[PHP-DB] Count database-values
Hi everybody,
I've got a little problem, which I can't solve. It is as follow:
In my database I made a table with different fields.
One of the fields is named 'bedrag' and contains a numeric value like 15.47 or 78.16
and so on.
If I want a value of the table I use in most cases the following code:
?
$squery = mysql_query(SELECT * FROM finance WHERE posneg = 'af',$db);
while($row = mysql_fetch_array($squery)) {
$bedrag = $row[bedrag];
echo $bedragbr;
}
?
In this case I receive a list like:
15.47
78.16
and so on...
So far no problems, but I want to count all these values. I tried with SUM but with
the following code it doesn't work:
$squery = SELECT SUM(bedrag) AS Total FROM finance WHERE posneg = 'af';
any suggestions how to do? I want to store it in a variabele ($total) so I can echo
it. (E.g.: $total = 93.63 in this case)
Thanx in advance,
Robert van der Mast
Re: [PHP-DB] Count database-values
So far no problems, but I want to count all these values.
I tried with SUM but with the following code it doesn't work:
$squery = SELECT SUM(bedrag) AS Total FROM finance WHERE posneg = 'af';
any suggestions how to do? I want to store it in a variabele ($total)
so I can echo it. (E.g.: $total = 93.63 in this case)
If you want the total, you need to do something like the following:
$query = SELECT SUM(bedrag) AS Total FROM finance WHERE posneg = 'af';
$result = mysql_query( $query );
$total = mysql_fetch_array( $result );
$total = $total['Total'];
You had forgotten to pull the data out of your result set. It's a common
mistake. :)
Edward Dudlik
Becoming Digital
www.becomingdigital.com
- Original Message -
From: Robert Wanadoo [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Monday, 09 June, 2003 11:40
Subject: [PHP-DB] Count database-values
Hi everybody,
I've got a little problem, which I can't solve. It is as follow:
In my database I made a table with different fields.
One of the fields is named 'bedrag' and contains a numeric value like 15.47 or
78.16 and so on.
If I want a value of the table I use in most cases the following code:
?
$squery = mysql_query(SELECT * FROM finance WHERE posneg = 'af',$db);
while($row = mysql_fetch_array($squery)) {
$bedrag = $row[bedrag];
echo $bedragbr;
}
?
In this case I receive a list like:
15.47
78.16
and so on...
So far no problems, but I want to count all these values. I tried with SUM but
with the following code it doesn't work:
$squery = SELECT SUM(bedrag) AS Total FROM finance WHERE posneg = 'af';
any suggestions how to do? I want to store it in a variabele ($total) so I can
echo it. (E.g.: $total = 93.63 in this case)
Thanx in advance,
Robert van der Mast
--
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Re: [PHP-DB] Count database-values
I screwed up my own code. Silly me. It should read:
$query = SELECT SUM(bedrag) AS Total FROM finance WHERE posneg = 'af';
$result = mysql_query( $query );
while ( $total = mysql_fetch_array( $result ) )
{
$total = $total['Total'];
}
Edward Dudlik
Becoming Digital
www.becomingdigital.com
- Original Message -
From: Becoming Digital [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Tuesday, 10 June, 2003 06:43
Subject: Re: [PHP-DB] Count database-values
So far no problems, but I want to count all these values.
I tried with SUM but with the following code it doesn't work:
$squery = SELECT SUM(bedrag) AS Total FROM finance WHERE posneg = 'af';
any suggestions how to do? I want to store it in a variabele ($total)
so I can echo it. (E.g.: $total = 93.63 in this case)
If you want the total, you need to do something like the following:
$query = SELECT SUM(bedrag) AS Total FROM finance WHERE posneg = 'af';
$result = mysql_query( $query );
$total = mysql_fetch_array( $result );
$total = $total['Total'];
You had forgotten to pull the data out of your result set. It's a common
mistake. :)
Edward Dudlik
Becoming Digital
www.becomingdigital.com
- Original Message -
From: Robert Wanadoo [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Monday, 09 June, 2003 11:40
Subject: [PHP-DB] Count database-values
Hi everybody,
I've got a little problem, which I can't solve. It is as follow:
In my database I made a table with different fields.
One of the fields is named 'bedrag' and contains a numeric value like 15.47 or
78.16 and so on.
If I want a value of the table I use in most cases the following code:
?
$squery = mysql_query(SELECT * FROM finance WHERE posneg = 'af',$db);
while($row = mysql_fetch_array($squery)) {
$bedrag = $row[bedrag];
echo $bedragbr;
}
?
In this case I receive a list like:
15.47
78.16
and so on...
So far no problems, but I want to count all these values. I tried with SUM but
with the following code it doesn't work:
$squery = SELECT SUM(bedrag) AS Total FROM finance WHERE posneg = 'af';
any suggestions how to do? I want to store it in a variabele ($total) so I can
echo it. (E.g.: $total = 93.63 in this case)
Thanx in advance,
Robert van der Mast
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
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RE: [PHP-DB] Count database-values
-Original Message-
From: Becoming Digital [mailto:[EMAIL PROTECTED]
Sent: 10 June 2003 12:15
I screwed up my own code. Silly me. It should read:
Surely your first attempt is the right one? There's only ever going to be 1 Total, so
why waste a while loop trying to read more than one result row?
