Re: [PHP] helping people...
No that wasnt a ddos threat you idiot, i dont play them games. And when you keep sending spam is when it starts to piss people off. - Original Message - From: Jochem Maas [EMAIL PROTECTED] To: [php] PHP General List php-general@lists.php.net Sent: Wednesday, June 21, 2006 1:55 AM Subject: [PHP] helping people... helping some people will get you no end of trouble. and so it seems as though I'm going to be DoSSed by someone who uses Outlook Express as their mail client. I guess it's monday somewhere. Original Message From: - Wed Jun 21 01:47:39 2006 X-Mozilla-Status: 0001 X-Mozilla-Status2: Return-Path: [EMAIL PROTECTED] X-Original-To: [EMAIL PROTECTED] Delivered-To: [EMAIL PROTECTED] Received: from localhost (localhost [127.0.0.1]) by mx1.moulin.nl (Postfix) with ESMTP id EECE119EB29 for [EMAIL PROTECTED]; Wed, 21 Jun 2006 01:47:00 +0200 (CEST) Received: from mx1.moulin.nl ([127.0.0.1]) by localhost (mx1.moulin.nl [127.0.0.1]) (amavisd-new, port 10024) with ESMTP id 01046-16 for [EMAIL PROTECTED]; Wed, 21 Jun 2006 01:46:57 +0200 (CEST) Received: from mail.fiberuplink.com (newyork.hardlink.com [140.186.181.161]) by mx1.moulin.nl (Postfix) with SMTP id A4FBB19EAF0 for [EMAIL PROTECTED]; Wed, 21 Jun 2006 01:46:56 +0200 (CEST) Received: (qmail 86220 invoked by uid 1013); 20 Jun 2006 23:46:57 - Received: from 208.107.101.135 by eclipse.fiberuplink.com (envelope-from [EMAIL PROTECTED], uid 1011) with qmail-scanner-1.25-st-qms (clamdscan: 0.87/1102. spamassassin: 3.0.1. perlscan: 1.25-st-qms. Clear:RC:0(208.107.101.135):SA:0(-1.9/4.5):. Processed in 3.519899 secs); 20 Jun 2006 23:46:57 - X-Antivirus-INetKing-Mail-From: [EMAIL PROTECTED] via eclipse.fiberuplink.com X-Antivirus-INetKing: 1.25-st-qms (Clear:RC:0(208.107.101.135):SA:0(-1.9/4.5):. Processed in 3.519899 secs Process 86212) Received: from host-135-101-107-208.midco.net (HELO rob) ([EMAIL PROTECTED]@208.107.101.135) by mail.fiberuplink.com with SMTP; 20 Jun 2006 23:46:52 - Message-ID: [EMAIL PROTECTED] From: Rob W. [EMAIL PROTECTED] To: [EMAIL PROTECTED] Subject: Date: Tue, 20 Jun 2006 19:18:02 -0500 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary==_NextPart_000_003E_01C6949E.403D9880 X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 6.00.2900.2869 X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.2869 X-Virus-Scanned: amavisd-new at moulin.nl Unless you wanna keep your server up, I suggest you quit sending me shit. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Still trying to figure this out...
I still have not yet found anybody to help me with this. All the code that
people have given me has not worked at all. I still get no output. So to add
a little more to this, I'm gonna define how the database looks.
--
|switchid|switchport |
--
|1 | 1 |
|1 | 2 |
|1 | 3 |
|1 | 4 |
|1 | 7 |
|1 | 10|
--
ect so on and so forth. There are other switch id's in there as well along
with their switch port's which sorting the other switchid's is not a
problem. I also know the maximum amount for each switchid so that's not a
problem as well. I need to be able to display the missing numbers in there.
I have no extra spaces in there for blank numbers. After a server has been
entered in to the database, it take's that number that was not shown
previously in the database and enters it in, there for displaying it and not
making it available for the next time another server is added. I have no way
to alter the database as I am trying to clone a program and this is the only
thing haning me up as of right now. If I can find someone to solve this
problem for me, I will be glad to send money via paypal for helping out
solving this problem as I have been on it for 2 days now and this is the
last bit of code for this program that is holding me up. Please read below
for more information on what I am doing.
- Original Message -
From: Rob W. [EMAIL PROTECTED]
To: php-general@lists.php.net
Sent: Monday, June 19, 2006 3:39 PM
Subject: [PHP] Still trying to figure this out...
Ok, I am still trying to get this figured out from what I tried doing last
night. If anybody wants to try it, please be my guest. This is the deal.
Create an INTEGER table. Put the numbers 1 - 24 in it. Leave out like number
8, 9, 22, 23 ect.. Now create a php statement that will take and pull that
out of the database, find out the missing numbers and display them as an
html drop down selection.
And tedd in response to why, is because the values that are already located
in the database represent in use already. They represent a server port that
a server is sitting on, so that's not the problem. I could have done that
along time ago.
Any help is appreciated but here is the current code that I have.
$query=SELECT switchport FROM network;
$result=mysql_query($query);
$sql_range=mysql_fetch_array($result);
$true_range=range(1,24);
$next_range=array_diff($true_range,$sql_range);
foreach ($next_range as $final_range) {
echo option value='$final_range'$final_range\n;
}
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Re: [PHP] Still trying to figure this out...
OMG Chris, Thankyou very much. That is exactally what I needed. If you wish
for me to pay you some amount, I will be very happy to. Please send me a
private email where I can paypal you. If not, Thank you again very much for
the help.
Sorry everyone for the hastle for those of you wernt able to help with what
I wanted.
- Rob
- Original Message -
From: Chris [EMAIL PROTECTED]
To: Rob W. [EMAIL PROTECTED]
Cc: php-general@lists.php.net
Sent: Tuesday, June 20, 2006 1:39 AM
Subject: Re: [PHP] Still trying to figure this out...
Rob W. wrote:
I still have not yet found anybody to help me with this. All the code
that people have given me has not worked at all. I still get no output.
So to add a little more to this, I'm gonna define how the database looks.
--
|switchid|switchport |
--
|1 | 1 |
|1 | 2 |
|1 | 3 |
|1 | 4 |
|1 | 7 |
|1 | 10|
--
I'm sure the code that people have posted will work if you do more than
just copy/paste it and try to understand it and match it to what you want.
$switch_ports_taken = array();
$query = select switchport from table where switchid=1;
$result = mysql_query($query);
if (!$result) die(error: . mysql_error());
while($row = mysql_fetch_assoc($result)) {
array_push($switch_ports_taken, $row['switchport']);
}
echo Got these ports: . print_r($switch_ports_taken, true) . br/;
$all_ports = range(1,24);
$unused_ports = array_diff($all_ports, $switch_ports_taken);
echo Unused ports: . print_r($unused_ports, true) . br/;
If that doesn't work tell us what doesn't work and don't make us guess
what's going on.
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Re: [PHP] Still trying to figure this out...
Ya know, after working on this for 2 days straight, I have tried my own
work. I never said I was an expert at php, that's why this list is here, so
the next time you critisize someone, keep it to your self cause no one cares
to hear about it. If you dont wanna help, then stfu.
- Original Message -
From: Jochem Maas [EMAIL PROTECTED]
To: Rob W. [EMAIL PROTECTED]
Cc: php-general@lists.php.net
Sent: Tuesday, June 20, 2006 3:54 AM
Subject: Re: [PHP] Still trying to figure this out...
