Re: [PHP] PHP Doesn't Crash but show blank page

2002-01-03 Thread Steve Cayford

Two other lines to look for in your php.ini file:

display_errors = On
log_errors = Off

If display_errors is on the error will be displayed on the web page, if 
log_errors is on the error will be logged--assuming you're using Linux 
check /var/log/messages, but you can change this with the 
error_log = ... option.

-Steve


On Wednesday, January 2, 2002, at 05:11  AM, [@-!-%] wrote:

>
> Hello everyone!
>
> I'm having issues with my php pages. When I have errors, Instead of
> crashing or showing  errors, the browser shows a blank page.
>
> For example, if I leave the left bracket in my if statement, like
>
> if(blah blah)
><---missing left bracket '{'
>
> do this
>
> }else {
>  do that
> }
>
> or, I leave the semiconlon at the end of a line, like
>
> $myvar= " this is my line" <-- missing ';'
>
> PHP does not report the error.
>
> My php.ini is configured as
>
> error_reporting =
> E_COMPILE_ERROR|E_ERROR|E_CORE_ERROR|E_USER_ERROR|E_USER_WARNING|E_USER_NOTICE
>
> My system has
> PHP 4.0.6, MySQL 3.23.46-nt, Win2000
>
> Anyone knows why it's doing that?
>
> -john
>
> __John Monfort_
> _+---+_
>  P E P I E  D E S I G N S
>www.pepiedesigns.com
> "The world is waiting, are you ready?"
> -+___+-
>
>
>
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Re: [PHP] Regular Expression

2002-01-03 Thread Steve Cayford

You can do as Jim says here or go back and read the manual section again 
for eregi and ereg. What you're trying to do should be written more like 
this:

';
$numMatches = eregi('()',$str,$results);
print("numMatches: $numMatches \n");
print("body contents: $results[2] \n");
print("all results: ");
print_r($results);
?>

-Steve


On Thursday, January 3, 2002, at 12:15  PM, Jim Lucas [php] wrote:

> I have seen this question reposted for the past week.  now why don't you
> just work with the entire thing.
>
> get the ""
>
> now once you have that, do this
>
> $str = preg_replace(" $str = preg_replace(">", "", $str)
>
> now your $str var will only have the properties.
>
> Jim
> - Original Message -
> From: "[-^-!-%-" <[EMAIL PROTECTED]>
> To: <[EMAIL PROTECTED]>
> Sent: Wednesday, January 02, 2002 12:45 PM
> Subject: [PHP] Regular Expression
>
>
>>
>> Hello everyone!
>>
>> I'm trying to get the text inside the  tag, using regular
>> expression.
>>
>> $area = eregi('()',$str);
>>
>> Where $str is the string containing
>> 
>>
>> When I print  $area, the string contains the entire content of $str. I
>> get something like:
>>
>> 
>> 
>> .
>> .
>> .
>> 
>> 
>>
>>
>>
>> __John Monfort_
>> _+---+_
>>  P E P I E  D E S I G N S
>>www.pepiedesigns.com
>> "The world is waiting, are you ready?"
>> -+___+-
>>
>>
>>
>>
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>>
>>
>
>
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Re: [PHP] counting with dates (help!)

2002-01-07 Thread Steve Cayford

Well, I'll chime in as well. I'd recommend doing all your calculations 
in timestamps in seconds, then convert the results into days, dates, or 
whatever. If you only have a date to start with then convert to a 
timestamp, do the calculation, and convert back. You could wrap it in a 
function like this:

\n");
print("last_week: $last_week \n");
print("diff in days: " . dateDiffInDays($last_week, $today) . "\n");
?>


On Thursday, February 7, 2002, at 04:21  PM, Sander Peters wrote:

> Hello,
>
>
> This is my problem:
>
> $today = date("Ymd", mktime(0,0,0, date(m),date(d),date(Y)));
> $last_week = date("Ymd", mktime(0,0,0, date(m),date(d)-7,date(Y)));
> echo ($today - $last_week);
> The result is a number like 8876 (20020107-20011231 = 8876)
> But in date thinking it should be 7!
>
> How can I let php count in real days/month/years in stead of numbers?
>
> Maybe this is a silly question, but anyone who has the answer would help
> me very much!
>
> Thanks in advance!
>
>
> --
> Met vriendelijke groet / With Greetings,
>
> Sander Peters
>
>site: http://www.visionnet.nl/
>   email: mailto:[EMAIL PROTECTED]
> webmail: mailto:[EMAIL PROTECTED]
>
>
>
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Re: [PHP] Error

2002-01-08 Thread Steve Cayford


On Tuesday, January 8, 2002, at 09:37  AM, Dean Ouellette wrote:

> Hi I am learning php with Sams leanr php in 24 hours.
>
> This is one example
>function addNums($firstnum, $secondnum)
>   {
>   $result = $firstnum + $secondnum;
>   return $result;11:  } // this is line 13

that should be:
  return $result; }

what's the 11: doing in there?

-Steve

>   print addNums (3,5);
>   
> ?>
>
> When I run it get error line 13
>
> any  ideas?
>
>
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Re: [PHP] Is there Any way to call Non-Existent function in PHP

2002-01-17 Thread Steve Cayford

You could wrap your function calls, maybe. Like

function my_Draw_Image() {
if(function_exists('Draw_Image')) {
return Draw_Image(func_get_args());
} else {
// do whatever...
}
}

Haven't tested this at all, but it seems plausible. Might be possible to 
wrap generic function calls like this using call_user_func() or eval(). 
Again, I haven't tested it.

-Steve


On Thursday, January 17, 2002, at 12:07  AM, S. Murali Krishna wrote:

> Hi!
>   Thanks for your kind response. The solution given by u is fine.
> But Iam asking just more than that. See whenever I call a function
> like
>   Draw_Image();
>
> ( In language construct level )
> Is there anything to checks this non-existence  of Draw_Image() and call
> some default function.
>
> I don't want to manually check if it exist or not.
>
> Thanks for Ur Solution.
>
>
> On Wed, 16 Jan 2002, Neil Freeman wrote:
>
>> You can use function_exists() to check whether a function actually 
>> exists:
>>
>> eg:
>> if (function_exists('imap_open')) {
>> echo "IMAP functions are available.\n";
>> } else {
>> echo "IMAP functions are not available.\n";
>> }
>>
>> HTH
>>
>> Neil
>>
>> "S. Murali Krishna" wrote:
>>
>>> Hai ALL
>>> Is there any way to capture function call in PHP
>>> and redirect to some other function if that function doesn't exists.
>>>
>>> Perl programmers remind   AUTOLOAD method in a package.
>>>
>>> S.Murali Krishna
>>> [EMAIL PROTECTED]
>>> =
>>> We grow slow trying to be great
>>>
>>>- E. Stanley Jones
>>> -
>>>
>>> --
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>>>
>>> ***
>>>  This message was virus checked with: SAVI 3.52
>>>  last updated 8th January 2002
>>> ***
>>
>> --
>> 
>>  Email:  [EMAIL PROTECTED]
>>  [EMAIL PROTECTED]
>> 
>>
>>
>
> S.Murali Krishna
> [EMAIL PROTECTED]
> =
> We grow slow trying to be great
>   
>  - E. Stanley Jones
> -
>
>
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Re: [PHP] variable variables

2002-01-17 Thread Steve Cayford


On Thursday, January 17, 2002, at 10:11  AM, Mike Krisher wrote:

> I can not wrap my head around variable variables today, not awake yet or
> something.
>
> For instance I trying something like this:
>
> while ($i<$loopcounter) {
>   $temp = "size";
>   $valueofsize = $$temp$i;
>
try $valueofsize = ${$temp$i};

>   $i++;
> }
>
> this doesn't work obviously, $valueofsize ends up with a literal value 
> of
> "$size1". But I need it to equal the value of a variable named $size1. 
> Do I
> need to use a eval() or something?
>
> Thanks in advance,
> » Michael Krisher
>   [EMAIL PROTECTED]
>
>
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Re: [PHP] calculating US holidays

2002-04-15 Thread Steve Cayford

You could port Date::Calc from perl... or just call a perl script to do 
the calculation. Probably the easiest way to figure out Easter (for the 
extreme example) that I can think of.