$query = SELECT SUM(bedrag) AS Total FROM finance WHERE
posneg = 'af';
$result = mysql_query( $query );
while ( $total = mysql_fetch_array( $result ) )
{
$total = $total['Total'];
}
Edward Dudlik
Becoming Digital
www.becomingdigital.com
- Original Message -
From: Becoming Digital [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Tuesday, 10 June, 2003 06:43
Subject: Re: [PHP-DB] Count database-values
So far no problems, but I want to count all these values.
I tried with SUM but with the following code it doesn't work:
$squery = SELECT SUM(bedrag) AS Total FROM finance WHERE
posneg = 'af';
any suggestions how to do? I want to store it in a
variabele ($total)
so I can echo it. (E.g.: $total = 93.63 in this case)
If you want the total, you need to do something like the following:
$query = SELECT SUM(bedrag) AS Total FROM finance WHERE
posneg = 'af';
$result = mysql_query( $query );
$total = mysql_fetch_array( $result );
$total = $total['Total'];
Cheers!
Mike
-
Mike Ford, Electronic Information Services Adviser,
Learning Support Services, Learning Information Services,
JG125, James Graham Building, Leeds Metropolitan University,
Beckett Park, LEEDS, LS6 3QS, United Kingdom
Email: [EMAIL PROTECTED]
Tel: +44 113 283 2600 extn 4730 Fax: +44 113 283 3211
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Count database-values
Unfortunately, I can't, despite my best efforts, get the data to display unless
it's put inside a loop. If anyone can tell me how to, I'd just for joy.
Edward Dudlik
Becoming Digital
www.becomingdigital.com
- Original Message -
From: Ford, Mike [LSS] [EMAIL PROTECTED]
To: 'Becoming Digital' [EMAIL PROTECTED]; [EMAIL PROTECTED]
Sent: Tuesday, 10 June, 2003 11:54
Subject: RE: [PHP-DB] Count database-values
-Original Message-
From: Becoming Digital [mailto:[EMAIL PROTECTED]
Sent: 10 June 2003 12:15
I screwed up my own code. Silly me. It should read:
Surely your first attempt is the right one? There's only ever going to be 1
Total, so why waste a while loop trying to read more than one result row?
$query = SELECT SUM(bedrag) AS Total FROM finance WHERE
posneg = 'af';
$result = mysql_query( $query );
while ( $total = mysql_fetch_array( $result ) )
{
$total = $total['Total'];
}
Edward Dudlik
Becoming Digital
www.becomingdigital.com
- Original Message -
From: Becoming Digital [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Tuesday, 10 June, 2003 06:43
Subject: Re: [PHP-DB] Count database-values
So far no problems, but I want to count all these values.
I tried with SUM but with the following code it doesn't work:
$squery = SELECT SUM(bedrag) AS Total FROM finance WHERE
posneg = 'af';
any suggestions how to do? I want to store it in a
variabele ($total)
so I can echo it. (E.g.: $total = 93.63 in this case)
If you want the total, you need to do something like the following:
$query = SELECT SUM(bedrag) AS Total FROM finance WHERE
posneg = 'af';
$result = mysql_query( $query );
$total = mysql_fetch_array( $result );
$total = $total['Total'];
Cheers!
Mike
-
Mike Ford, Electronic Information Services Adviser,
Learning Support Services, Learning Information Services,
JG125, James Graham Building, Leeds Metropolitan University,
Beckett Park, LEEDS, LS6 3QS, United Kingdom
Email: [EMAIL PROTECTED]
Tel: +44 113 283 2600 extn 4730 Fax: +44 113 283 3211
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
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RE: [PHP-DB] Count database-values
Edward,
Have you considered trying mysql_result()?
It appears that your query is going to always return a single piece of data:
Total. So you essentially have an array with only a single element. If you
did this:
$total = mysql_result($result,0);
You'll get the value of the item at index 0 (the first position in an array)
of the $result array stored in the $total variable.
Details are here:
http://us3.php.net/manual/en/function.mysql-result.php
The page also makes reference to other high-performance options, but since
you're only grabbing a single item and if you use the index number instead
of the column name as the documentation recommends, the performance should
be just fine.
Hope this helps.
Rich
-Original Message-
From: Becoming Digital [mailto:[EMAIL PROTECTED]
Sent: Tuesday, June 10, 2003 12:50 PM
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Count database-values
Unfortunately, I can't, despite my best efforts, get the data
to display unless
it's put inside a loop. If anyone can tell me how to, I'd
just for joy.
Edward Dudlik
Becoming Digital
www.becomingdigital.com
- Original Message -
From: Ford, Mike [LSS] [EMAIL PROTECTED]
To: 'Becoming Digital' [EMAIL PROTECTED];
[EMAIL PROTECTED]
Sent: Tuesday, 10 June, 2003 11:54
Subject: RE: [PHP-DB] Count database-values
-Original Message-
From: Becoming Digital [mailto:[EMAIL PROTECTED]
Sent: 10 June 2003 12:15
I screwed up my own code. Silly me. It should read:
Surely your first attempt is the right one? There's only
ever going to be 1
Total, so why waste a while loop trying to read more than one
result row?