Rob W. wrote:
OMG Chris, Thankyou very much. That is exactally what I needed. If you
wish for me to pay you some amount, I will be very happy to. Please send
me a private email where I can paypal you. If not, Thank you again very
much for the help.
just do us all a favor and only come back when you have grasped the
concept
of debugging/investigating a problem (any problem) and doing your *own*
work.
Sorry everyone for the hastle for those of you wernt able to help with
['wernt'? that is not a word.]
what Chris wrote is a tiny variation of code examples that have been
posted
by at least 4 different people. the fact is that you didn't bother to
study the code and just pasted into your script expecting it to work.
what I wanted.
- Rob
- Original Message - From: Chris [EMAIL PROTECTED]
To: Rob W. [EMAIL PROTECTED]
Cc: php-general@lists.php.net
Sent: Tuesday, June 20, 2006 1:39 AM
Subject: Re: [PHP] Still trying to figure this out...
Rob W. wrote:
I still have not yet found anybody to help me with this. All the code
that people have given me has not worked at all. I still get no
output. So to add a little more to this, I'm gonna define how the
database looks.
--
|switchid|switchport |
--
|1 | 1 |
|1 | 2 |
|1 | 3 |
|1 | 4 |
|1 | 7 |
|1 | 10|
--
I'm sure the code that people have posted will work if you do more
than just copy/paste it and try to understand it and match it to what
you want.
$switch_ports_taken = array();
$query = select switchport from table where switchid=1;
$result = mysql_query($query);
if (!$result) die(error: . mysql_error());
while($row = mysql_fetch_assoc($result)) {
array_push($switch_ports_taken, $row['switchport']);
}
echo Got these ports: . print_r($switch_ports_taken, true) . br/;
$all_ports = range(1,24);
$unused_ports = array_diff($all_ports, $switch_ports_taken);
echo Unused ports: . print_r($unused_ports, true) . br/;
If that doesn't work tell us what doesn't work and don't make us guess
what's going on.
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Re: [PHP] Still trying to figure this out...
Yes, I do have it setup right to display my errors, the reason why I wasnt getting any input is because when the array was matched to range(1,24) and the other array had nothing in it, I was either getting a list of numbers 1 though 24 or nothing at all. - Rob - Original Message - From: Stut [EMAIL PROTECTED] To: Rob W. [EMAIL PROTECTED] Cc: php-general@lists.php.net Sent: Tuesday, June 20, 2006 4:08 AM Subject: Re: [PHP] Still trying to figure this out... Rob W. wrote: Ya know, after working on this for 2 days straight, I have tried my own work. I never said I was an expert at php, that's why this list is here, so the next time you critisize someone, keep it to your self cause no one cares to hear about it. If you dont wanna help, then stfu. Just one more suggestion before we all STFU and leave you to work it out for yourself... you do have PHP set up to display errors right? You've said several times that you get no output which leads me to believe you're trying to code with one hand tied behind your back and a red-hot poker in each eye. -Stut -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Still trying to figure this out...
- Original Message -
From: Jochem Maas [EMAIL PROTECTED]
To: Rob W. [EMAIL PROTECTED]
Cc: php-general@lists.php.net
Sent: Tuesday, June 20, 2006 4:13 AM
Subject: Re: [PHP] Still trying to figure this out...
Rob W. wrote:
Ya know, after working on this for 2 days straight, I have tried my own
so what.
Why the heck you bitching then?
work. I never said I was an expert at php,
neither did I.
Dont complain about people asking for help when that is what this list is
for.
that's why this list is here,
really, could you please quote your support contract number.
If it's not, dont reply then idiot.
so the next time you critisize someone, keep it to your self cause no
one cares to hear about it. If you dont wanna help, then stfu.
your right about not wanting to help you - you silly little toad.
Good, ignore my posts and shut the hell up about it, no one want's your
opinion.
- Original Message - From: Jochem Maas [EMAIL PROTECTED]
To: Rob W. [EMAIL PROTECTED]
Cc: php-general@lists.php.net
Sent: Tuesday, June 20, 2006 3:54 AM
Subject: Re: [PHP] Still trying to figure this out...
Rob W. wrote:
OMG Chris, Thankyou very much. That is exactally what I needed. If you
wish for me to pay you some amount, I will be very happy to. Please
send
me a private email where I can paypal you. If not, Thank you again very
much for the help.
just do us all a favor and only come back when you have grasped the
concept
of debugging/investigating a problem (any problem) and doing your
*own* work.
Sorry everyone for the hastle for those of you wernt able to help with
['wernt'? that is not a word.]
what Chris wrote is a tiny variation of code examples that have been
posted
by at least 4 different people. the fact is that you didn't bother to
study the code and just pasted into your script expecting it to work.
what I wanted.
- Rob
- Original Message - From: Chris [EMAIL PROTECTED]
To: Rob W. [EMAIL PROTECTED]
Cc: php-general@lists.php.net
Sent: Tuesday, June 20, 2006 1:39 AM
Subject: Re: [PHP] Still trying to figure this out...
Rob W. wrote:
I still have not yet found anybody to help me with this. All the code
that people have given me has not worked at all. I still get no
output. So to add a little more to this, I'm gonna define how the
database looks.
--
|switchid|switchport |
--
|1 | 1 |
|1 | 2 |
|1 | 3 |
|1 | 4 |
|1 | 7 |
|1 | 10|
--
I'm sure the code that people have posted will work if you do more
than just copy/paste it and try to understand it and match it to what
you want.
$switch_ports_taken = array();
$query = select switchport from table where switchid=1;
$result = mysql_query($query);
if (!$result) die(error: . mysql_error());
while($row = mysql_fetch_assoc($result)) {
array_push($switch_ports_taken, $row['switchport']);
}
echo Got these ports: . print_r($switch_ports_taken, true) .
br/;
$all_ports = range(1,24);
$unused_ports = array_diff($all_ports, $switch_ports_taken);
echo Unused ports: . print_r($unused_ports, true) . br/;
If that doesn't work tell us what doesn't work and don't make us guess
what's going on.
--
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[PHP] Array
$query=SELECT switchport FROM network; $result=mysql_query($query); $sql_range=array($result['switchport']); Anybody tell me what i'm doing wrong and why this isnt going in to an array?
[PHP] Still trying to figure this out...
Ok, I am still trying to get this figured out from what I tried doing last
night. If anybody wants to try it, please be my guest. This is the deal. Create
an INTEGER table. Put the numbers 1 - 24 in it. Leave out like number 8, 9, 22,
23 ect.. Now create a php statement that will take and pull that out of the
database, find out the missing numbers and display them as an html drop down
selection.
And tedd in response to why, is because the values that are already located in
the database represent in use already. They represent a server port that a
server is sitting on, so that's not the problem. I could have done that along
time ago.
Any help is appreciated but here is the current code that I have.
$query=SELECT switchport FROM network;
$result=mysql_query($query);
$sql_range=mysql_fetch_array($result);
$true_range=range(1,24);
$next_range=array_diff($true_range,$sql_range);
foreach ($next_range as $final_range) {
echo option value='$final_range'$final_range\n;
}
Re: [PHP] Still trying to figure this out...
This also didnt work, Still giving me no output just like the last try.
- Original Message -
From: Xavier Casto
To: Rob W.
