;
	}

nephish

This is what i am getting

Warning: mysql_numrows(): supplied argument is not a valid MySQL result resource in /var/www/phptest.php on line 39

this is line 39

$num=mysql_numrows($result);

is it seeing the variable as an integer ?
could i force it to see it as a string?

thanks

nephish

oh, if i replace the $Hour with 14 it runs correctly.

nephish

i changed all the field names from 01 to H01, 02 to H02, 03 to..... (you get the idea)
still getting an error

$Hour = strftime("H%H");

$connection = mysql_connect('localhost', 'piv', 'crayon99'); 
mysql_select_db('DDS', $connection);


function OneDay($Unit){
	$result = mysql_query("SELECT `VOffset`, `$Hour` FROM Rain_Gauges WHERE Sensor_ID = $Unit");
	$Row = mysql_fetch_row($result);
	//subtract the Hour Value from the VOffset value
	$Unit =  ($Row[0] - $Row[1]);
	return $Unit;
}

gives me

Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in /var/www/phptest.php on line 30

i have no idea why

leatherback

what does the mysql_error() give you?
And echo out the query you put into the mysql_query). Maybe $Unit is not set?

nephish

THat part is fixed,
the strftime('H%H') has to come first INSIDE the function
thanks all

nephish

OK sorry, it isnt fixed,
looked like it for a while.
i have tried with static numbers instead of variables and it works then
going about nuts here