$query = SELECT SUM(bedrag) AS Total FROM finance WHERE
posneg = 'af';
$result = mysql_query( $query );
while ( $total = mysql_fetch_array( $result ) )
{
$total = $total['Total'];
}
Edward Dudlik
Becoming Digital
www.becomingdigital.com
- Original Message -
From: Becoming Digital [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Tuesday, 10 June, 2003 06:43
Subject: Re: [PHP-DB] Count database-values
So far no problems, but I want to count all these values.
I tried with SUM but with the following code it doesn't work:
$squery = SELECT SUM(bedrag) AS Total FROM finance WHERE
posneg = 'af';
any suggestions how to do? I want to store it in a
variabele ($total)
so I can echo it. (E.g.: $total = 93.63 in this case)
If you want the total, you need to do something like the following:
$query = SELECT SUM(bedrag) AS Total FROM finance WHERE
posneg = 'af';
$result = mysql_query( $query );
$total = mysql_fetch_array( $result );
$total = $total['Total'];
Cheers!
Mike
-
Mike Ford, Electronic Information Services Adviser,
Learning Support Services, Learning Information Services,
JG125, James Graham Building, Leeds Metropolitan University,
Beckett Park, LEEDS, LS6 3QS, United Kingdom
Email: [EMAIL PROTECTED]
Tel: +44 113 283 2600 extn 4730 Fax: +44 113 283 3211
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
--
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To unsubscribe, visit: http://www.php.net/unsub.php
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Re: [PHP-DB] Count database-values
I love when things go completely over my head for no apparent reason. I have an
incredible talent for missing the obvious.
Edward Dudlik
Becoming Digital
www.becomingdigital.com
- Original Message -
From: Hutchins, Richard [EMAIL PROTECTED]
To: 'Becoming Digital' [EMAIL PROTECTED]; [EMAIL PROTECTED]
Sent: Tuesday, 10 June, 2003 13:14
Subject: RE: [PHP-DB] Count database-values
Edward,
Have you considered trying mysql_result()?
It appears that your query is going to always return a single piece of data:
Total. So you essentially have an array with only a single element. If you
did this:
$total = mysql_result($result,0);
You'll get the value of the item at index 0 (the first position in an array)
of the $result array stored in the $total variable.
Details are here:
http://us3.php.net/manual/en/function.mysql-result.php
The page also makes reference to other high-performance options, but since
you're only grabbing a single item and if you use the index number instead
of the column name as the documentation recommends, the performance should
be just fine.
Hope this helps.
Rich
-Original Message-
From: Becoming Digital [mailto:[EMAIL PROTECTED]
Sent: Tuesday, June 10, 2003 12:50 PM
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Count database-values
Unfortunately, I can't, despite my best efforts, get the data
to display unless
it's put inside a loop. If anyone can tell me how to, I'd
just for joy.
Edward Dudlik
Becoming Digital
www.becomingdigital.com
- Original Message -
From: Ford, Mike [LSS] [EMAIL PROTECTED]
To: 'Becoming Digital' [EMAIL PROTECTED];
[EMAIL PROTECTED]
Sent: Tuesday, 10 June, 2003 11:54
Subject: RE: [PHP-DB] Count database-values
-Original Message-
From: Becoming Digital [mailto:[EMAIL PROTECTED]
Sent: 10 June 2003 12:15
I screwed up my own code. Silly me. It should read:
Surely your first attempt is the right one? There's only
ever going to be 1
Total, so why waste a while loop trying to read more than one
result row?
$query = SELECT SUM(bedrag) AS Total FROM finance WHERE
posneg = 'af';
$result = mysql_query( $query );
while ( $total = mysql_fetch_array( $result ) )
{
$total = $total['Total'];
}
Edward Dudlik
Becoming Digital
www.becomingdigital.com
- Original Message -
From: Becoming Digital [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Tuesday, 10 June, 2003 06:43
Subject: Re: [PHP-DB] Count database-values
So far no problems, but I want to count all these values.
I tried with SUM but with the following code it doesn't work:
$squery = SELECT SUM(bedrag) AS Total FROM finance WHERE
posneg = 'af';
any suggestions how to do? I want to store it in a
variabele ($total)
so I can echo it. (E.g.: $total = 93.63 in this case)
If you want the total, you need to do something like the following:
$query = SELECT SUM(bedrag) AS Total FROM finance WHERE
posneg = 'af';
$result = mysql_query( $query );
$total = mysql_fetch_array( $result );
$total = $total['Total'];
Cheers!
Mike
-
Mike Ford, Electronic Information Services Adviser,
Learning Support Services, Learning Information Services,
JG125, James Graham Building, Leeds Metropolitan University,
Beckett Park, LEEDS, LS6 3QS, United Kingdom
Email: [EMAIL PROTECTED]
Tel: +44 113 283 2600 extn 4730 Fax: +44 113 283 3211
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
--
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