Cc: php-general@lists.php.net
Sent: Monday, June 19, 2006 4:33 PM
Subject: Re: [PHP] Still trying to figure this out...
Have you ever thought of doing something like this:
for($i = 0; $i = $maxport; $i++)
{
$query=SELECT switchport FROM network WHERE switchport = .$i;
$result=mysql_query($query);
if(empty($result))
{
echo p$i/p;
}
}
It should only return the ones that aren't in the table and they will
automatically be sorted in ascending order.
On 6/19/06, Rob W. [EMAIL PROTECTED] wrote:
Ok, I am still trying to get this figured out from what I tried doing last
night. If anybody wants to try it, please be my guest. This is the deal. Create
an INTEGER table. Put the numbers 1 - 24 in it. Leave out like number 8, 9, 22,
23 ect.. Now create a php statement that will take and pull that out of the
database, find out the missing numbers and display them as an html drop down
selection.
And tedd in response to why, is because the values that are already located
in the database represent in use already. They represent a server port that a
server is sitting on, so that's not the problem. I could have done that along
time ago.
Any help is appreciated but here is the current code that I have.
$query=SELECT switchport FROM network;
$result=mysql_query($query);
$sql_range=mysql_fetch_array($result);
$true_range=range(1,24);
$next_range=array_diff($true_range,$sql_range);
foreach ($next_range as $final_range) {
echo option value='$final_range'$final_range\n;
}
Re: [PHP] Still trying to figure this out...
Yep, and still nothing output's.
- Original Message -
From: Xavier Casto [EMAIL PROTECTED]
To: 'Rob W.' [EMAIL PROTECTED]
Cc: php-general@lists.php.net
Sent: Monday, June 19, 2006 6:20 PM
Subject: RE: [PHP] Still trying to figure this out...
You did make sure to assign a numerical value to $maxport?
-Original Message-
From: Rob W. [mailto:[EMAIL PROTECTED]
Sent: Monday, June 19, 2006 6:15 PM
To: php-general@lists.php.net
Subject: Re: [PHP] Still trying to figure this out...
This also didnt work, Still giving me no output just like the last try.
- Original Message -
From: Xavier Casto
To: Rob W.
Cc: php-general@lists.php.net
Sent: Monday, June 19, 2006 4:33 PM
Subject: Re: [PHP] Still trying to figure this out...
Have you ever thought of doing something like this:
for($i = 0; $i = $maxport; $i++)
{
$query=SELECT switchport FROM network WHERE switchport = .$i;
$result=mysql_query($query);
if(empty($result))
{
echo p$i/p;
}
}
It should only return the ones that aren't in the table and they will
automatically be sorted in ascending order.
On 6/19/06, Rob W. [EMAIL PROTECTED] wrote:
Ok, I am still trying to get this figured out from what I tried doing
last night. If anybody wants to try it, please be my guest. This is the
deal. Create an INTEGER table. Put the numbers 1 - 24 in it. Leave out
like
number 8, 9, 22, 23 ect.. Now create a php statement that will take and
pull
that out of the database, find out the missing numbers and display them as
an html drop down selection.
And tedd in response to why, is because the values that are already
located in the database represent in use already. They represent a server
port that a server is sitting on, so that's not the problem. I could have
done that along time ago.
Any help is appreciated but here is the current code that I have.
$query=SELECT switchport FROM network;
$result=mysql_query($query);
$sql_range=mysql_fetch_array($result);
$true_range=range(1,24);
$next_range=array_diff($true_range,$sql_range);
foreach ($next_range as $final_range) {
echo option value='$final_range'$final_range\n;
}
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Re: [PHP] Still trying to figure this out...
Nope, I still dont get anything back from that.
So i'm still stuck in the mud.
- Original Message -
From: Jochem Maas [EMAIL PROTECTED]
To: Rob W. [EMAIL PROTECTED]
Cc: php-general@lists.php.net
Sent: Monday, June 19, 2006 4:05 PM
Subject: Re: [PHP] Still trying to figure this out...
Rob W. wrote:
Ok, I am still trying to get this figured out from what I tried doing
last night. If anybody wants to try it, please be my guest. This is the
deal.
Create an INTEGER table. Put the numbers 1 - 24 in it.
Leave out like number 8, 9, 22, 23 ect..
I'll assume the DB creation and filling is something you have already
covered/dealt with.
Now create a php statement that will take and pull that out of the
database,
this you have already (as per your example code).
find out the missing numbers and display them as an html drop down
selection.
And tedd in response to why, is because the values that are already
located in the
database represent in use already. They represent a server port that a
server is sitting on
, so that's not the problem. I could have done that along time ago.
huh? whats not a problem, what could have done a long time ago? (and why
didn't you?)
Any help is appreciated but here is the current code that I have.
try the apples below
$query=SELECT switchport FROM network;
$result=mysql_query($query);
$sql_range=mysql_fetch_array($result);
$true_range=range(1,24);
$next_range=array_diff($true_range,$sql_range);
foreach ($next_range as $final_range) {
echo option value='$final_range'$final_range\n;
}
$sqlRange = array();
if ($res = mysql_query('SELECT switchport FROM network')) {
while ($row = mysql_fetch_array($res)) $sqlRange[] = $row['switchport'];
}
echo 'select name=port';
foreach (array_diff(range(1,24), $sqlRange) as $port) {
echo 'option value=',$port,'',$port,'/option';
}
echo '/select';
/*
NOTE:
1. you have to loop your mysql result to get each port that is already
used in the DB into you array
2. try to output something that resembles proper HTML (an option without
a closing tag sucks)
*/
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Re: [PHP] Still trying to figure this out...
Yeah, that's what i'm trying to print is the missing numbers in the db.
That's what I've been trying to figure out this whole time.
- Original Message -
From: Xavier Casto [EMAIL PROTECTED]
To: 'Rob W.' [EMAIL PROTECTED]
Cc: php-general@lists.php.net
Sent: Monday, June 19, 2006 6:34 PM
Subject: RE: [PHP] Still trying to figure this out...
Are there any numbers missing in your database table between 0 and the
number you assigned to $maxport?
-Original Message-
From: Rob W. [mailto:[EMAIL PROTECTED]
Sent: Monday, June 19, 2006 6:26 PM
To: php-general@lists.php.net
Subject: Re: [PHP] Still trying to figure this out...
Yep, and still nothing output's.
- Original Message -
From: Xavier Casto [EMAIL PROTECTED]
To: 'Rob W.' [EMAIL PROTECTED]
Cc: php-general@lists.php.net
Sent: Monday, June 19, 2006 6:20 PM
Subject: RE: [PHP] Still trying to figure this out...
You did make sure to assign a numerical value to $maxport?
-Original Message-
From: Rob W. [mailto:[EMAIL PROTECTED]
Sent: Monday, June 19, 2006 6:15 PM
To: php-general@lists.php.net
Subject: Re: [PHP] Still trying to figure this out...
This also didnt work, Still giving me no output just like the last try.
- Original Message -
From: Xavier Casto
To: Rob W.
Cc: php-general@lists.php.net
Sent: Monday, June 19, 2006 4:33 PM
Subject: Re: [PHP] Still trying to figure this out...
Have you ever thought of doing something like this:
for($i = 0; $i = $maxport; $i++)
{
$query=SELECT switchport FROM network WHERE switchport = .$i;
$result=mysql_query($query);
if(empty($result))
{
echo p$i/p;
}
}
It should only return the ones that aren't in the table and they will
automatically be sorted in ascending order.