-Steve

On Monday, April 15, 2002, at 03:05  PM, Tom Beidler wrote:

> That's fine for fixed dates, and I have that figured out, but presidents
> day, memorial day, labor day and thanksgiving fall on different days 
> each
> year. Presidents day is the third monday of February. How do I calculate
> that?
>
>> From: "Tyler Longren" <[EMAIL PROTECTED]>
>> Organization: Captain Jack Communications
>> Date: Mon, 15 Apr 2002 14:48:15 -0500
>> To: "Tom Beidler" <[EMAIL PROTECTED]>, "php list"
>> <[EMAIL PROTECTED]>
>> Subject: Re: [PHP] calculating US holidays
>>
>> > $month = date("m");
>> $day = date("d");
>> if ($month == "10" && $day == "31") {
>> print "It's Halloween!";
>> }
>> ?>
>>
>> you could do something similar to that.
>>
>> tyler
>>
>> - Original Message -
>> From: "Tom Beidler" <[EMAIL PROTECTED]>
>> To: "php list" <[EMAIL PROTECTED]>
>> Sent: Monday, April 15, 2002 2:17 PM
>> Subject: [PHP] calculating US holidays
>>
>>
>>> I need to calculate the date to see if it's a holiday, i.e. is today
>>> presidents day, which happens to be the third monday of february. Can
>>> someone point me to some code that can do that?
>>>
>>>
>>> --
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>>>
>>>
>>
>>
>
>
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Re: [PHP] Please Help

2002-04-16 Thread Steve Cayford

Check out the bottom of each message...

-Steve

On Tuesday, April 16, 2002, at 10:43  AM, Omland Christopher m wrote:

> Can anyone tell me how to unsubscribe. There is just a bit too much 
> volume
> for me.
> Thanks.
> -Chris
>
>
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Re: [PHP] Array Question

2002-04-18 Thread Steve Cayford


On Thursday, April 18, 2002, at 04:38  PM, Jason Lam wrote:

> $arr2 is a 2d array.
>
> $arr1[0] = 1;
> $arr1[1] = 10;
> $arr2[0] = $arr1;
> print $arr2[0][1];
>
> Result will be 10
>
> But,
>
> $arr1[0] = 1;
> $arr1[1] = 10;
> $arr2[0] = $arr1;
> $arr3 = each($arr2);
> print $arr3[1];

What are you expecting? Check out the documentation for each() at 
http://www.php.net/manual/en/function.each.php

At this point $arr3 should look like this (I think, but try 
print_r($arr3) to be sure):

{
0 => 0,
1 => array (
0 => array (
0 => 1,
1 => 10
)
),
key => 0,
value => array (
0 => array (
0 => 1,
1 => 10
)
)
}

>
> Result is not 10. So, function "each" is not taking the whole $arr2[0]
> out..
>
> My question is what function should I use to iterate array elements 
> (with
> arrays in it)?
>

How about foreach()?

-Steve

> Jay
>
>
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Re: [PHP] Cannot add header information

2002-04-29 Thread Steve Cayford


On Monday, April 29, 2002, at 12:02  PM, Bo Pritchard wrote:

> I know without the accompanying code there's no way to help me...But 
> without
> having to get real specific what does the following message tell me is
> wrong?
>
> Thanks
>
> Warning: Cannot add header information - headers already sent by (output
> started at /home/omnidevi/omnidevices-www/s-cart/form.phtml:4) in
> /home/omnidevi/omnidevices-www/s-cart/shop-head.phtml on line 44
>

In shop-head.phtml on line 44 you're apparently trying to do something 
that sends an http header (like starting a session, setting a cookie, 
etc.), however, all http headers must be sent before any html is goes 
out and the error says html output already started in form.phtml on line 
4.

Note this is about http headers which are different than the html  
stuff.

-Steve


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Re: [PHP] parse error

2002-04-30 Thread Steve Cayford

Looks like your problem is on line 24. See below.

-Steve

On Tuesday, April 30, 2002, at 12:48  PM, Jule wrote:

> Hey guys, i'm writing this guestbook script for my site, and i'm 
> getting a
> random parse error where i think everything is normal, it gives it on 
> line 26
> which is
>
> echo "Your entry will be posted!";
>
> my script follows
>
> Jule
>
> --SCRIPT--
>
>$Guestbook["dateadd"] = date("F j, Y H:i:s");
>   $Guestbook["name"] = trim($Guestbook["name"]);
>   $Guestbook["town"] = trim($Guestbook["town"]);
>   $Guestbook["email"] = trim($Guestbook["email"]);
>   $Guestbook["website"] = trim($Guestbook["website"]);
>   $Guestbook["favsong"] = trim($Guestbook["favsong"]);
>   $Guestbook["comments"] = trim($Guestbook["comments"]);
>   $Guestbook["mailinglist"] = trim($Guestbook["mailinglist"]);
>
>   $Host = "localhost";
>   $User = "";
>   $Password = "*";
>   $DBName = "blindtheory";
>   $TableName = "guestbook";
>   $TableName2 = "mailinglist";
>   $Pattern = ".+@.+..+";
>   $Pattern2 = "(http://)?([^[:space:]]+)([[:alnum:]\.,-_?/&=])";
>
>   $Link = mysql_connect ($Host, $User, $Password);
>   $Query = "INSERT into $TableName values('0', '$Guestbook[dateadd]',
> '$Guestbook[name]', '$Guestbook[town]', '$Guestbook[email]',
> '$Guestbook[website]','$Guestbook[favsong]', '$Guestbook[comments]')";
>   $Link2 = mysql_connect ($Host, $User, $Password);
>   $Query2 = "INSERT into $TableName2 values('0', '$Guestbook[dateadd]',
> '$Guestbook[name]', '$Guestbook[email]');

You're missing a final double quote here ^.

>
>   if (mysql_db_query ($DBName, $Query, $Link)) {
>   echo "Your entry will be added";
>   } else {
>   echo "There was an error in during the posting, please contact 
> href=\"mailto:[EMAIL PROTECTED]\";>me and I will fix the
> problem.";
>   }
>   
>   if (isset($Guestbook[mailinglist])) {
>
>   if (mysql_db_query ($DBName, $Query2, $Link2)) {
>   echo "Your e-mail address was sucessfully added to 
> our mailinglist";
>   } else {
>   echo "There was an error in during the posting, please 
>contact  href=\"mailto:[EMAIL PROTECTED]\";>me and I will fix the
> problem.";
>   }
>   }
>   mysql_close ($Link);
>   mysql_close ($Link2);
> ?>
> --
> Jule Slootbeek
> [EMAIL PROTECTED]
> http://blindtheory.cjb.net
>
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Re: [PHP] contents of fetch array into a string

2002-03-06 Thread Steve Cayford


On Wednesday, March 6, 2002, at 12:13  PM, Kris Vose wrote:

> I have a problem with reading the contents of $r[1] into a string 
> called = $mail.  Does anyone have any suggestions?  Any help is greatly 
> appreciated.
>
> Kris Vose
>
>
> if ($czero != "")
> {
>
> $t = mysql_query("SELECT * FROM AddExisting");

$t is now a resource variable, not an array. If you want to know the 
number of rows returned
use mysql_num_rows($t) not count($t)

>
> $number_of_customers = count($t);
>
> while($r = mysql_fetch_array($t))

Note: you'll loop through this section $number_of_customer times, 
pulling one table row into $r
each time through.

> {
> extract($r);

Why are you using extract here?

>   for ($i = 0; $i<$number_of_customers; $i++)

You're looping through $number_of_customers columns on each row in the 
table. Is this what you want?  Is your table really $number_of_customers 
wide?

>   {
> echo $r[1].", ";
Do you mean $r[$i] ?
>   }
>
> }
>
Now you've dropped out of both loops. $r will be holding the last row 
you retrieved from the database table.

> $mail = ' ';
> foreach($r[1] as $mailline)

$r[1] is not an array here. What are you trying to do?

> {
> $mail.=$mailline;
> }
>
> }
>


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Re: [PHP] Perl and PHP

2002-03-06 Thread Steve Cayford

If you're running a perl script on the command line you would use
/path/to/perl/script.pl value1 value2 value3 ...

In your perl script $ARGV[1] should hold value1, $ARGV[2] should hold 
value2, etc.

-Steve

On Wednesday, March 6, 2002, at 02:48  PM, [EMAIL PROTECTED] wrote:

> I need to pass a variable to Perl from a PHP script.  I am somewhat know
> PHP but I do NOT know perlI am running my perl script off of the
> command line.  I tried /path/to/somewhere/script.pl?var=var
>
> but it did not work.
>
> I would love any help you could provide me
>
> Thanks,
> Michael
>
>
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Re: [PHP] Perl and PHP

2002-03-06 Thread Steve Cayford

If the value you want to send back to php is an integer you could just 
say

exit $number;

from perl and that will be put into your $var.

Otherwise use exec() instead of system.

exec("/path/to/perlscript.pl", $scriptOutputArray, $scriptExitValue);

Anything that gets printed from the perl script should show up in the 
$scriptOuputArray.

