displaying the result of a COUNT query in MySQL

**Kimmo Laine** · Dec 19 '05, 07:35 PM

Re: displaying the result of a COUNT query in MySQL

<briansmccabe@g mail.com> kirjoitti
viestissä:11350 16969.513258.27 6220@g44g2000cw a.googlegroups. com...[color=blue]
> Does anyone have a good approach to displaying in PHP a simple COUNT
> query that is performed on a table in a MySQL db?
>[/color]

$result = mysql_query('SE LECT COUNT(*) AS foo FROM table') or
die(mysql_error ());
$bar = mysql_fetch_arr ay($result);
echo $bar['foo'];

--
SETI @ Home - Donate your cpu's idle time to science.
Further reading at <http://setiweb.ssl.ber keley.edu/>
Kimmo Laine <antaatulla.sik anautaa@gmail.c om.NOSPAM.inval id>

**briansmccabe@gmail.com** · Dec 19 '05, 08:05 PM

Re: displaying the result of a COUNT query in MySQL

my code:

$query6 = "SELECT COUNT (movie_id) as quant FROM movies WHERE divx = 1
AND format = 'reg'";
$result6 = mysql_fetch_arr ay($query6);

later on the page:

echo $result6[quant]

result:

Warning: mysql_fetch_arr ay(): supplied argument is not a valid MySQL
result resource in /home/tgupc/public_html/admin/divxreport.php on line
19

Any suggestions?

thanks

**Andy Hassall** · Dec 19 '05, 08:05 PM

Re: displaying the result of a COUNT query in MySQL

On 19 Dec 2005 11:50:43 -0800, briansmccabe@gm ail.com wrote:
[color=blue]
>my code:
>
>$query6 = "SELECT COUNT (movie_id) as quant FROM movies WHERE divx = 1
>AND format = 'reg'";
>$result6 = mysql_fetch_arr ay($query6);
>
>later on the page:
>
>echo $result6[quant]
>
>result:
>
>Warning: mysql_fetch_arr ay(): supplied argument is not a valid MySQL
>result resource in /home/tgupc/public_html/admin/divxreport.php on line
>19[/color]

Kimmo posted code with basic error handling, which you've removed. Put it back
in again and it'll tell you why it failed.

--
Andy Hassall :: andy@andyh.co.u k :: http://www.andyh.co.uk
http://www.andyhsoftware.co.uk/space :: disk and FTP usage analysis tool

**briansmccabe@gmail.com** · Dec 19 '05, 08:15 PM

Re: displaying the result of a COUNT query in MySQL

I put it back in, and the exact same result occurs.

FYI, line 19 is as follows:

$result6 = mysql_fetch_arr ay($query6);

**Andy Hassall** · Dec 19 '05, 08:25 PM

Re: displaying the result of a COUNT query in MySQL

On 19 Dec 2005 12:01:34 -0800, briansmccabe@gm ail.com wrote:
[color=blue]
>[/color]

Please quote some context when replying. The default "reply" at the bottom in
Google Groups cuts out all previous text; this is not the accepted way to post
to Usenet. Click "show options" next to the author/date for the message, then
use the "Reply" option there; this quotes and attributes the previous message.
Then follow:

Error 404 (Not Found)!!1

http://groups.google.com/googlegroups/posting_style.html#summarize

... i.e. don't just quote the lot, unless the message is small and the whole
message is relevant to your reply.
[color=blue]
>I put it back in, and the exact same result occurs.
>
>FYI, line 19 is as follows:
>
>$result6 = mysql_fetch_arr ay($query6);[/color]

It can't have got here and produced the same error, if you put in the error
handling Kimmo posted.

Post your revised code for the lines between mysql_query and
mysql_fetch_arr ay.

--
Andy Hassall :: andy@andyh.co.u k :: http://www.andyh.co.uk
http://www.andyhsoftware.co.uk/space :: disk and FTP usage analysis tool

**briansmccabe@gmail.com** · Dec 19 '05, 08:35 PM

Re: displaying the result of a COUNT query in MySQL

$query6 = "SELECT COUNT (movie_id) as quant FROM movies WHERE divx = 1
AND format = 'reg'" or die(mysql_error ());
$result6 = mysql_fetch_arr ay($query6);

later on the page:

echo $result6[quant]

result:

Warning: mysql_fetch_arr ay(): supplied argument is not a valid MySQL
result resource in /home/tgupc/public_html/admin/divxreport.php on line
19

see for yourself: http://www.tgupc.com/admin/divxreport.php

Andy Hassall wrote:[color=blue]
> On 19 Dec 2005 12:01:34 -0800, briansmccabe@gm ail.com wrote:
> It can't have got here and produced the same error, if you put in the error
> handling Kimmo posted.
>
> Post your revised code for the lines between mysql_query and
> mysql_fetch_arr ay.[/color]

**Andy Hassall** · Dec 19 '05, 08:55 PM

Re: displaying the result of a COUNT query in MySQL

On 19 Dec 2005 12:24:35 -0800, briansmccabe@gm ail.com wrote:

Kimmo originally wrote:
[color=blue][color=green]
>>$result = mysql_query('SE LECT COUNT(*) AS foo FROM table') or
>>die(mysql_err or());[/color][/color]

But you've used:
[color=blue]
>$query6 = "SELECT COUNT (movie_id) as quant FROM movies WHERE divx = 1
>AND format = 'reg'" or die(mysql_error ());[/color]

You are missing the call to mysql_query().
[color=blue]
>$result6 = mysql_fetch_arr ay($query6);[/color]

... so $query6 just contains the SQL string, and not a MySQL result set
resource identifier. You're not actually executing the SQL anywhere, or if you
are in the rest of the code, you haven't done the error checking there.
[color=blue]
>later on the page:
>
>echo $result6[quant][/color]

This should be: $result6['quant'].