On 6/19/06, Rob W. [EMAIL PROTECTED] wrote:
Ok, I am still trying to get this figured out from what I tried doing
last night. If anybody wants to try it, please be my guest. This is the
deal. Create an INTEGER table. Put the numbers 1 - 24 in it. Leave out
like
number 8, 9, 22, 23 ect.. Now create a php statement that will take and
pull
that out of the database, find out the missing numbers and display them
as
an html drop down selection.
And tedd in response to why, is because the values that are already
located in the database represent in use already. They represent a server
port that a server is sitting on, so that's not the problem. I could have
done that along time ago.
Any help is appreciated but here is the current code that I have.
$query=SELECT switchport FROM network;
$result=mysql_query($query);
$sql_range=mysql_fetch_array($result);
$true_range=range(1,24);
$next_range=array_diff($true_range,$sql_range);
foreach ($next_range as $final_range) {
echo option value='$final_range'$final_range\n;
}
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Re: [PHP] Still trying to figure this out...
Nope, still does nothing.
- Original Message -
From: Xavier Casto [EMAIL PROTECTED]
To: 'Rob W.' [EMAIL PROTECTED]
Cc: php-general@lists.php.net
Sent: Monday, June 19, 2006 6:49 PM
Subject: RE: [PHP] Still trying to figure this out...
I think the problem is I forgot to reset $result to 0
for($i = 0; $i = $maxport; $i++)
{
$query=SELECT switchport FROM network WHERE switchport = .$i;
$result=mysql_query($query);
if(empty($result))
{
echo p$i/p;
$result=0;
}
}
Though I still haven't tested it, I don't have a current installation of
MySQL...
-Original Message-
From: Xavier Casto [mailto:[EMAIL PROTECTED]
Sent: Monday, June 19, 2006 6:34 PM
To: 'Rob W.'
Cc: php-general@lists.php.net
Subject: RE: [PHP] Still trying to figure this out...
Are there any numbers missing in your database table between 0 and the
number you assigned to $maxport?
-Original Message-
From: Rob W. [mailto:[EMAIL PROTECTED]
Sent: Monday, June 19, 2006 6:26 PM
To: php-general@lists.php.net
Subject: Re: [PHP] Still trying to figure this out...
Yep, and still nothing output's.
- Original Message -
From: Xavier Casto [EMAIL PROTECTED]
To: 'Rob W.' [EMAIL PROTECTED]
Cc: php-general@lists.php.net
Sent: Monday, June 19, 2006 6:20 PM
Subject: RE: [PHP] Still trying to figure this out...
You did make sure to assign a numerical value to $maxport?
-Original Message-
From: Rob W. [mailto:[EMAIL PROTECTED]
Sent: Monday, June 19, 2006 6:15 PM
To: php-general@lists.php.net
Subject: Re: [PHP] Still trying to figure this out...
This also didnt work, Still giving me no output just like the last try.
- Original Message -
From: Xavier Casto
To: Rob W.
Cc: php-general@lists.php.net
Sent: Monday, June 19, 2006 4:33 PM
Subject: Re: [PHP] Still trying to figure this out...
Have you ever thought of doing something like this:
for($i = 0; $i = $maxport; $i++)
{
$query=SELECT switchport FROM network WHERE switchport = .$i;
$result=mysql_query($query);
if(empty($result))
{
echo p$i/p;
}
}
It should only return the ones that aren't in the table and they will
automatically be sorted in ascending order.
On 6/19/06, Rob W. [EMAIL PROTECTED] wrote:
Ok, I am still trying to get this figured out from what I tried doing
last night. If anybody wants to try it, please be my guest. This is the
deal. Create an INTEGER table. Put the numbers 1 - 24 in it. Leave out
like
number 8, 9, 22, 23 ect.. Now create a php statement that will take and
pull
that out of the database, find out the missing numbers and display them
as
an html drop down selection.
And tedd in response to why, is because the values that are already
located in the database represent in use already. They represent a server
port that a server is sitting on, so that's not the problem. I could have
done that along time ago.
Any help is appreciated but here is the current code that I have.
$query=SELECT switchport FROM network;
$result=mysql_query($query);
$sql_range=mysql_fetch_array($result);
$true_range=range(1,24);
$next_range=array_diff($true_range,$sql_range);
foreach ($next_range as $final_range) {
echo option value='$final_range'$final_range\n;
}
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Re: [PHP] Still trying to figure this out...
$range = array_flip( range( 1, 24 ) );
$query = SELECT switchport FROM network ;
if( ($result = mysql_query( $query )) !== false )
{
while( ($row = mysql_fetch_array( $result )) !== false )
{
unset( $range[$row[0]] );
}
}
foreach( $range as $switchport = $foo )
{
echo 'option
value='.$switchport.''.$switchport.'/option'.\n;
$result = 0;
}
With that, I dont get anything, Not sure what you mean by empty to is_null
ect..
- Rob
- Original Message -
From: Xavier Casto [EMAIL PROTECTED]
To: 'Rob W.' [EMAIL PROTECTED]
Cc: php-general@lists.php.net
Sent: Monday, June 19, 2006 7:37 PM
Subject: RE: [PHP] Still trying to figure this out...
Ok Rob,
I tested my code in Oracle (obviously changed for that database type)
and discovered I needed to change empty to is_null I had forgotten
that
empty represents zero but we aren't getting a count so we needed to check
for a null.
it worked me for once I made that change and didn't need to do the other
change I had previously mentioned.
-Original Message-
From: Rob W. [mailto:[EMAIL PROTECTED]
Sent: Monday, June 19, 2006 6:50 PM
To: php-general@lists.php.net
Subject: Re: [PHP] Still trying to figure this out...
Yeah, that's what i'm trying to print is the missing numbers in the db.
That's what I've been trying to figure out this whole time.
- Original Message -
From: Xavier Casto [EMAIL PROTECTED]
To: 'Rob W.' [EMAIL PROTECTED]
Cc: php-general@lists.php.net
Sent: Monday, June 19, 2006 6:34 PM
Subject: RE: [PHP] Still trying to figure this out...
Are there any numbers missing in your database table between 0 and the
number you assigned to $maxport?
-Original Message-
From: Rob W. [mailto:[EMAIL PROTECTED]
Sent: Monday, June 19, 2006 6:26 PM
To: php-general@lists.php.net
Subject: Re: [PHP] Still trying to figure this out...
Yep, and still nothing output's.
- Original Message -
From: Xavier Casto [EMAIL PROTECTED]
To: 'Rob W.' [EMAIL PROTECTED]
Cc: php-general@lists.php.net
Sent: Monday, June 19, 2006 6:20 PM
Subject: RE: [PHP] Still trying to figure this out...
You did make sure to assign a numerical value to $maxport?
-Original Message-
From: Rob W. [mailto:[EMAIL PROTECTED]
Sent: Monday, June 19, 2006 6:15 PM
To: php-general@lists.php.net
Subject: Re: [PHP] Still trying to figure this out...
This also didnt work, Still giving me no output just like the last try.
- Original Message -
From: Xavier Casto
To: Rob W.
Cc: php-general@lists.php.net
Sent: Monday, June 19, 2006 4:33 PM
Subject: Re: [PHP] Still trying to figure this out...