-Steve

On Wednesday, March 6, 2002, at 04:19  PM, Michael Hess wrote:

>
>
> Thanks,
>
> Ok,
> I got it to work.now how do I take a varaible back into PHP
>
>
> I do a $var = system("/path/to/cgi $varforCGI")
> and in the CGI script I do a print $varforPHP
>
> however it prints the varforPHP to the broswer, I need it saved it var
>
> Any (more) help would be great!!
>
> Michael
>
> On
> Wed, 6
> Mar 2002, Steve Cayford wrote:
>
>> If you're running a perl script on the command line you would use
>> /path/to/perl/script.pl value1 value2 value3 ...
>>
>> In your perl script $ARGV[1] should hold value1, $ARGV[2] should hold
>> value2, etc.
>>
>> -Steve
>>
>> On Wednesday, March 6, 2002, at 02:48  PM, [EMAIL PROTECTED] wrote:
>>
>>> I need to pass a variable to Perl from a PHP script.  I am somewhat 
>>> know
>>> PHP but I do NOT know perlI am running my perl script off of the
>>> command line.  I tried /path/to/somewhere/script.pl?var=var
>>>
>>> but it did not work.
>>>
>>> I would love any help you could provide me
>>>
>>> Thanks,
>>> Michael
>>>
>>>
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>>>
>>
>


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Re: [PHP] Driving me nuts, need one second of your time

2002-03-15 Thread Steve Cayford

It looks like your sql query failed, so the result is invalid.

Assuming that this is really the whole script, then you're counting on 
the mysql_db_query function to open a connection to the database "db" 
using the default connection values which (according to the manual) are 
host: localhost, user: (whatever user is running the script, probably 
'www', 'apache', or 'nobody' depending on your system), and password: 
(blank). Do you really have your mysql database setup to allow access 
with these defaults?

I'd recommend trying mysql_connect() first, then if that works continue 
on to the mysql_query(). At least you'll know better which aspect failed.

-Steve

On Friday, March 15, 2002, at 05:14  PM, cosmin laslau wrote:

>  $query = "SELECT * from mytable";
> $result = mysql_db_query("db", $query);
>
> while ($myarray = mysql_fetch_array($result))
> {
> $title = $myarray[title];
> echo "$title";
> }
> ?>
>
> Can someone PLEASE tell me why the coding above gives the following 
> error:
>
>
> Warning: Supplied argument is not a valid MySQL result resource in 
> /var/web/somesite.com/html/index.php on line 5
>
>
> It's absurd. It's driving me nuts. I'm about to introduce my computer 
> to the pavement 40 feet below.
>
> Thanks in advance for whoever sees what I am sure is a glaring and 
> obvious flaw in the coding. I've been looking at it for an hours and 
> just can't get anything from where I'm standing, maybe a different 
> perpective will help.
>
>
>
> _
> Get your FREE download of MSN Explorer at 
> http://explorer.msn.com/intl.asp.
>
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Re: [PHP] Where is php.ini on Mac OS X?

2002-03-27 Thread Steve Cayford

On mine it's in /usr/local/lib, but I seem to recall discussion earlier 
about some OSX installations lacking the ini file.

-Steve

On Tuesday, March 26, 2002, at 04:53  PM, Chuck "PUP" Payne wrote:

> Can some one please tell me where php.ini is located on Mac OS X?
>
> Thanks,
>
> Chuck Payne
> Magi Design and Support
>
>
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Re: [PHP] Parsing error

2002-04-01 Thread Steve Cayford

It looks like you've got a closing curly bracket "}" for your switch 
statement, but not for your while statement.

Also, have you read up on the switch statement? Keep in mind that if 
something matches your case 1, it will also fall through and execute 
case 2, case 3, etc... unless you include break statements. e.g. try 
this:

switch ($row)
{
   case 1:
 {
   $vignette = $data[$c];
   break;
 }
   case 2:
 {
   $photo = $data[$c];
   break;
 }
...
...
...

-Steve


On Monday, April 1, 2002, at 10:44  AM, news.php.net wrote:

> Hi,
>
> I'm real new in php and trying to read a txt file
>
> this is my code :
>
>   $row = 1;
>  $fp = fopen ("lecteurs.txt","r");
>   while ($data = fgetcsv ($fp, 1000, ";"))
>   {
> $num = count ($data);
> $row++;
> for ($c=0; $c<$num; $c++)
> switch ($row)
>  {
>  case 1 :
>   {
>   $vignette = $data[$c];
>   }
>  case 2 :
>{
>$photo= $data[$c];
>}
>case 3 :
>{
>$marque= $data[$c];
>}
>   case 4 :
>{
>$nom = $data[$c];
>}
>   case 5 :
>{
>$pdfproduit= $data[$c];
>}
>   case 6 :
>{
>$commprod = $data[$c];
>}
> }
>   fclose ($fp);
>   echo $vignette;
>   echo $photo;
>   echo $marque;
>   echo $nom;
>   echo $pdfproduit;
>   echo $commprod;
> ?>
>
> and when i run it i get : Parse error: parse error in essai.php on line 
> 45
>
> here' my txt file (just 1 line for probe)
>
> petitimage;photo;marque;Nom;lepdf;commproduit
>
> Can someone help me to go thru
>
> TIA
>
> Hubert
>
>
>
>
>
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Re: [PHP] session

2002-04-02 Thread Steve Cayford

Try putting session_start() in your retief.php script as well as 
piet.php.

-Steve

On Wednesday, April 3, 2002, at 01:06  PM, R. Lindeman wrote:

> okay i've posted something before here are my scripts either you tell me
> what i do wrong or i'll go beserk
>
> you can check the outcome of the code on the following adress
>
> http://www.filenexus.com/piet.php
>
> here's the code for piet.php :
>
> 
> // open session and begin registering values
>
> session_start();
>
> session_register("fubar1");
> session_register("fubar2");
> session_register("fubar3");
>
> $fubar1="Remko";
>
> $fubar2="Lindeman";
>
> $fubar3="I3M1I";
>
> header("Location: retief.php?".SID);
>
> ?>
>
> here's the code for retief.php :
>
> 
> // begin the retrieval of propagated values
>
> $remko   = $HTTP_SESSION_VARS["fubar1"];
> $lindeman = $HTTP_SESSION_VARS["fubar2"];
> $klas   = $HTTP_SESSION_VARS["fubar3"];
>
> echo $remko;
> echo $lindeman;
> echo $klas;
>
> ?>
>
>
> the problem is that the values don't get registered
>
>
>
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Re: [PHP] Loop, array, life....

2002-04-03 Thread Steve Cayford

On your first email you loop through all the news in the $newsfetch 
resource. I think you need to then do a mysql_data_seek() call against 
that to reset it for the next time through the loop. Otherwise you'll 
just get a null result because you're already at the end of the data.

-Steve

On Wednesday, April 3, 2002, at 02:30  PM, Gerard Samuel wrote:

> (made a correction in sudo code)
>
> Ok, this one is breaking my back all day.
> First some sudo-code ->
>
> fuction email_to_user() {
>  $sql = 'select distinct(email) from user';
>  $emailfetch = mysql_query($sql);
>
>  $sql = 'select news, date from news order by sid desc limit 10';
>  $newsfetch = mysql_query($sql);
>
>  while ($data = mysql_fetch_row($emailfetch)) {
>  $container = array();
>  $container[] = '';
>
>  $message = 'Have a nice day';
>  $container[] = $message;
> // PROBLEM IN THIS WHILE LOOP MAYBE //
>  while (list($news, $date) = mysql_fetch_row($newsfetch)) {
>  $container[] = '' . $news . '';
>  $container[] = '';
>  }
>  $container[] = '';
>  $message2 = '';
>  foreach($container as $foo) {
>  $message2 .= $foo;
>  }
>  mail(Send mail to $data[0], $message2);
>  unset($container);
>  }
> }
>
> Basically it grabs all the user's email addresses, then loop them.
> On each loop grab all news items.
> Then emails results to the user and moves on to the next user.
>
> Im running this on my test box, that only has two users, but the 2nd
> user never gets the expected results.
> The first user get the message and the news.
> The second only gets the message.
> The code structure is pretty much unchanged from a working example till
> I started using $container to hold array elements.
>
> Could anyone see bad logic in the above code??
> Thanks
>
>
>
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Re: [PHP] Classes??

2002-04-04 Thread Steve Cayford

Yeah, sure.

-Steve

On Thursday, April 4, 2002, at 10:42  AM, Gerard Samuel wrote:

> Maybe a simple question.
> But can one file contain 2 or more classes??
> Thanks
>
>
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Re: [PHP] English/Arabic Mysql problem...

2002-04-04 Thread Steve Cayford

Are you using unicode? I don't know the answer for you - maybe check the 
mysql site - but I'd be interested in hearing an answer as well if 
anyone has one.

