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correct sql argument

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Oct 20.2004

Hello!!

I am using the sql argument:

$total_results = mysql_result(mysql_query(“SELECT COUNT(*) as Num FROM reply”),0);

but I need the following :

$total_results = mysql_result(mysql_query(“SELECT COUNT(*) as Num FROM reply WHERE MEMBER_ID=’$MEMBER_ID'”),0);

It says is not a valid argument. How can I get a spefic number of results for this?

thanks for any help

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PHP

12 Comments(s) ↴

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@JonaOct 20.2004 — [font=trebuchet ms]Specific [i]number of results[/i] or do you just want the second query to [i]work[/i]?[/font]
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@sandro27authorOct 20.2004 — Hi

I want the second query to work.

You see, I am getting the total number of results in the first argument but I need to get only the results for a specific member.

Cheers
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@JonaOct 20.2004 — [font=trebuchet ms]Are you sure you have all the field names correct and $MEMBER_ID is defined? What is the exact error you get from MySQL? Try...[/font]

[code=php]
$total_results = mysql_result(mysql_query("SELECT COUNT(*) as Num FROM reply WHERE MEMBER_ID = '$MEMBER_ID'"),0);
[/code]
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@sandro27authorOct 20.2004 — The error says:

Warning: mysql_result(): supplied argument is not a valid MySQL result resource in c:sitesmembersmember_account.php on line 36

That´s strange, because When I do $total_results = mysql_result(mysql_query("SELECT COUNT(*) as Num FROM reply"),0); it works

Sandro27
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@JonaOct 20.2004 — [font=trebuchet ms]Are you sure $MEMBER_ID is defined? Tell me if this works, even if it isn't what you want.[/font]

[code=php]
$total_results = mysql_result(mysql_query("SELECT * FROM reply WHERE MEMBER_ID = '$MEMBER_ID'"),0);
[/code]
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@sandro27authorOct 20.2004 — It doesn´t work. The whole script as follows:

<?

include ("intro.php");

$id_member=$_GET['member_id'];

$sql2="select * from members where member_id='$id_member'";

$total_results = mysql_result(mysql_query("SELECT COUNT(*) as Num FROM reply"),0);

$busca=mysql_query($sql2,$conexao) or die(mysql_error());

while($lista=mysql_fetch_array($busca)){

?>

}

This script is working but not when I try:

$total_results = mysql_result(mysql_query("SELECT COUNT(*) as Num FROM reply WHERE member_id='$id_member'"),0);
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@JonaOct 20.2004 — [font=trebuchet ms]Maybe PHP is being fussy about the quotes.[/font]

[code=php]
$total_results = mysql_result(mysql_query("SELECT COUNT(*) as Num FROM reply WHERE MEMBER_ID='".$MEMBER_ID."'"),0);
[/code]
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@sandro27authorOct 20.2004 — Dear Jona,

Thanks for you help. I have tested the argument on mysql front and it showed me where I was wrong. The columm member_id has another name!! That´s why I coudn´t get the results I wanted.

I am new with this stuff so sometimes I oversee silly mistakes.

Cheers
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@JonaOct 20.2004 — [font=trebuchet ms]lol. Well, thanks for wasting my time. ? Anyway, I'm glad you got it fixed. ? [/font]
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@sandro27authorOct 20.2004 — I am sorry for your wasted time. I can send you a post card if you want.
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@JonaOct 20.2004 — [i]Originally posted by sandro27 [/i]

[B]I am sorry for your wasted time. I can send you a post card if you want. [/B][/QUOTE]

[font=trebuchet ms]Haha, it's no problem, really.[/font]
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@sandro27authorOct 20.2004 — uffff.(sigh of relief)
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