Have you ever thought of doing something like this:
for($i = 0; $i = $maxport; $i++)
{
$query=SELECT switchport FROM network WHERE switchport = .$i;
$result=mysql_query($query);
if(empty($result))
{
echo p$i/p;
}
}
It should only return the ones that aren't in the table and they will
automatically be sorted in ascending order.
On 6/19/06, Rob W. [EMAIL PROTECTED] wrote:
Ok, I am still trying to get this figured out from what I tried doing
last night. If anybody wants to try it, please be my guest. This is the
deal. Create an INTEGER table. Put the numbers 1 - 24 in it. Leave out
like
number 8, 9, 22, 23 ect.. Now create a php statement that will take and
pull
that out of the database, find out the missing numbers and display them
as
an html drop down selection.
And tedd in response to why, is because the values that are already
located in the database represent in use already. They represent a
server
port that a server is sitting on, so that's not the problem. I could
have
done that along time ago.
Any help is appreciated but here is the current code that I have.
$query=SELECT switchport FROM network;
$result=mysql_query($query);
$sql_range=mysql_fetch_array($result);
$true_range=range(1,24);
$next_range=array_diff($true_range,$sql_range);
foreach ($next_range as $final_range) {
echo option value='$final_range'$final_range\n;
}
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[PHP] mysql ORDER BY problem
Ok, here's what i got in my mysql db. I got a table listed with numbers as follows 1 2 3 4 5 6 7 10 11 12 13 14 15 16 17 18 19 20 21 22 25 These numbers I can display fine. I'm using .. $query=SELECT * FROM db ORDER BY numbers ASC; Right now it displays it as 1 10 11 12 . 2 22 23 25 3 4 5 6 7 Is there a way with my mysql query so that I can list the numbers in correct order? Any help is appricated. - Rob
Re: [PHP] Re: mysql ORDER BY problem
It's not in general to mysql, it's how php and mysql is displaying it. - Original Message - From: Michael Rasmussen [EMAIL PROTECTED] To: php-general@lists.php.net Sent: Sunday, June 18, 2006 4:51 PM Subject: [PHP] Re: mysql ORDER BY problem On Sun, 18 Jun 2006 15:55:14 -0500, Rob W. wrote: Is there a way with my mysql query so that I can list the numbers in correct order? In what way is this problem related to PHP? Try a MySQL group instead. -- Hilsen/Regards Michael Rasmussen http://keyserver.veridis.com:11371/pks/lookup?op=getsearch=0xE3E80917 -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] mysql ORDER BY problem
Yeah, I got this problem fixed. The datbase was listed under [chr] instead of [integer]. Thanks for the help though. - Original Message - From: Xavier Casto To: Rob W. Cc: php-general@lists.php.net Sent: Sunday, June 18, 2006 7:30 PM Subject: Re: [PHP] mysql ORDER BY problem Rob, Your problem is the data type you are trying to sort. You have a few solutions that can be used depending on your skills, access, and intent of the data. As some had suggested, the problem may be best handled at the database level, but it all depends on what your intent is. When sorting a datatype of Character (or string) 10 will always come before 2, you would either need to convert the string to a number or left pad the string with 0 to get the result your are looking for. Some good information for your issue can be found here: http://dev.mysql.com/doc/refman/5.0/en/cast-functions.html try the CAST or CONVERT functions in your query: $query=SELECT * FROM db ORDER BY CONVERT(numbers, DECIMAL) ASC; On 6/18/06, Rob W. [EMAIL PROTECTED] wrote: Ok, here's what i got in my mysql db. I got a table listed with numbers as follows 1 2 3 4 5 6 7 10 11 12 13 14 15 16 17 18 19 20 21 22 25 These numbers I can display fine. I'm using .. $query=SELECT * FROM db ORDER BY numbers ASC; Right now it displays it as 1 10 11 12 . 2 22 23 25 3 4 5 6 7 Is there a way with my mysql query so that I can list the numbers in correct order? Any help is appricated. - Rob
[PHP] Ok next php problem
I got the previous question answered, Now here's my next problem.
With the numbers displaying correctly again I got:
1
2
3
4
5
6
7
10
11
12
13
14
15
16
17
18
19
20
21
22
25
listed in that order in the database
Now i'm trying to figure out how to write a syntax saying that if like number 8
isnt listed, display it.
I've tried doing a
if ($count != $data) {
echo $data;
}
$count++;
But when I get to like id number 9 it dont work right because the next entrie
is displayed as 10 in the db. So that's my problem is to try and display only
them numbers that are not in there. I have also tried putting the numbers in to
an array and matching from there but it still come's up as the same as above.
- Rob
Re: [PHP] Re: Ok next php problem
No, but ok here goes the whole thing and sorry if I didnt explain this to
everyone before.
I am trying to duplicate a program that keep's track of server information,
switch ports, apc ports ect ect...
In the one column is switchports
This lists a total of 24 or 48 or how many ever ports are on that switch.
With in there is numbers like this for example:
1
2
3
4
5
6
7
10
11
12
13
14
15
ect... Hopefully that help's so far.
As far as listing the numbers is ok, I can pull it from the db and list all
the numbers just fine in numerical order as they are listed.
My problem is inside this script, when a new server is added, it pulls the
information from the database and finds an empty number, which is why,
though the whole list of 1 though 24 i need to get the information of the
missing numbers. These missing numbers HAFT to be dynamic and can not be
hard coded in because servers are always moved around and changed and added.
I hope this information was helpful for those of you trying to help solve my
problem.
- Rob
- Original Message -
From: João Cândido de Souza Neto [EMAIL PROTECTED]
To: php-general@lists.php.net
Sent: Sunday, June 18, 2006 8:38 PM
Subject: [PHP] Re: Ok next php problem
If i understood right, you want to list a sequence of numbers and follow
some data like:
1 - data of number 1
2 - data of number 2
...
If it's right, you can do it.
$query=SELECT * FROM db ORDER BY numbers DESC limit 1;
$maxnum=$numbers;
for ($num=1;$num=$maxnum; $num++) {
$query=SELECT * FROM db where numbers=.$num;
echo $num. - .$data.br;
}
Rob W. [EMAIL PROTECTED] escreveu na mensagem
news:[EMAIL PROTECTED]
I got the previous question answered, Now here's my next problem.
With the numbers displaying correctly again I got:
1
2
3
4
5
6
7
10
11
12
13
14
15
16
17
18
19
20
21
22
25
listed in that order in the database
Now i'm trying to figure out how to write a syntax saying that if like
number 8 isnt listed, display it.
I've tried doing a
if ($count != $data) {
echo $data;
}
$count++;
But when I get to like id number 9 it dont work right because the next
entrie is displayed as 10 in the db. So that's my problem is to try and
display only them numbers that are not in there. I have also tried putting
the numbers in to an array and matching from there but it still come's up
as the same as above.