-Steve

On Thursday, April 4, 2002, at 12:11  AM, Dhaval Desai wrote:

> Hello people,
>
> I am making a bilingual website English/Arabic. I am facing some 
> problem with this. The problem is that I am not able to insert arabic 
> lanaguage characters ionto Mysql databse. It gets junk 
> characters...when I try to display it...
>
> Thank You
>
> Best Regards,
> Dhaval Desai
>
>
>
> _
> Send and receive Hotmail on your mobile device: http://mobile.msn.com
>
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Re: [PHP] objects handling objects

2002-04-04 Thread Steve Cayford


On Thursday, April 4, 2002, at 04:21  PM, Erik Price wrote:

> I looked in the manual, but didn't see anything about this.  I've read 
> that PHP isn't a true object-oriented language, but rather simulates 
> elements of object-oriented programming.  Can I write a class that 
> performs operations and manipulates objects?  Can objects be placed 
> into arrays etc?
>
> Erik
>
The real hindrance I've come up against is that you can't do chained 
method calls, otherwise objects perform pretty well as expected. Keep in 
mind the difference between passing by reference and by copy, though, or 
you'll find yourself updating a copy of an object somewhere instead of 
the original.

-Steve

Here's some code I was just playing around with:

method1();
 }
 function otherMethod2($object) {
 return $object->method2();
 }
}

$a = new classA;
$b = new classB;
$c = new classC;
$d = new classD;

print("a1: " . $a->method1() . "\n");
print("a2: " . $a->method2() . "\n");
print("b1: " . $b->method1($a->method1()) . "\n");
print("b2: " . $b->method2($a->method2()) . "\n");
print("d1: " . $d->otherMethod1($b) . "\n");
print("d2: " . $d->otherMethod2($c->echoMethod($a)) . "\n");

// this next line would generate a parse error
// print("c:a1: " . $c->echo($a)->method1() . "\n");

// this outputs:
// a1: 1
// a2: 2
// b1: 11
// b2: 22
// d1: 10
// d2: 2

?>

>
>
>
>
> 
>
> Erik Price
> Web Developer Temp
> Media Lab, H.H. Brown
> [EMAIL PROTECTED]
>
>
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Re: [PHP] objects handling objects

2002-04-04 Thread Steve Cayford

Doh, typo:

// this next line would generate a parse error
// print("c:a1: " . $c->echo($a)->method1() . "\n");

should read

// this next line would generate a parse error
// print("c:a1: " . $c->echoMethod($a)->method1() . "\n");


On Thursday, April 4, 2002, at 05:25  PM, Steve Cayford wrote:

>
> On Thursday, April 4, 2002, at 04:21  PM, Erik Price wrote:
>
>> I looked in the manual, but didn't see anything about this.  I've read 
>> that PHP isn't a true object-oriented language, but rather simulates 
>> elements of object-oriented programming.  Can I write a class that 
>> performs operations and manipulates objects?  Can objects be placed 
>> into arrays etc?
>>
>> Erik
>>
> The real hindrance I've come up against is that you can't do chained 
> method calls, otherwise objects perform pretty well as expected. Keep 
> in mind the difference between passing by reference and by copy, 
> though, or you'll find yourself updating a copy of an object somewhere 
> instead of the original.
>
> -Steve
>
> Here's some code I was just playing around with:
>
> 
> class classA {
> function method1() {
> return 1;
> }
> function method2() {
> return 2;
> }
> }
>
> class classB {
> function method1($num = 0) {
> return $num + 10;
> }
> function method2($num = 0) {
> return $num+20;
> }
> }
>
> class classC {
> function echoMethod($object) {
> return $object;
> }
> }
>
> class classD {
> function otherMethod1($object) {
> return $object->method1();
> }
> function otherMethod2($object) {
> return $object->method2();
> }
> }
>
> $a = new classA;
> $b = new classB;
> $c = new classC;
> $d = new classD;
>
> print("a1: " . $a->method1() . "\n");
> print("a2: " . $a->method2() . "\n");
> print("b1: " . $b->method1($a->method1()) . "\n");
> print("b2: " . $b->method2($a->method2()) . "\n");
> print("d1: " . $d->otherMethod1($b) . "\n");
> print("d2: " . $d->otherMethod2($c->echoMethod($a)) . "\n");
>
> // this next line would generate a parse error
> // print("c:a1: " . $c->echo($a)->method1() . "\n");
>
> // this outputs:
> // a1: 1
> // a2: 2
> // b1: 11
> // b2: 22
> // d1: 10
> // d2: 2
>
> ?>
>
>>
>>
>>
>>
>> 
>>
>> Erik Price
>> Web Developer Temp
>> Media Lab, H.H. Brown
>> [EMAIL PROTECTED]
>>
>>
>> -- PHP General Mailing List (http://www.php.net/)
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>>
>
>
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Re: [PHP] Set Global Variables

2002-04-11 Thread Steve Cayford

Keep in mind that there is no common memory space for all the users. 
Each request is being handled by a different process with its own memory 
space that's forked off of the original web server. (Assuming apache, I 
don't know how IIS works).

-Steve


On Thursday, April 11, 2002, at 01:52  PM, [EMAIL PROTECTED] wrote:

> If I do that, won't the data that I fetch be duplicated in memory for 
> each
> user?  I'm trying to have just one space in memory allocated for these
> variables, sort of like the HTTP_SERVER_VARS.  It seems that even if I 
> do an
> autoprepend that file is going to run each time a user visits the site 
> and a
> new memory space will be created for that variable.  What I'm asking is 
> how
> to set those constant server-wide global variables.
>
> Thanks
> Roger
>
> -Original Message-
> From: Rasmus Lerdorf [mailto:[EMAIL PROTECTED]]
> Sent: Thursday, April 11, 2002 10:15 AM
> To: [EMAIL PROTECTED]
> Cc: [EMAIL PROTECTED]
> Subject: Re: [PHP] Set Global Variables
>
>
> Set up an auto_prepend file that either simply sets these constant 
> globals
> ot fetches them from the DB.
>
> -Rasmus
>
> On Thu, 11 Apr 2002 [EMAIL PROTECTED] wrote:
>
>> Is there a way to set a global variable that all users can access?  I 
>> have
>> some values in a database that I'm reading at the beginning of the 
>> session
>> and carrying through out the session.  The thing is, those values are 
>> the
>> same for every user and instead of having a memory space for each users
>> values I'd rather just use one memory space for all users.
>>
>> Thanks
>>
>> Roger Ramirez
>> Web Developer
>> LifeFiles.com Inc.
>>
>>
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Re: [PHP] How to sort by 3rd row in an 2d-Array

2002-04-12 Thread Steve Cayford

Or try the usort() function?



-Steve

On Friday, April 12, 2002, at 12:26  PM, Kevin Stone wrote:

> This is a terribly inneficient way of handling your situation but I 
> believe
> it would work.
>
> // First we're going to make a list of all company values
> for($i=0; $i {
> $company_values[] = $myarray[$i][company];
> }
>
> // Then we're going to order the array using a standard function.
> natsort($company_values);
>
> // Then we're going to go through $myarray  looking for each matching
> // value and build a new array based on the order of the sorted list.
> for($i=0; $i {
> for ($j=0; $j= {
> if ($company_values[$i] == $myarray[$j][company])
> {
> $mynewarray[] = $myarray[$j];
> }
> }
> }
>
> // bada bing bada boom you gots your company sorted list... I think
> $myarray = $mynewarray;
>
> I wrote this off the top of my head in like three minutes so any and all
> corrections would be most welcome.  ;)
> -Kevin
>
> - Original Message -
> From: "SED" <[EMAIL PROTECTED]>
> To: <[EMAIL PROTECTED]>
> Sent: Friday, April 12, 2002 10:25 AM
> Subject: [PHP] How to sort by 3rd row in an 2d-Array
>
>
>> Hi,
>>
>> I'm trying to sort an array like following
>>
>> myArray[1][firstname] = "Joe";
>> myArray[1][lastname] = "Smith";
>> myArray[1][company] = "Bullock";
>> myArray[1][email] = "[EMAIL PROTECTED]";
>> myArray[2][firstname] = "Jim";
>> myArray[2][lastname] = "Cords";
>> myArray[2][company] = "Jamen";
>> myArray[2][email] = "[EMAIL PROTECTED]";
>> etc...
>>
>> by the company name. How can I do it?
>>
>> I found the solution on php.net
>> (http://www.php.net/manual/en/function.array-multisort.php) but it 
>> sorts
>> only the first row:
>>
>> foreach ($myArray as $val) {
>> $sortarray[] = $val['nafn'];
>> echo $val['stadur'];
>> }
>> array_multisort($myArray, $sortarray);
>>
>> Thanks in advance!
>>
>> SED
>>
>>
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>>
>
>
>
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Re: [PHP] Wanting a better understanding of classes in PHP...

