RE: [PHP] Confused

2002-09-23 Thread M . A . Bond 

Nope,

Do a select like so:

Select user_name, group_name from users,groups where
users.group_id=groups_id order by group_name;


Then do a php loop (you'll need to find the eact php commands, this is just
an example):

While (!$results-EOF)
{

If ($results[group_name]!=$old_groupname)
{
$old_groupname=$results[group_name];
print TRTD$old_groupname/tdtdnbsp;/td/tr;
}else
{
print TRTDnbsp;/tdTD$results[user_name]/td/tr;
}
$results=mysql_fetch_array($sql);
}

You'll obviously need to get exact mysql commands etc, and open and close
your table outside the loop, but this should be basically it.


Thanks

Mark


-Original Message-
From: Rankin, Randy [mailto:[EMAIL PROTECTED]] 
Sent: 23 September 2002 13:37
To: '[EMAIL PROTECTED]'
Subject: [PHP] Confused


I have two MySQL tables, groups and users:

groups: group_id, group_name

users: user_id, user_name, group_id, etc.

I would like to produce one table for each group which will list all the

members for that particular group. For example:

Blue (group_id 1)

Tom (group_id 1)

Nancy (group_id 1)

Jim (group_id 1)

Red (group_id 2)

Bob (group_id 1)

Susan (group_id 1)

James (group_id 1)

...

My question is, do I need to run 2 queries? The first to select all the

groups and the second to select all the users in each group based on the

group_id? How would I loop through to create a table for each group? 

Thanks in advance for any help.

Randy Rankin


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RE: [PHP] mysql_num_rows error

2002-09-23 Thread M . A . Bond 


 I am new to php and that the folowing error: Warning:
mysql_num_rows():
 supplied argument is not a valid MySQL result resource in 
 /home/tbonestu/public_html/smallimages.php
 
 i dont know what i am doing wrong here is the code:
 
   $db = mysql_pconnect(connect info);
 
  mysql_select_db(images);
  $query = select * from images where type =.$type.;  $result = 
 mysql_query($query);  $num_results = mysql_num_rows($result);

That generally means your query failed. Try to echo mysql_error() after
your query to see what the problem is. You probably need quotes around $type
or $type has no value.

You also have too many quotes on the $query line, try echoing your $query
variable to see if it reads properly.

Mark



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RE: [PHP] How do i assign Integers only and not real numbers?

2002-09-23 Thread M . A . Bond 

You could cast it to an int, although the manual warns that you can get
spurious errors upon casting an unknown fraction, so be careful.

$year= (int) ($years);

Mark


-Original Message-
From: Tom [mailto:[EMAIL PROTECTED]] 
Sent: 23 September 2002 15:47
To: [EMAIL PROTECTED]
Subject: [PHP] How do i assign Integers only and not real numbers?


Hi all,

I have a line of code that assigns a new year number: -

$years = $years + ( $themonth / 12 );

but sometimes $years == 1998.08

or

$year == 2002.75

etc...

I cant find anything like a round() or floor() function in PHP so that year
would be 1998 or 2002 only.

Does anyone know what function would do this?  Sorry if I`m wasting your
time if its a really obvious one!

thanks in advance,
Tom



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RE: [PHP] MS SQL Problem

2002-09-13 Thread M . A . Bond 


?PHP
$hostname = LocalServer;
^^

This should be LocalHost (unless you are trying to connect to a server
with a DNS alias of LocalServer).

Mark

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RE: [PHP] Re: Mail problem

2002-08-30 Thread M . A . Bond 

I actually had to set the SMTP setting to localhost on our solaris boxes to
get this to work. I discovered this because we have a few older machines
without php.ini files altogether, and the default is to have this set (the
older machines worked).

Thanks

Mark Bond


-Original Message-
From: Mark Colvin [mailto:[EMAIL PROTECTED]] 
Sent: 30 August 2002 13:40
To: 'Erwin'
Cc: Php (E-mail)
Subject: RE: [PHP] Re: Mail problem


Still not resolved. I checked my local php.ini and the sendmail_path
variable is commented out. If I do a phpinfo() on the same server,
sendmail_path has a value of /usr/sbin/sendmail -t -i ? I don't know where
it pick up this value from but I can send emails from this server. The
production server that is situated with my isp also has the sendmail_path
set to /usr/sbin/sendmail -t -i but this won't send emails. Both web servers
are on linux 7.2 boxes. Am I correct in assuming that the SMTP and
sendmail_from variables are only relevant for webservers on windows hosts
and all I need to be concerned with is sendmail_path? I have searched
various places for an answer as to why I can't mail from my production
server without much success.



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RE: [PHP] Exiting from an include or required file

2002-08-21 Thread M . A . Bond 

Or just do a return();

?php
if (empty($_POST['search_criteria'])) 
{
echo You must provide search criteria;
  return;
}



-Original Message-
From: Roedel, Mark [mailto:[EMAIL PROTECTED]] 
Sent: 21 August 2002 15:48
To: Henry; php-general
Subject: RE: [PHP] Exiting from an include or required file



Seems like the easy way, at least in this case, would be to make the second
part of your File A an else to your if.

?php
if (empty($_POST['search_criteria'])) {
echo You must provide search criteria;
} else {
echo You provided search criteria;
}


---
Mark Roedel   | Blessed is he who has learned to laugh
Systems Programmer|  at himself, for he shall never cease
LeTourneau University |  to be entertained.
Longview, Texas, USA  |  -- John Powell


 -Original Message-
 From: Henry [mailto:[EMAIL PROTECTED]]
 Sent: Wednesday, August 21, 2002 9:40 AM
 To: [EMAIL PROTECTED]
 Subject: [PHP] Exiting from an include or required file
 
 
 Hi All,
 
 I would like to exit from an include file and continue in the calling 
 script!
 
 As an example
 
 FILE A:
 
 ?php
 if (empty($_POST['search_criteria'))
 {
   echo You must provide search criteria;
 
   // exit the include file here But how?
  }
 
 echo You have provided serach criteria;
 ?
 
 
 FILE B
 
 #include A;
 
 form method=post
 input name=search_criteria type=text
 /form
 
 TIA
 
 Henry
 
 
 
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