- Rob
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Re: [PHP] Ok next php problem
Ok but my problem is is that in the process of doing that, numbers can be released so they pretty much haft to be dynamic. Any idea how I do it with that. IE: 1 2 3 6 9 10 ... So if them numbers change, which they can, because they are assigned port numbers for servers, How do I make it so they are not scripted statically. - Original Message - From: Chris [EMAIL PROTECTED] To: Rob W. [EMAIL PROTECTED] Cc: php-general@lists.php.net Sent: Sunday, June 18, 2006 8:38 PM Subject: Re: [PHP] Ok next php problem Rob W. wrote: I got the previous question answered, Now here's my next problem. With the numbers displaying correctly again I got: 1 2 3 4 5 6 7 10 11 12 13 14 15 16 17 18 19 20 21 22 25 listed in that order in the database Now i'm trying to figure out how to write a syntax saying that if like number 8 isnt listed, display it. Get them both into arrays and compare: $good_list = range(1,10); $db_list = array(4,7,10); $missing = array_diff($good_list, $db_list); http://php.net/array_diff -- Postgresql php tutorials http://www.designmagick.com/ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Javascript PHP
Ok. I am trying to figure out if this would be the correct thing to do. I start
off with Java script that looks like this.
/*
function getType(what,available,owner) {
var serverid = serveradd.serverid.value;
var cabinetid = serveradd.cabinetid.value;
var rowid = serveradd.rowid.value;
var dcid = serveradd.dcid.value;
var colour = serveradd.colour.value;
var spec = serveradd.spec.value;
var mainip = serveradd.mainip.value;
var oldip = serveradd.oldip.value;
var switchid = serveradd.switchid.value;
var switchport = serveradd.switchport.value;
var apcid = serveradd.apcid.value;
var apcport = serveradd.apcport.value;
document.location=('admin.php?page=serveraddaction=' + add +
'myid=serverid=' + serverid + 'cabinetid=' +
cabinetid + 'rowid=' + rowid + 'dcid=' + dcid + 'colour=' + colour +
'spec=' + spec + 'mainip=' + mainip +
'oldip=' + oldip + 'switchid=' + switchid + 'switchport=' + switchport +
'apcid=' + apcid + 'apcport=' + apcport +
'available=' + available + 'owner=' + owner);
}*/
Do I, Before that exec's, add in my mysql get to set the variables to the java
variables or how do I enter that in?
Any help is appricated.
-Rob
Re: [PHP] Javascript PHP
Well i'm trying to figure out how to bind the php variables to the
javascript variables after I pull them out of the mysql database.
- Rob
- Original Message -
From: Jochem Maas [EMAIL PROTECTED]
To: Rob W. [EMAIL PROTECTED]
Cc: php-general@lists.php.net
Sent: Friday, June 16, 2006 4:13 AM
Subject: Re: [PHP] Javascript PHP
Rob W. wrote:
Ok. I am trying to figure out if this would be the correct thing to do. I
start off with Java script that looks like this.
/*
function getType(what,available,owner) {
var serverid = serveradd.serverid.value;
var cabinetid = serveradd.cabinetid.value;
var rowid = serveradd.rowid.value;
var dcid = serveradd.dcid.value;
var colour = serveradd.colour.value;
var spec = serveradd.spec.value;
var mainip = serveradd.mainip.value;
var oldip = serveradd.oldip.value;
var switchid = serveradd.switchid.value;
var switchport = serveradd.switchport.value;
var apcid = serveradd.apcid.value;
var apcport = serveradd.apcport.value;
document.location=('admin.php?page=serveraddaction=' + add +
'myid=serverid=' + serverid + 'cabinetid=' +
cabinetid + 'rowid=' + rowid + 'dcid=' + dcid + 'colour=' + colour +
'spec=' + spec + 'mainip=' + mainip +
'oldip=' + oldip + 'switchid=' + switchid + 'switchport=' + switchport
+ 'apcid=' + apcid + 'apcport=' + apcport +
'available=' + available + 'owner=' + owner);
}*/
Do I, Before that exec's, add in my mysql get to set the variables to the
java variables or how do I enter that in?
I read that line 3 times - I know its english, but the meaning escapes
me... maybe others have the same problem.
could you rephrase the question?
Any help is appricated.
-Rob
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Re: [PHP] Javascript PHP
Well for example
when I pull my mysql db variable
$serverid=mysql_result($result,$i,serverid);
How do I bind it so this ^^ to the java var below?
var serverid = serveradd.serverid.value;
- Rob
- Original Message -
From: Jochem Maas [EMAIL PROTECTED]
To: Rob W. [EMAIL PROTECTED]
Cc: php-general@lists.php.net
Sent: Friday, June 16, 2006 4:26 AM
Subject: Re: [PHP] Javascript PHP
Rob W. wrote:
Well i'm trying to figure out how to bind the php variables to the
javascript variables after I pull them out of the mysql database.
something like this? ...
$data = getServerTypeDataOrSomething();
?
form name=serveradd
input type=text name=serverid value=? echo $data['serverid']; ?
... etc
or am I missing the point completely?
- Rob
- Original Message - From: Jochem Maas [EMAIL PROTECTED]
To: Rob W. [EMAIL PROTECTED]
Cc: php-general@lists.php.net
Sent: Friday, June 16, 2006 4:13 AM
Subject: Re: [PHP] Javascript PHP
Rob W. wrote:
Ok. I am trying to figure out if this would be the correct thing to
do. I start off with Java script that looks like this.
/*
function getType(what,available,owner) {
var serverid = serveradd.serverid.value;
var cabinetid = serveradd.cabinetid.value;
var rowid = serveradd.rowid.value;
var dcid = serveradd.dcid.value;
var colour = serveradd.colour.value;
var spec = serveradd.spec.value;
var mainip = serveradd.mainip.value;
var oldip = serveradd.oldip.value;
var switchid = serveradd.switchid.value;
var switchport = serveradd.switchport.value;
var apcid = serveradd.apcid.value;
var apcport = serveradd.apcport.value;
document.location=('admin.php?page=serveraddaction=' + add +
'myid=serverid=' + serverid + 'cabinetid=' +
cabinetid + 'rowid=' + rowid + 'dcid=' + dcid + 'colour=' + colour
+ 'spec=' + spec + 'mainip=' + mainip +
'oldip=' + oldip + 'switchid=' + switchid + 'switchport=' +
switchport + 'apcid=' + apcid + 'apcport=' + apcport +
'available=' + available + 'owner=' + owner);
}*/
Do I, Before that exec's, add in my mysql get to set the variables to
the java variables or how do I enter that in?
I read that line 3 times - I know its english, but the meaning escapes
me... maybe others have the same problem.
could you rephrase the question?
Any help is appricated.
-Rob
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Re: [PHP] Javascript PHP
Yeah, I think something like that might work.
( I also know it's javascript and not java - Little tired being up @ 5am and
havnt slept for over 24 hrs already. )
- Rob
- Original Message -
From: Jochem Maas [EMAIL PROTECTED]
To: Rob W. [EMAIL PROTECTED]
Cc: php-general@lists.php.net
Sent: Friday, June 16, 2006 4:55 AM
Subject: Re: [PHP] Javascript PHP
Rob W. wrote:
Well for example
when I pull my mysql db variable
$serverid=mysql_result($result,$i,serverid);
How do I bind it so this ^^ to the java var below?
(I hope english is not you first language - or maybe you
write in shorthand *and* you don't bother to reread what you
have written)
btw it's not JAVA but JAVASCRIPT.
var serverid = serveradd.serverid.value;
have you thought of something like this:
script type=text/javascript
!--
var serverid = ? echo $serverid ?;
//--
/script
- Rob
- Original Message - From: Jochem Maas [EMAIL PROTECTED]
To: Rob W. [EMAIL PROTECTED]
Cc: php-general@lists.php.net
Sent: Friday, June 16, 2006 4:26 AM
Subject: Re: [PHP] Javascript PHP
Rob W. wrote:
Well i'm trying to figure out how to bind the php variables to the
javascript variables after I pull them out of the mysql database.
something like this? ...