See:

PHP: Arrays - Manual

http://www.php.net/manual/en/language.types.array.php#language.types.array.donts

Arrays

p.s. You've got the posting style nearly right - however you should put your
new message _under_ the old one, not above, so the whole message makes sense
read on its own. You have "top posted".

--
Andy Hassall :: andy@andyh.co.u k :: http://www.andyh.co.uk
http://www.andyhsoftware.co.uk/space :: disk and FTP usage analysis tool

**briansmccabe@gmail.com** · Dec 19 '05, 10:25 PM

Re: displaying the result of a COUNT query in MySQL

Andy Hassall wrote:[color=blue]
> On 19 Dec 2005 12:24:35 -0800, briansmccabe@gm ail.com wrote:
>
> Kimmo originally wrote:
>[color=green][color=darkred]
> >>$result = mysql_query('SE LECT COUNT(*) AS foo FROM table') or
> >>die(mysql_err or());[/color][/color]
>
> But you've used:
>[color=green]
> >$query6 = "SELECT COUNT (movie_id) as quant FROM movies WHERE divx = 1
> >AND format = 'reg'" or die(mysql_error ());[/color]
>
> You are missing the call to mysql_query().
>[color=green]
> >$result6 = mysql_fetch_arr ay($query6);[/color]
>
> ... so $query6 just contains the SQL string, and not a MySQL result set
> resource identifier. You're not actually executing the SQL anywhere, or if you
> are in the rest of the code, you haven't done the error checking there.
>[color=green]
> >later on the page:
> >
> >echo $result6[quant][/color]
>
> This should be: $result6['quant'].
>
> See:
>
> http://www.php.net/manual/en/languag...es.array.donts
>
> p.s. You've got the posting style nearly right - however you should put your
> new message _under_ the old one, not above, so the whole message makes sense
> read on its own. You have "top posted".
>
> --
> Andy Hassall :: andy@andyh.co.u k :: http://www.andyh.co.uk
> http://www.andyhsoftware.co.uk/space :: disk and FTP usage analysis tool[/color]

**briansmccabe@gmail.com** · Dec 19 '05, 10:25 PM

Re: displaying the result of a COUNT query in MySQL

> ... so $query6 just contains the SQL string, and not a MySQL result set[color=blue]
> resource identifier. You're not actually executing the SQL anywhere, or if you
> are in the rest of the code, you haven't done the error checking there.[/color]

I have this part working now, thanks to your help.
[color=blue]
>[color=green]
> >later on the page:
> >
> >echo $result6[quant][/color]
>
> This should be: $result6['quant'].
>
> See:
>
> http://www.php.net/manual/en/languag...es.array.donts
>[/color]

So if I am echoing the result in between strings of HTML, I am having a
hard time knowing / remembering how to "escape" the single-quote marks
inside the result brackets. Does that make sense?

Thanks

Brian

**briansmccabe@gmail.com** · Dec 19 '05, 10:25 PM

Re: displaying the result of a COUNT query in MySQL

>[color=blue]
> ... so $query6 just contains the SQL string, and not a MySQL result set
> resource identifier. You're not actually executing the SQL anywhere, or if you
> are in the rest of the code, you haven't done the error checking there.[/color]

I have this part working now. Thanks for your help.
[color=blue][color=green]
> >later on the page:
> >
> >echo $result6[quant][/color]
>
> This should be: $result6['quant'].
>
> See:
>
> http://www.php.net/manual/en/languag...es.array.donts[/color]

I have tried to figure out how to single-quote the contents of the []
but when it is concatenated in between strings of HTML, it goofs things
up. I cannot remember how to do that. I tried a few things I noticed in
the page you link to above, but I am not getting very far.

**Andy Hassall** · Dec 19 '05, 11:05 PM

Re: displaying the result of a COUNT query in MySQL

On 19 Dec 2005 14:08:15 -0800, briansmccabe@gm ail.com wrote:
[color=blue][color=green]
>> This should be: $result6['quant'].
>>
>> See:
>>
>> http://www.php.net/manual/en/languag...es.array.donts[/color]
>
> I have tried to figure out how to single-quote the contents of the []
>but when it is concatenated in between strings of HTML, it goofs things
>up. I cannot remember how to do that. I tried a few things I noticed in
>the page you link to above, but I am not getting very far.[/color]

Use the "curly brace" format within a string:

echo "{$result6['quant']}";

PHP: Strings - Manual

http://www.php.net/manual/en/language.types.string.php#language.types.string.parsing.complex

Strings

--
Andy Hassall :: andy@andyh.co.u k :: http://www.andyh.co.uk
http://www.andyhsoftware.co.uk/space :: disk and FTP usage analysis tool

**william.clarke@gmail.com** · Dec 20 '05, 10:05 PM

Re: displaying the result of a COUNT query in MySQL

briansmcc try your SQL like this...

$result6 = mysql_query('SE LECT COUNT (movie_id) as quant FROM movies
WHERE divx = 1 AND format = 'reg') or die(mysql_error ());
$movie_count = mysql_fetch_arr ay($result6 );

later on the page:
echo $movie_count['quant'];

displaying the result of a COUNT query in MySQL

displaying the result of a COUNT query in MySQL

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