2002-04-12 Thread Steve Cayford

Well, start here probably:

http://www.php.net/manual/en/language.oop.php

-Steve

On Friday, April 12, 2002, at 12:50  PM, Chuck "PUP" Payne wrote:

> Hi, I was up on freshmeat and I saw a TON of php classes. I like to 
> know how
> can I use them? And is a class a bit of code that you are always using? 
> If
> so, how can I create my own classes.
>
> Chuck "PUP" Payne
> Sr. System Administrator
> GDI Engineering, Inc.
> 2075-E West Park Place Blvd.
> Stone Mountain, GA 30087
> ---
> (678) 476-0747 ext. 18 (Phone)
> (770) 498-1590 (Fax)
> (404) 451-3579 (Mobile)
> (800) 631-1371 (Alpha Page)
> [EMAIL PROTECTED]  (Text Pager)
> [EMAIL PROTECTED] (E-Mail)
> ---
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> http://www.gditelecommunication.com
> http://www.softmeta.com
>
>
>
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Re: [PHP] "What date/time was it 7 hours ago"?

2002-04-15 Thread Steve Cayford

You could convert it to a timestamp using strtotime(), subtract 7*60*60, 
then convert it back to a string with strftime().

-Steve

On Monday, April 15, 2002, at 09:50  AM, Torkil Johnsen wrote:

> "What date/time was it 7 hours ago"?
>
> I'm just trying to make a log using mysql/php
> This log is kinda supposed to show the time, local to the user.
>
> So I store datetime on the format -MM-DD HH:MM:SS in a datetime 
> field.
> Using the php date function to get the date.
>
> This will store the time that right then ON THE SERVER.
>
> Now, when fetching data from the mysql table, I want to display what the
> time WAS, LOCALLY (where I'm at, not the server) when the log entry was
> made.
>
> So. How do I make a function that takes in -MM-DD HH:MM:SS and 
> spits out
> the -MM-DD HH:MM:SS minus... for instance, 7 hours?
>
> "What date/time was it 7 hours ago"?
>
>
>
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Re: [PHP] require & include

2001-10-22 Thread Steve Cayford

http://www.php.net/manual/en/function.require.php

require() pulls in the target file when the source file is 
parsed/compiled, include() pulls in the target file when the source file 
is executed. So an include() nested in an if statement will only be 
included if the if statement evaluates to true, a require() nested in an 
if statement will be included regardless.

Also check out include_once() and require_once().

-Steve

On Monday, October 22, 2001, at 05:15  PM, jtjohnston wrote:

> Coverting from perl ...
> What's the differencw between require and include? Where, when, why?
>
> John?
>
> 
> Email post & reply always appreciated
>
>
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Re: [PHP] Re: require & include

2001-10-23 Thread Steve Cayford

So both include() and require() *are* subject to conditional statements 
in the code? Guess I missed that.

Thanks.

-Steve

On Tuesday, October 23, 2001, at 01:00  AM, Rasmus Lerdorf wrote:

> That's outdated.  The only difference today is that if a file can't be
> included/required for some reason it is a fatal error with require and a
> warning with include.
>
> -Rasmus
>
> On Tue, 23 Oct 2001, Jason G. wrote:
>
>>  From the manual:
>>
>> Unlike include(), require() will always read in the target file, even 
>> if
>> the line it's on never executes. If you want to conditionally include a
>> file, use include(). The conditional statement won't affect the 
>> require().
>> However, if the line on which the require() occurs is not executed, 
>> neither
>> will any of the code in the target file be executed.
>>
>> Similarly, looping structures do not affect the behaviour of require().
>> Although the code contained in the target file is still subject to the
>> loop, the require() itself happens only once.
>>
>>
>>
>>
>>
>> At 08:48 AM 10/23/2001 +0900, Yasuo Ohgaki wrote:
>>> Jtjohnston wrote:
>>>
 Coverting from perl ...
 What's the differencw between require and include? Where, when, why?
>>>
>>>
>>> I forgot from which version, but current PHP's require/include works 
>>> the
>>> same way except
>>>
>>> - require() raise fatal error, if it can't find file
>>> - include() raise warning, if it can't find file
>>>
>>>
>>> requrie_once()/include_once() works almost the same as 
>>> require()/include()
>>> except they include file only once. (Hash table is used to determine 
>>> if
>>> files are included or not)
>>>
>>> See also get_{required|included}_files()
>>>
>>> --
>>> Yasuo Ohgaki
>>>
>>>
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>>> [EMAIL PROTECTED]
>>>
>>
>
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[PHP] Confused object in session variable.

2001-10-25 Thread Steve Cayford

Well, it's probably me that's confused. I have an authenticate() 
function which should start a session and if the user is not logged in 
then show the login screen otherwise return after storing and 
registering a user object in a session variable. This object has 
accessor methods to get the login name, access level, etc... This seems 
to work okay--within the authenticate function I can access the object 
in the HTTP_SESSION_VARS array. But as soon as I return to the main 
script it's gone. I must be doing something wrong with the scoping, but 
I can't see what. Any thoughts? Here's the code:

getLogin() . "\n");
print("From index HTTP_SESSION_VARS['ucAuthoUser']: ");
print($HTTP_SESSION_VARS['ucAuthoUser']->getLogin() . "\n"); //this 
is line 12

?>

isValid()) {
 return;
 } else {
 if (isset($HTTP_POST_VARS['authoSubmit'])) {
 $HTTP_SESSION_VARS['ucAuthoUser'] =&
 new 
UcAuthoUser($HTTP_POST_VARS['authoLogin'],$HTTP_POST_VARS['authoPword']);
 if ($HTTP_SESSION_VARS['ucAuthoUser']->isValid()) {
 session_register('ucAuthoUser');
 $testUser = $HTTP_SESSION_VARS['ucAuthoUser'];
 print("From authenticate testUser: " . 
$testUser->getLogin() . "\n");
 print("From authenticate 
HTTP_SESSION_VARS['ucAuthoUser']: ");
 print($HTTP_SESSION_VARS['ucAuthoUser']->getLogin() . 
"\n");
 return;
 }
 }
 showLogin($appName);
 }
}

/* more functions and the class declaration snipped */

?>

Here's what I get when I login as 'steve' with a good password:

 From authenticate testUser: steve
 From authenticate HTTP_SESSION_VARS['ucAuthoUser']: steve
 From index testUser: steve
 From index HTTP_SESSION_VARS['ucAuthoUser']:
Fatal error: Call to a member function on a non-object in 
/home/httpd/html/ucdamage/index.php on line 12

Note the testUser works in both instances, the session var only works 
inside the function.

-Steve


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Re: [PHP] Confused object in session variable.

2001-10-25 Thread Steve Cayford

An update...
I had register_globals turned off. Now that I've turned register_globals 
on and changed my references from $HTTP_SESSION_VARS['ucAuthoUser'] to 
$ucAuthoUser it works as I expect.

I'm not real happy with having register_globals on, though, and I'd like 
to figure out what I was doing wrong.

Thanks for any suggestions.

-Steve

On Thursday, October 25, 2001, at 12:07  PM, Steve Cayford wrote:

> Well, it's probably me that's confused. I have an authenticate() 
> function which should start a session and if the user is not logged in 
> then show the login screen otherwise return after storing and 
> registering a user object in a session variable. This object has 
> accessor methods to get the login name, access level, etc... This seems 
> to work okay--within the authenticate function I can access the object 
> in the HTTP_SESSION_VARS array. But as soon as I return to the main 
> script it's gone. I must be doing something wrong with the scoping, but 
> I can't see what. Any thoughts? Here's the code:
>
>  /* - */
> /* index.php */
> /* - */
>
> require_once('ucautho/ucautho.inc');
>
> authenticate();
>
> print("From index testUser: " . $testUser->getLogin() . "\n");
> print("From index HTTP_SESSION_VARS['ucAuthoUser']: ");
> print($HTTP_SESSION_VARS['ucAuthoUser']->getLogin() . "\n"); //this 
> is line 12
>
> ?>
>
>  /* --- */
> /* ucautho/ucautho.inc */
> /* --- */
>
> function authenticate($appName="") {
> global $HTTP_SESSION_VARS, $HTTP_POST_VARS;
> global $testUser;
> session_name("UCAutho");
> session_start();
>
> if (isset($HTTP_SESSION_VARS['ucAuthoUser']) && 
> $HTTP_SESSION_VARS['ucAuthoUser']->isValid()) {
> return;
> } else {
> if (isset($HTTP_POST_VARS['authoSubmit'])) {
> $HTTP_SESSION_VARS['ucAuthoUser'] =&
> new 
> UcAuthoUser($HTTP_POST_VARS['authoLogin'],$HTTP_POST_VARS['authoPword']);
> if ($HTTP_SESSION_VARS['ucAuthoUser']->isValid()) {
> session_register('ucAuthoUser');
> $testUser = $HTTP_SESSION_VARS['ucAuthoUser'];
> print("From authenticate testUser: " . 
> $testUser->getLogin() . "\n");
> print("From authenticate 
> HTTP_SESSION_VARS['ucAuthoUser']: ");
> print($HTTP_SESSION_VARS['ucAuthoUser']->getLogin() . 
> "\n");
> return;
> }
> }
> showLogin($appName);
> }
> }
>
> /* more functions and the class declaration snipped */
>
> ?>
>
> Here's what I get when I login as 'steve' with a good password:
>
> From authenticate testUser: steve
> From authenticate HTTP_SESSION_VARS['ucAuthoUser']: steve
> From index testUser: steve
> From index HTTP_SESSION_VARS['ucAuthoUser']:
> Fatal error: Call to a member function on a non-object in 
> /home/httpd/html/ucdamage/index.php on line 12
>
> Note the testUser works in both instances, the session var only works 
> inside the function.
>
> -Steve
>
>
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Re: [PHP] function names