$data = getServerTypeDataOrSomething();
?
form name=serveradd
input type=text name=serverid value=? echo $data['serverid'];
?
... etc
or am I missing the point completely?
- Rob
- Original Message - From: Jochem Maas [EMAIL PROTECTED]
To: Rob W. [EMAIL PROTECTED]
Cc: php-general@lists.php.net
Sent: Friday, June 16, 2006 4:13 AM
Subject: Re: [PHP] Javascript PHP
Rob W. wrote:
Ok. I am trying to figure out if this would be the correct thing to
do. I start off with Java script that looks like this.
/*
function getType(what,available,owner) {
var serverid = serveradd.serverid.value;
var cabinetid = serveradd.cabinetid.value;
var rowid = serveradd.rowid.value;
var dcid = serveradd.dcid.value;
var colour = serveradd.colour.value;
var spec = serveradd.spec.value;
var mainip = serveradd.mainip.value;
var oldip = serveradd.oldip.value;
var switchid = serveradd.switchid.value;
var switchport = serveradd.switchport.value;
var apcid = serveradd.apcid.value;
var apcport = serveradd.apcport.value;
document.location=('admin.php?page=serveraddaction=' + add +
'myid=serverid=' + serverid + 'cabinetid=' +
cabinetid + 'rowid=' + rowid + 'dcid=' + dcid + 'colour=' + colour
+ 'spec=' + spec + 'mainip=' + mainip +
'oldip=' + oldip + 'switchid=' + switchid + 'switchport=' +
switchport + 'apcid=' + apcid + 'apcport=' + apcport +
'available=' + available + 'owner=' + owner);
}*/
Do I, Before that exec's, add in my mysql get to set the variables to
the java variables or how do I enter that in?
I read that line 3 times - I know its english, but the meaning escapes
me... maybe others have the same problem.
could you rephrase the question?
Any help is appricated.
-Rob
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Re: [PHP] Automatic email sending
If you have a unix box, Use cron to run a script to exec a php file to do it. If you have a windows box, you can setup a scheduled time to run a php file as well. - Rob - Original Message - From: Murtaza Chang [EMAIL PROTECTED] To: php-general-digest@lists.php.net; php-general@lists.php.net Sent: Friday, June 16, 2006 5:15 AM Subject: [PHP] Automatic email sending Hi everyone, my website need to automatically send an email to people when thier account is about to expire, but iam not sure how to do it. In other words its easy to put a check if an account is about to expire and send automatic email when admin logs in to system, but how is it possible to do it in background. -- Murtaza Chang http://dirtinme.blogspot.com/ A Place to throw personal trash http://milddreamerz.blogspot.com For Photoshopers And Digital Art Enthusiasts -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Getting totals
Yeah, it counts how many times block is in $address, so if there's 254 ip's
listed in the database, it will incriment it 254 times and display that.
- Original Message -
From: Rabin Vincent [EMAIL PROTECTED]
To: Rob W. [EMAIL PROTECTED]
Cc: php-general@lists.php.net
Sent: Wednesday, June 07, 2006 2:03 AM
Subject: Re: [PHP] Getting totals
On 6/7/06, Rob W. [EMAIL PROTECTED] wrote:
I got the fix, strstr didnt work right because it was relaying more than
just what I was thinking.
Here is the fix.
$value=array(strstr($block, $address));
foreach ($value as $var) {
$block_total_ip++;
}
How can this work? $value will have 1 element no matter what,
so $block_total_ip will always be incremented once. The above
code is equivalent to just:
$block_total_ip++;
Rabin
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[PHP] Removing an aspect of a variable...
Say I have a variable setting an ip address of 192.168.100.0 I want to be able to remove the last to chr's of that variable ie: .0 What would be my best solution to do that?
Re: [PHP] Removing an aspect of a variable...
Tnx. - Original Message - From: rich gray [EMAIL PROTECTED] To: Rob W. [EMAIL PROTECTED] Sent: Tuesday, June 06, 2006 2:44 AM Subject: Re: [PHP] Removing an aspect of a variable... substr($variable,0,-2); Rob W. wrote: Say I have a variable setting an ip address of 192.168.100.0 I want to be able to remove the last to chr's of that variable ie: .0 What would be my best solution to do that? -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Getting totals
Ok, Here is my next problem.
Inside my database, I have a list of ip's of about 10 blocks
192.168.100.0 all the way though 255
along with
192.168.101.0 though 255
and
192.168.102.0 though 255
and soforth
My problem is, is i'm trying to figure out a pattern to match so I can count
how many ip's are in each block and display them.
So far what I have gotten is a stristr match but it's not working correctly. I
have a variable that basically weed's out the last digits of the ip it's self
from help previously
So my code so far is:
if (stristr($block,$address)) {
$count_ip++;
}
$block would == 192.168.100
$address would == 192.168.100.0 - 255
Any help would be appricated.
Re: [PHP] Getting totals
Acutall no, Because in my database is is all them blocks but each ip in all
them blocks has a seperate entry in a column. So in listing the whole column
in my mysql request, i need a line that will weed out and count how many
ip's are in a block.
- Original Message -
From: [EMAIL PROTECTED]
To: Rob W. [EMAIL PROTECTED]
Sent: Tuesday, June 06, 2006 1:16 PM
Subject: Re: [PHP] Getting totals
i may be missing something in your description, but does turning the
first and last ipnumber in a block into its integer representation, and
then doing the math (subtraction) to get the ipnumber count per block,
accomplish what you're after?
- Rick
Original Message
Date: Tuesday, June 06, 2006 12:15:27 PM -0500
From: Rob W. [EMAIL PROTECTED]
To: php-general@lists.php.net
Subject: [PHP] Getting totals
Ok, Here is my next problem.
Inside my database, I have a list of ip's of about 10 blocks
192.168.100.0 all the way though 255
along with
192.168.101.0 though 255
and
192.168.102.0 though 255
and soforth
My problem is, is i'm trying to figure out a pattern to match so I
can count how many ip's are in each block and display them.
So far what I have gotten is a stristr match but it's not working
correctly. I have a variable that basically weed's out the last
digits of the ip it's self from help previously
So my code so far is:
if (stristr($block,$address)) {
$count_ip++;
}
$block would == 192.168.100
$address would == 192.168.100.0 - 255
Any help would be appricated.
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Re: [PHP] Getting totals
if (strstr(192.168.100,192.168.100.10)) {
$inc++;
}
echo $inc;
That returns nothing. What am i still doing wrong?
- Original Message -
From: Rabin Vincent [EMAIL PROTECTED]
To: Rob W. [EMAIL PROTECTED]
Cc: php-general@lists.php.net
Sent: Tuesday, June 06, 2006 1:36 PM
Subject: Re: [PHP] Getting totals
On 6/6/06, Rob W. [EMAIL PROTECTED] wrote:
So far what I have gotten is a stristr match but it's not working
correctly. I have a variable that basically weed's out the last digits of
the ip it's self from help previously
So my code so far is:
if (stristr($block,$address)) {
$count_ip++;
}
You've got the parameters mixed up. strstr is (haystack,
needle) so you need strstr($address, $block). php.net/stristr.
