[PHP-DB] mysql_connect(): Lost connection to MySQL server during query

2008-08-11 Thread Manoj Singh
Hello all,
I am getting this error when connecting to the database through
mysql_connect function.

Error: mysql_connect() [function.mysql-connect]: Lost connection
to MySQL server during query

I am getting this error when the db server has the high load.

Please help me out to fix this error.

Best Regards,
Manoj Kumar SIngh


Re: [PHP-DB] Pull Down Menu with ODBC query

2008-08-11 Thread kitfox69
Thanks everyone for your help so far.

The last two responses I got were great and worked. However I stuck with my 
original script with  the small $row fix presented earlier and it works great.

The only problem I see with the last two are that there is no scripting to push 
to the next page like I have set.

Saying that here is what I currently have:

";
echo  "";
while($row = @odbc_fetch_array($result))
{
echo  "$row[sls_his_cust_id]";
}
echo  "";
?>

I need to know how to name the selection so I can push it to the next page 
using the submit button.

The next page so far is as follows:



I need to be able to pass the selected CUSTID on to this page so I can display 
all the data pertinent to that customer... also I would like to be able to use 
the billdate that was present earlier but am not sure how to push it through 
again...

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Re: [PHP-DB] Pull Down Menu with ODBC query

2008-08-11 Thread Dan Shirah
Try something like this:


   
   --SELECT--\n";
 foreach ($menu as $m)
 {
   if ($m['sls_his_cust_id'] == $_POST['menu'])
  echo "{$m['sls_his_cust_id']}\n";
   else
  echo "{$m['sls_his_cust_id']}\n";
 }
 ?>
   
   


RE: [PHP-DB] Pull Down Menu with ODBC query

2008-08-11 Thread kitfox69
BTW I want to make sure I am doing this correctly...

with that current script (edited below with your fix) will I be able to pass 
that selection on into a further PHP page? I do not see how without assigning a 
variable to it and passing it with POST... how would I go about this?
 -- Original message --
From: "Simcha" <[EMAIL PROTECTED]>
> 
> odbc_fetch_array() returns an associative array, so will you need to use
> $row[' sls_his_cust_id'] instead of $row[0].
> 
>  
> 
> -Original Message-
> From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] 
> Sent: Monday, August 11, 2008 10:14 PM
> To: php-db@lists.php.net
> Subject: [PHP-DB] Pull Down Menu with ODBC query
> 
> I am having trouble getting this script to populate the query data into a
> pull down menu.
> 
> The SQL script works fine and displays the data with odbc_result_all but
> will not put it in the menu.
> 
> Script is as follows:
> 
>  $conn = odbc_connect("HOMES", "", "");
> $billdate = $_POST[ 'billdate' ];
> $query = ("SELECT DISTINCT sls_his_cust_id FROM sls_his where
> sls_his_prchdat_alt = $billdate");
> $result = odbc_exec($conn, $query);
> 
> echo  "";
> echo  "";
> while($row = @odbc_fetch_array($result))
> {
> echo  " VALUE=\"$row[sls_his_cust_id]\">$row[sls_his_cust_id]";
> }
> echo  " Results\">";
> ?>
> 
> Any ideas?
> 
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> PHP Database Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
> 
> No virus found in this incoming message.
> Checked by AVG - http://www.avg.com 
> Version: 8.0.138 / Virus Database: 270.6.0/1604 - Release Date: 11/08/2008
> 05:50
> 


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[PHP-DB] Pull Down Menu with ODBC query

2008-08-11 Thread kitfox69
I am having trouble getting this script to populate the query data into a pull 
down menu.

The SQL script works fine and displays the data with odbc_result_all but will 
not put it in the menu.

Script is as follows:

";
echo  "";
while($row = @odbc_fetch_array($result))
{
echo  "$row[0]";
}
echo  "";
?>

Any ideas?

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Re: [PHP-DB] Help to improve MySQL query

2008-08-11 Thread Dee Ayy
On Sat, Aug 9, 2008 at 1:32 AM, Niel Archer <[EMAIL PROTECTED]> wrote:
>  Hi
>
> You do not say how you identify the last call (there is no date/time
> field for example), so a complete answer is not really possible

With the "id (auto incremented int)", the last record of either table
would be the largest id in a particular group.


> Do not use "NOT LIKE 'Completed'", it's an inefficient way of doing "!=
> 'Completed'"

I'll modify that as you said.

My real concern is that $theHugeListOfIncidentIds keeps growing and is
used in the "Calls.id IN ($theHugeListOfIncidentIds)".

Even if the $theHugeListOfIncidentIds was replaced with an actual
MySQL query instead of first being processed by PHP, I think that is a
bad approach (but maybe that is the solution if I only change this
query and do not change/add columns or states).

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Re: [PHP-DB] Help to improve MySQL query

2008-08-11 Thread Micah Gersten
Use an appropriate status.

Thank you,
Micah Gersten
onShore Networks
Internal Developer
http://www.onshore.com



Dee Ayy wrote:
> On Fri, Aug 8, 2008 at 5:25 PM, Micah Gersten <[EMAIL PROTECTED]> wrote:
>   
>> How about "select Incidents.* from Incidents inner join Calls on
>> Incidents.id=Calls.incidentid where Calls.status='Open'"?
>> 
> ...
>   
>> Dee Ayy wrote:
>> 
> ...
>   
>>> The status column never has the text "Open".
>>>   
> ...
>   

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Re: [PHP-DB] Help to improve MySQL query

2008-08-11 Thread Dee Ayy
On Fri, Aug 8, 2008 at 5:25 PM, Micah Gersten <[EMAIL PROTECTED]> wrote:
> How about "select Incidents.* from Incidents inner join Calls on
> Incidents.id=Calls.incidentid where Calls.status='Open'"?
...
>
> Dee Ayy wrote:
...
>> The status column never has the text "Open".
...

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