"Marcus Bointon" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
> According to the docs, MySQL has a particular optimisation that means
> that it should be:
>
> $sql = "SELECT COUNT(*) FROM tbvara WHERE Varunamn LIKE '$checkLev%'";
>
It is not just a MySql optimization but this ho
On 7 Aug 2005, at 05:04, TalkativeDoggy wrote:
be coz this way is slow and costs more overload.
$sql = "SELECT COUNT(IDVara) cn FROM tbvara WHERE Varunamn LIKE
'$checkLev%'";
$querys = mysql_query($sql);
//Count products in db
//
$dbArray = mysql_fetch_row($querys);
$nrOfP
27;re right, I can add an index, but Is that the only way to get a
faster solution?
/mr G
@varupiraten.se
- Original Message ----- From: "Sebastian"
<[EMAIL PROTECTED]>
To: "Gustav Wiberg" <[EMAIL PROTECTED]>
Sent: Saturday, August 06, 2005 11:08 PM
Subject: R
ster solution?
/mr G
@varupiraten.se
- Original Message - From: "Sebastian"
<[EMAIL PROTECTED]>
To: "Gustav Wiberg" <[EMAIL PROTECTED]>
Sent: Saturday, August 06, 2005 11:08 PM
Subject: Re: [PHP] Fast count of recordset in php...
how many records in the t
rds.
You're right, I can add an index, but Is that the only way to get a
faster solution?
/mr G
@varupiraten.se
- Original Message - From: "Sebastian"
<[EMAIL PROTECTED]>
To: "Gustav Wiberg" <[EMAIL PROTECTED]>
Sent: Saturday, August 06, 2005 11:08
$num=mysql_num_rows($sql);
Gustav Wiberg wrote:
Hello there!
How do i get a fast count of a recordset in php?
Look at this code:
$sql = "SELECT COUNT(IDVara) cn FROM tbvara WHERE Varunamn LIKE
'$checkLev%'";
$querys = mysql_query($sql);
//Count products in db
//
$dbArray =
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