2001-10-25 Thread Steve Cayford


On Thursday, October 25, 2001, at 02:08  PM, Martín Marqués wrote:

> On Jue 25 Oct 2001 15:36, you wrote:
>> Hello php-general,
>>
>>   I have such code:
>>
>>   class A
>>   {
>> var $xxx;
>>
>> function print()
>> {
>>  echo $xxx;
>
> $xxx is internal to the print function. Instead you need $this->xxx 
> which
> will give you the value of the $xxx of the A class.
>
>> }
>>   }
>>
>>   And that's what I get:
>> "Parse error: parse error, expecting `T_STRING' in xxx.php on line 
>> nn"
>>
>>   Php doesn't let any function or class member have a name which is
>>   already "used" by another function (or only function from library),
>>   am I right? Or maybe "print" has special status. Maybe that's
>>   because print() is actually not a function? Can anyone tell me
>>   something about that, please?
>
> Th print function of PHP has nothing to do with this, just because 
> print is
> internal to the A class, and has nothing with the PHPs internal print
> function.

Hmm. I think you're wrong here. I made this test script:
a = "hello";
 }

 function print() { // this is line 10
 echo $this->a;
 }
}

$obj = new test;
$obj->print();

?>
Which gives this:

Parse error: parse error, expecting `T_STRING' in 
/home/httpd/html/ucdamage/test.php on line 10

If I change the name of the print() method it works okay.

-Steve


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Re: [PHP] Setting A MySQL Column to NO DUPLICATES

2001-10-25 Thread Steve Cayford

I think you want UNIQUE.

create table sometable (
somecolumn char(40),
someothercolum int,
unique somecolumn
)

...something like that. I haven't messed with it much.

-Steve

On Thursday, October 25, 2001, at 04:31  PM, Jeff Gannaway wrote:

> Does anyone know how to set a column in a MySQL table to disallow any
> duplicate entries.
>
> I've already got a PRIMARY KEY defined for this table.  I tried setting 
> the
> other field to a MUL KEY, hoping that it would not permit duplicates, 
> but
> it does.
>
> I've searched the MySQL docs without success. I wish their site was a 
> good
> as PHP.net.
>
> Later,
> Jeff Ganaway
> ___
>
> Save 15% on 2002 Calendars and Holiday Cards!
>
> http://www.AvantGifts.com
> Discount Code: hopper22ct
> Offer Good Through October 31, 2001
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>
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Re: [PHP] Re: header("Location:);

2001-10-30 Thread Steve Cayford


On Tuesday, October 30, 2001, at 11:41  AM, Henrik Hansen wrote:

> [EMAIL PROTECTED] (Alberto) wrote:
>
>> Ok, it works fine for me, i can redirection, but I don't want the 
>> browser to
>> load another page, I want to IF A THEN RUN A.PHP, IF B THEN RUN 
>> B.PHP, I
>> know i can use include('a.php') or require('a.php') but I'm sure 
>> theres a
>> way to say php parser to run one or another script without "changing 
>> URL" to
>> the client.
>
> use include as you say or html frames.
>

Yeah, why not
if($A) {
include('a.php');
} elseif ($B) {
include('b.php');
}

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>
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Re: [PHP] Sessions & Header PROBLEM again!

2001-11-02 Thread Steve Cayford

Don't know offhand what the problem is, but a couple thoughts:

1. Why are you using session_name(mysession) instead of 
session_name('mysession')?

2. Assuming you have register_globals on, you're trying to pass $count 
both as a session variable and a post variable. One of these is going to 
get overwritten by the other if I'm not mistaken.

3. Why not have the page 1 form action point directly at page 2 instead 
of being redirected through page 1? Do posted variables follow a 
redirect?

4. If posted variables do follow a redirect then page 1 will see that 
$submit is set and redirect to page 2, which will redirect to page 3, 
which will redirect to page 1, etc... Can you get an infinite 
redirection loop?

5. The form on page 1 includes a hidden count variable, page 2 and page 
3 don't.

What happens when you run this?

-Steve

On Friday, November 2, 2001, at 08:49  AM, Alessandro BOSSI wrote:

> Why in these 3 linked page the var "count"
> does not go through the page if I have the browser sessions disabled???
>
> Where is the mistake?
>
> I see the session_id, changing at every page!
>
> Someone can help me?
> Many thank
> Alessandro Bossi
>
> -
>  session_name(mysession);
> session_start();
> session_register ("count");
>
> if (isset($submit)){
>  header("Location: Seconda.php?".SID);
> }else{
>  $count++;
> }
>
> print "session_encode=>> ".session_encode()."";
>
> ?>
>
> 
>  First Page
> 
>
> First Page
> Hello visitor, you have seen this page  times.
> count = 
> 
>
> 
> 
> 
> 
>
> 
> -
> 
> session_name(mysession);
> session_start();
> session_register('count');
>
> if (isset($submit)){
>  header("Location: Third.php?".SID);
> }
> print "session_encode=>> ".session_encode()."";
>
> ?>
>
> 
> Second Page
> 
>
> Second Page
> Hello visitor, you have seen this page  times.
> count = 
> 
> 
> 
> 
>
> 
> 
> -
> 
> session_name(mysession);
> session_start();
> session_register ("count");
>
>
> if (isset($submit)){
>   header("Location: First.php?".SID);
> }
> print "session_encode=>> ".session_encode()."";
> ?>
>
> 
> 
> 
> Third Page
> 
>
> 
> Third Page
> Hello visitor, you have seen this page  times.
> Count = 
> 
>
> 
> 
> 
>
> 
> 
>
>
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Re: [PHP] very weird PHP behaviour...

2001-11-14 Thread Steve Cayford

Sure sounds like you're hitting this function twice by accident. Are you 
sure you're only calling it once? That would explain why you only get 
one output with the die(), but two without it.

-Steve

On Wednesday, November 14, 2001, at 10:34  AM, Christian Dechery wrote:

> I don't know what it is... but PHP is acting very weird today... or I 
> am becoming crazy... can anyone tell me which one?
>
> I have this function:
> function 
> Upload($source_file,$source_filename,$dest_dir,$allowed_types=array())
> {
> global $upload_errmsg;
>
> if( count($allowed_types) )
> {
>   $file_ext=ArquivoExt($source_filename);
>   $ext_found=in_array($file_ext,$allowed_types);
>   //die("source=$source_filename");
>   //var_dump($ext_found);
>   //die;
>   if( !$ext_found )
>   {
> $upload_errmsg="O arquivo \"$source_filename\" não está entre os 
> tipos autorizados";
> //die("errmsg=$upload_errmsg");
> return false;
>   }
> }
> ... some more code here ..
> return true;
> }
>
> the first crazy behaviour is... I use a test file called something.jpg, 
> ArquivoExt() returns me the extension of a given filename, in this case 
> "jpg". $allowed_types contains "gif","jpg" and "png" so that in_array() 
> should return true... and it does... but get this... if I uncomment 
> that var_dump() guess what it outputs?
>
> bool(true) bool(false)
>
> very weird... how can one var_dump() call gives me two outputs?
> This double output doesn't occur if I uncomment the "die" after the 
> var_dump(). It only outputs bool(true) in that case.
>
> Now for the second weird behaviour... since no matter what, $ext_found 
> will always be false (due to the previous weirdness) it will always 
> enter the IF... but the weird thing is when the caller of Upload() 
> prints $upload_errmsg it only outputs:
>
> O arquivo "" não está entre os tipos autorizados
>
> where is the filename? I've checked it tons of times... If I uncomment 
> the second die, right after the $upload_errmsg assignment, it outputs 
> ok, with the filename.
>
> Now, can someone explain me this VERY WEIRD behaviour??
>
>
>
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Re: [PHP] Anyone ever done this - sort a multi-dimensional array

2001-11-15 Thread Steve Cayford

You can roll your own fairly easily:

function mycmp($a,$b) {
 return strcmp($a[1],$b[1]);
}
usort($array, 'mycmp');

...something like that at any rate.