Rabin
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Re: [PHP] Getting totals
Sorry for the miss understanding, That's the way the viarable will look, i'm
putting it in as a viariable.
if (strstr($block,$address)) {
$inc++;
}
- Original Message -
From: Richard Lynch [EMAIL PROTECTED]
To: Rob W. [EMAIL PROTECTED]
Sent: Tuesday, June 06, 2006 8:58 PM
Subject: Re: [PHP] Getting totals
Put quotes or apostrophes on the strings...
if (strstr('192.168.100','192.168.100.10')) {
On Tue, June 6, 2006 8:46 pm, Rob W. wrote:
if (strstr(192.168.100,192.168.100.10)) {
$inc++;
}
echo $inc;
That returns nothing. What am i still doing wrong?
- Original Message -
From: Rabin Vincent [EMAIL PROTECTED]
To: Rob W. [EMAIL PROTECTED]
Cc: php-general@lists.php.net
Sent: Tuesday, June 06, 2006 1:36 PM
Subject: Re: [PHP] Getting totals
On 6/6/06, Rob W. [EMAIL PROTECTED] wrote:
So far what I have gotten is a stristr match but it's not working
correctly. I have a variable that basically weed's out the last
digits of
the ip it's self from help previously
So my code so far is:
if (stristr($block,$address)) {
$count_ip++;
}
You've got the parameters mixed up. strstr is (haystack,
needle) so you need strstr($address, $block). php.net/stristr.
Rabin
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Re: [PHP] Getting totals
I got the fix, strstr didnt work right because it was relaying more than
just what I was thinking.
Here is the fix.
$value=array(strstr($block, $address));
foreach ($value as $var) {
$block_total_ip++;
}
- Original Message -
From: Rabin Vincent [EMAIL PROTECTED]
To: Rob W. [EMAIL PROTECTED]
Cc: php-general@lists.php.net
Sent: Tuesday, June 06, 2006 11:38 PM
Subject: Re: [PHP] Getting totals
On 6/7/06, Rob W. [EMAIL PROTECTED] wrote:
Sorry for the miss understanding, That's the way the viarable will look,
i'm
putting it in as a viariable.
if (strstr($block,$address)) {
$inc++;
}
Like I said before, strstr's argument list is haystack (what to
search in) first and then needle (what to search for). Therefore,
you need to have:
strstr($address, $block);
Rabin
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[PHP] Displaying data a certian way.
Ok, here's my issue that I have. Inside my database I have something that look's like this idserveridcabinetidect... - 1 server11 2 server21 3 server31 I am trying to get it to display the servers 1 2 3 under a display bar of cabinet 1. Anybody have any solutions to this? I am a novis to php and mysql but i know the majority of the basics.
Re: [PHP] Displaying data a certian way.
Just to clarify a little more, I want the output to do something like this... -- | Cabinet 1 | -- Server 1 Server 2 Server 3 and so forth... - Original Message - From: Rob W. [EMAIL PROTECTED] To: php-general@lists.php.net Sent: Sunday, June 04, 2006 10:47 PM Subject: [PHP] Displaying data a certian way. Ok, here's my issue that I have. Inside my database I have something that look's like this idserveridcabinetidect... - 1 server11 2 server21 3 server31 I am trying to get it to display the servers 1 2 3 under a display bar of cabinet 1. Anybody have any solutions to this? I am a novis to php and mysql but i know the majority of the basics. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Displaying data a certian way.
That worked. Thanks.
- Original Message -
From: Chris [EMAIL PROTECTED]
To: Rob W. [EMAIL PROTECTED]
Cc: php-general@lists.php.net
Sent: Sunday, June 04, 2006 10:59 PM
Subject: Re: [PHP] Displaying data a certian way.
Rob W. wrote:
Ok, here's my issue that I have.
Inside my database I have something that look's like this
idserveridcabinetidect...
-
1 server11
2 server21
3 server31
I am trying to get it to display the servers 1 2 3 under a display bar
of cabinet 1.
Anybody have any solutions to this? I am a novis to php and mysql but i
know the majority of the basics.
Remember the previous cabinet and you can use that for checking whether to
display another heading:
$previous_cabinet = '';
while($row = mysql_fetch_assoc($query_result)) {
if ($row['cabinetid'] != $previous_cabinet) {
echo Cabinet id . (int)$row['cabinetid'];
$previous_cabinet = $row['cabinetid'];
}
echo Server . htmlentities($row['serverid']) . br/;
}
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RE: [PHP] Escaping Chars
Thanks for the reply, but I still can't seem to make the connection...
If I enter the value
123\/'
in a web form and put the form post value directly into the db (no
stripslashes or any other function), the value as reported by the db at a
command line query is
123\/'
(it LOOKS like the same value that was entered), but to get it to return
that value, at the command prompt, I have to enter
select * from users where password = 123/\';.
OK, that makes sense. You have to 'slash' or escape every escape or
delimiter character. So, the value is apparently getting into the db
properly. Now, when I enter that same value (minus the outside quotes) into
the form field and then compare that with the value in the db, they don't
match.
I've tried add and strip slashes in various combinations, but that makes no
difference. I suspect there are some HTML entities or some other odd URL
encoding problem??? My app has a feature that will remind a user of their
password. This returns in an email exactly what I'd expect, that is,
123\/'
I can't see how to make the round trip from the original input into the db
and then back out again intact so it will 'match itself'...
That behavior doesn't seem to match the magic_quotes docs.
My current project is the first real app I have done for the Air Force in
PHP. Most of the PHP work I have done is for query only db interfaces,
counters, REMOTE_HOST tests for dynamic links or doing form-to-email type
stuff. Entering data INTO a db adds a whole new set of challenges.
I'd appreciate any other advice or clarification you could offer.
Thanks,
-Original Message-
From: John W. Holmes [mailto:[EMAIL PROTECTED]]
Sent: Friday, February 07, 2003 4:25 PM
To: 'Rob Walls'; [EMAIL PROTECTED]
Subject: RE: [PHP] Escaping Chars
I need to get a password value from a form, store it in a database and
then
later be able to compare a login password to the one stored in the db.
This works great unless the password contains the '\' char.
magic_quotes_gpc is ON and magic_quotes_runtime is OFF.
As a klude, I tried just removing slashes from the input password
using
stripslashes() before storing it in the db and then testing to see if
stripslashes(val from db)=stripslashes(val from form) in the login
test to
see if they match. (the user shouldn't even know that slashes are
being
striped, so I have to strip them on each input). They still don't
match
if
a slash is input for the original password storage, but I don't know
why.
Okay... you want the slash or escape character there when you insert
it into the database. But, since it's an escape character, it doesn't
actually go into the data of the database. If you put O'Kelly into your
form, magic_quotes_gpc will turn it into O\'Kelly. If you insert that
into the database, it'll use the \ as an escape character and the data
in the database will actually be just O'Kelly. With magic_quotes_runtime
OFF, that's exactly what you'll draw out of the database, too. So, if
you want to compare a form submitted value to a value drawn out of the
database, you have to use stripslashes() on the form data first.
A better option overall is to just do it in your query.
SELECT * FROM table WHERE user = '{$_POST['user']} and password =
'{$_POST['password']}'
Where your form is method=POST... If a row is returned, the username and
password matched. If no row is returned, then one or both didn't match.
---John Holmes...
PS: Just noticed the .af.mil address. Do you do any PHP programming for
the AirForce or is this on your own?
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