-Steve

On Thursday, November 15, 2001, at 10:07  AM, Richard S. Crawford wrote:

> Of the type...
>
> $array[0][0] = "!row of c's";
> $array[0][1] = "";
> $array[1][0] = "row of a's";
> $array[1][1] = "";
> $array[2][0] = "a row of b's";
> $array[2][1] = "";
>
>
> such that you sort on the value of the y-column?
>
> In this case, it would end up as:
>
> +--+--+
> | row of a's   |  |
> | a row of b's |  |
> | !row of c's  |  |
> +--+--+
>
> Strange that I've been working with PHP this long and have never come 
> across this problem.
>
>
>
> Sliante,
> Richard S. Crawford
>
> http://www.mossroot.com
> mailto:[EMAIL PROTECTED]
> AIM: Buffalo2K   ICQ: 11646404  Y!: rscrawford
> "It is only with the heart that we see rightly; what is essential is 
> invisible to the eye."  --Antoine de Saint Exupéry
>
> "Push the button, Max!"
>
>
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Re: [PHP] query works in mysql, but not from php

2001-11-19 Thread Steve Cayford

You don't need to put in the final semi-colon when running a query from 
php. Take that out and you should be fine.

-Steve

On Monday, November 19, 2001, at 04:58  PM, Olav Drageset wrote:

> Hi
>
> $sql = "SELECT user FROM persons WHERE user = '$firstName' and domain = 
> '$domainName' ; ";
> $result = mysql_query($sql,$connection ) or die(mysql_error());
>
> Calling above lines from php returns: You have an error in SQL 
> syntax near ';'  at line 1
>
> Issuing the command
>
> SELECT user FROM persons WHERE user = 'fred' AND domain = 
> 'company.net' ;
>
> in mysql give a proper result.
>
> Can anyone explain what might be causing the error???
> regards Olav
>
>
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[PHP] Why are slashes automatically stripped from db result?

2001-12-03 Thread Steve Cayford

Hey all. I'm storing some jpeg images in a mysql database using the PEAR 
classes. Before inserting the image into the db I call addslashes() on 
the data, I was, accordingly, calling stripslashes() on the data after 
pulling the image back out of the database, but the image was getting 
mangled. I finally realized that the slashes were already stripped from 
my query results so stripping them again was removing legitimate slashes.

The question is: why are the slashes already stripped out of the db 
results? I call set_magic_quotes_runtime(0) at the beginning of the 
scripts to turn off magic quoting. What else would cause this?

Thanks for any suggestions.

-Steve


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Re: [PHP] Why are slashes automatically stripped from db result?

2001-12-04 Thread Steve Cayford


On Monday, December 3, 2001, at 06:16  PM, Tyler Longren wrote:

> I believe you're correct Martin.  I think newer versions of MySQL
> automatically strip them out.  Just use php's stripslashes() and
> addslashes() functions when you need them.

Interesting. A bit unnerving at first, though.

Thanks.

-Steve


>
> Good luck Steve,
> Tyler Longren
>
> - Original Message -
> From: "Martin Towell" <[EMAIL PROTECTED]>
> To: "'Steve Cayford'" <[EMAIL PROTECTED]>;
> <[EMAIL PROTECTED]>
> Sent: Monday, December 03, 2001 6:04 PM
> Subject: RE: [PHP] Why are slashes automatically stripped from db 
> result?
>
>
>> maybe mysql is stripping the slashes and not php ??
>>
>> -Original Message-
>> From: Steve Cayford [mailto:[EMAIL PROTECTED]]
>> Sent: Tuesday, December 04, 2001 10:57 AM
>> To: [EMAIL PROTECTED]
>> Subject: [PHP] Why are slashes automatically stripped from db result?
>>
>>
>> Hey all. I'm storing some jpeg images in a mysql database using the 
>> PEAR
>> classes. Before inserting the image into the db I call addslashes() on
>> the data, I was, accordingly, calling stripslashes() on the data after
>> pulling the image back out of the database, but the image was getting
>> mangled. I finally realized that the slashes were already stripped from
>> my query results so stripping them again was removing legitimate 
>> slashes.
>>
>> The question is: why are the slashes already stripped out of the db
>> results? I call set_magic_quotes_runtime(0) at the beginning of the
>> scripts to turn off magic quoting. What else would cause this?
>>
>> Thanks for any suggestions.
>>
>> -Steve
>>
>>
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>> [EMAIL PROTECTED]
>>
>
>
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Re: [PHP] 'Running' a file from PHP script

2001-12-04 Thread Steve Cayford


On Tuesday, December 4, 2001, at 02:22  AM, George Pitcher wrote:

> Hi all,
>
> I'm a PHP newbie, my main scipring being done in Blueworld's Lasso.
>
> I have used php fro time to time with Lasso to do some file handling.
>
> My question is: 'Can I run a file from a PHP script?'
>
> The file in question would be a Filemaker database file. I want to be 
> able
> to open the file in the Filemaker application if the file is not already
> running.
>

If you mean can you open the Filemaker application on the client's 
machine using PHP, then no. Keep in mind that PHP is only running on the 
server(*), all the client machine sees is the HTML output from your PHP 
script. Maybe you could use javascript to do this, but I haven't used 
javascript much.

-Steve

(*) Assuming you're using PHP in its normal web-server scripting mode.


> MTIA
>
> George in Edinburgh
>
>
>
> _
> Do You Yahoo!?
> Get your free @yahoo.com address at http://mail.yahoo.com
>
>
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Re: [PHP] Cannot use a scalar value as an array

2001-12-04 Thread Steve Cayford


On Tuesday, December 4, 2001, at 06:04  AM, Xavier Antoviaque wrote:

>
> Hello all,
>
> I have a problem with a php script, which I want to use to fetch two 
> data
> (integer) from a MySQL table, divide the first by the second, and store 
> the
> value resulting in an double array. That seems not very difficult, but I
> always have the error 'Cannot use a scalar value as an array' when I 
> use my
> script.
>
> Here it is :
>
>   $result = mysql_query("SELECT timestamp_c,connectes,num FROM
> stats_serveurs WHERE timestamp_c LIKE '".$date."%' "AND serveur='".
> $serveur."'", $link);
>
>   $max = 0;
>   $reps = mysql_num_rows($result);
>   for($i = 1; $i<=$reps; $i++){
> $row = mysql_fetch_row($result);
> $serveur[$i]['heure'] = $row[0];
> $serveur[$i]['connectes'] = (int) ($row[1] / $row[2]); // Error!
> $max = max($max, $serveur[$i]['connectes']);
>   }
>

How does the operator precedence work in the statement ($row[1] / 
$row[2]) ? You might try (($row[1]) / ($row[2])) instead, or pull those 
into scalar variables before trying the division.

-Steve

> Any help is welcome ! :-)
>
> --
> Xavier.
>
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Re: [PHP] Stuck on array, need a little help.

2001-12-04 Thread Steve Cayford


On Tuesday, December 4, 2001, at 09:16  AM, Brian V Bonini wrote:

> I still can't get this to do what I want:
> $bikes = array(
>  "Road"  => array(
>   "Trek"  => array(
> "Trek 5200" => "road.php?brand=t5200"
> ),
>   "LeMond" => array(
> "Zurich" => "road.php?brand=zurich",
> "Chambery" => "road.php?brand=chambery",
> "Alpe d'Huez" => "road.php?brand=alpe",
> "BuenosAries" => "road.php?brand=bueno",
> "Tourmalet" => "road.php?brand=tourmalet"
> ),
>   "Moots" => array(
> "VaMoots"  => "road.php?brand=vamoots"
> )
>  ),
> );
> if ($cat == 'bikes') {
> while (list($key, $val)=each($bikes[$sub_cat])) {
>   echo " HEIGHT=\"1\"
> ALT=\"\" BORDER=\"0\">\n";
>   echo "$key\n";

(isn't $val here an array of bike models? If so you'll end up with )

> while (list($sub_key, $sub_val) = each($val)) {
> echo " CLASS=\"menu\">$sub_key\n";
> }
> }
> }
>
Are you looking for something more like this:

$modelString = "";
$makeString = "";
foreach($bikes['Road'] as $make => $models) {
$makeString .= " " . $make;
foreach($models as $model => $link) {
$modelString .= "$model";
}
}
print($makeString . "\n");
print($modelString . "\n");

(I don't get what you're trying to do with the table, though)

-Steve


> Will produce:
> Trek Trek 5200  LeMond Zurich Chambery Alpe d'Huez BuenosAries Tourmalet
> Moots VaMoots
> as it should...
>
> But, I need it to produce:
> Trek LeMond Moots
> Trek 5200  Zurich Chambery Alpe d'Huez BuenosAries Tourmalet VaMoots
>
> And I need to get the value of $sub_val in the nested "while" loop to 
> where
> $val is in the outer loop.
>
> I'm really stuck, any suggestions??
>
> -Brian
> **
>
>
>> -Original Message-
>> From: Jim Musil [mailto:[EMAIL PROTECTED]]
>> Sent: Friday, November 30, 2001 5:26 PM
>> To: [EMAIL PROTECTED]
>> Cc: [EMAIL PROTECTED]
>> Subject: RE: [PHP] Stuck on array, need a little help.
>>
>>
>>
>>
>> The script is still working right, you just need to nest another
>> while loop into your current while loop.
>>
>> Like so ...
>>
>> if ($cat == 'bikes' && $sub_cat != 'Road') {
>>   while (list($val, $key)=each($bikes[$sub_cat])) {
>>
>>  echo "$val";
>>
>>  while (list($sub_val, $sub_key) = each($key)) {
>>
>>
>>   echo ">   HEIGHT=\"1\"
>>   ALT=\"\" BORDER=\"0\">\n";
>>   echo "> CLASS=\"menu\">$sub_val\n";
>>
>>
>> }
>> }
>> }
>>
>> alternatively, if you know specifically what you want you could
>> do this ...
>>
>> if ($cat == 'bikes' && $sub_cat != 'Road') {
>>   while (list($val, $key)=each($bikes[$sub_cat]["Trek"])) {
>>
>>
>>
>>   echo ">   HEIGHT=\"1\"
>>   ALT=\"\" BORDER=\"0\">\n";
>>   echo "$val\n";
>>
>>
>>
>> }
>> }
>>
>>
>>
>>> No, all that will do is reverse the placement
>>> of the values. So now it prints out Array
>>> and puts the item in the URL. Still the same problem.
>>>
>>>
  -Original Message-
  From: Jim Musil [mailto:[EMAIL PROTECTED]]
  Sent: Friday, November 30, 2001 4:54 PM
  To: [EMAIL PROTECTED]
  Cc: [EMAIL PROTECTED]
  Subject: Re: [PHP] Stuck on array, need a little help.


  Your script is working like you are asking it to ...


  Change ...

   while (list($val, $key)=each($bikes[$sub_cat])) {

  To ...

   while (list($key, $val)=each($bikes[$sub_cat])) {

  and it should work like you WANT it to ...

> I'm stuck. $key returns "Array" how can I get at each
> level of this array?
>
> if ($cat == 'bikes' && $sub_cat != 'Road') {
> while (list($val, $key)=each($bikes[$sub_cat])) {
> echo ">>>  HEIGHT=\"1\"
> ALT=\"\" BORDER=\"0\">\n";
> echo "$val\n";
>
> $bikes = array(
>  "Road"  => array(
>   "Trek"  => array(
> "Trek 5200" => "road.php?brand=t5200"
> ),
>   "LeMond" => array(
> "Zurich" => "road.php?brand=zurich",
> "Chambery" => "road.php?brand=chambery",
> "Alpe d'Huez" => "road.php?brand=alpe",
> "BuenosAries" => "road.php?brand=bueno",
> "Tourmalet" => "road.php?brand=tourmalet"
> ),
>   "Moots" => array(
> "VaMoots"  => "road.php?brand=vamoots"
> )
>  ),
>  "Mountain"  => array(
>   "Trek"  => array(
> "Fuel 100" => "mountain.php?brand=tfuel90",
> "Fuel 90"  => "mountain.php?brand=schhg"
> ),
>   "Klein

Re: [PHP] 'Running' a file from PHP script

2001-12-04 Thread Steve Cayford

Ah, well, can't help you there, sorry.

-Steve

On Tuesday, December 4, 2001, at 09:40  AM, George Pitcher wrote:

> Steve,
>
> Thanks for the response, but it's a server-side 'run' I'm looking for, 
> not
> client-side (they get web access to the db).
>
> Basically I want to be able to test that the db is running and if it 
> isn't
> then I want to 'run' it. Platform is NT, server is IIS4
>
> George
>
> - Original Message -
> From: "Steve Cayford" <[EMAIL PROTECTED]>
> To: "George Pitcher" <[EMAIL PROTECTED]>
> Cc: <[EMAIL PROTECTED]>
> Sent: Tuesday, December 04, 2001 3:39 PM
> Subject: Re: [PHP] 'Running' a file from PHP script
>
>
>>
>> On Tuesday, December 4, 2001, at 02:22  AM, George Pitcher wrote:
>>
>>> Hi all,
>>>
>>> I'm a PHP newbie, my main scipring being done in Blueworld's Lasso.
>>>
>>> I have used php fro time to time with Lasso to do some file handling.
>>>
>>> My question is: 'Can I run a file from a PHP script?'
>>>
>>> The file in question would be a Filemaker database file. I want to be
>>> able
>>> to open the file in the Filemaker application if the file is not 
>>> already
>>> running.
>>>
>>
>> If you mean can you open the Filemaker application on the client's
>> machine using PHP, then no. Keep in mind that PHP is only running on 
>> the
>> server(*), all the client machine sees is the HTML output from your PHP
>> script. Maybe you could use javascript to do this, but I haven't used
>> javascript much.
>>
>> -Steve
>>
>> (*) Assuming you're using PHP in its normal web-server scripting mode.
>>
>>
>>> MTIA
>>>
>>> George in Edinburgh
>>>
>>>
>>>
>>> _
>>> Do You Yahoo!?
>>> Get your free @yahoo.com address at http://mail.yahoo.com
>>>
>>>
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>>>
>
>
> _
> Do You Yahoo!?
> Get your free @yahoo.com address at http://mail.yahoo.com
>
>
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Re: [PHP] Display BLOB Image

2001-12-06 Thread Steve Cayford

This is what I used to do what you're trying:

header("Content-Type: image/" . $imagetype); // $imagetype is jpeg or gif
header("Content-Length: " . strlen($image));
echo $image;

Note that "Type" is capitalized in "Content-Type", and include the 
"Content-Length" as well.

-Steve


On Thursday, December 6, 2001, at 02:45  PM, phantom wrote:

> I have successfully placed images (jpg,gif,png) into a MySQL database
> BLOB field, now I want to be able to pull the data out and diplay it.
>
> based on tutorial at
> http://www.zdnet.com/devhead/stories/articles/0,4413,2644827,00.html
>
> after Querying the DB I have the following lines:
>
> $ImgFile = mysql_result($Results,0,ImgFile); \\ image data;
> $ImgType = mysql_result($Results,0,ImgType);  \\ image type (image/jpeg,
> image/gif);
> header("Content-type: ${ImgType}");
> echo "$ImgFile";
>
> Problem: This method, when it pulls up an image, it shows no image but a
> bunch of greek code.  Example
> http://www.phantomcougar.com/strawberrypie_com/mem/show_img.php?PNum=8
> ((be sure to view page source)
>
>
>
>
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Re: [PHP] Finding num of days b/t two dates.

2001-12-07 Thread Steve Cayford

Here's one way to do it by converting dates into timestamps.
\n");
print(strftime("date 2 is %b %d, %Y", $date2) . "\n");
print("the difference in seconds is " . $timedif . "\n");
print("the difference in days is " . ($timedif / (60 * 60  * 24)) . 
"\n");
?>

-Steve

On Friday, December 7, 2001, at 04:48  PM, Alex Fritz wrote:

> If somebody could help me with this, it would save me a lot of 
> heartache.  I
> thought that this would be simple, but I can't seem to find a function
> anywhere in PHP that has this capability and I can't seem to find any
> external libraries for anything actually.  I need to be able to give 
> PHP a
> start date and an end date and have it return the number of days 
> between the
> dates.  If the first date is more recent than the second, I need it to 
> give
> me a negative number.  Can somebody please help?
>
> Alex
>
>
>
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[PHP] Class methods and inheritance...

2001-12-11 Thread Steve Cayford

Hi. Is there a way to find the class name of a method when called in the 
class::method() format? If called on an object (eg. object->method()) I 
could just ask for get_class($this), but when called as class::method(), 
$this should not be defined. Anyway around this?

Thanks.

-Steve


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Re: [PHP] Date formatting

2001-12-11 Thread Steve Cayford


On Monday, December 10, 2001, at 09:35  PM, phantom wrote:

> What would be an easy what to format a date value into month day year??
> I want to specially display a date stored in mysql in date format
> -MM-DD
>
> The manual says: string date (string format, int [timestamp])
>
> I tried $FormattedDate = date("F y, Y",${StoredDate})
>

try $FormattedDate = date("F y, Y",strtotime($StoredDate));

strtotime will convert your -MM-DD string into a unix timestamp.

> 1999-04-15 spit out December 31, 1969 when I had hoped for April 15,
> 1999.
>
> Apparently -MM-DD is not a valid timestamp... i tried mysql formats
> of timestamp(8) and timestamp(14) and that didn't work either
>
>
>
>
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Re: [PHP] Error while calling a function

2001-12-18 Thread Steve Cayford

I don't know all the details about how PHP compiles a program, but 
having your function definitions in an if-else statement that may not be 
executed looks suspicious to me.

You've got
if(!$Phone)
{
 do something
}
else
{
 function is_phone() { ...blah, blah...}
}

is_phone($Phone);

If you don't hit the else block, your is_phone function is probably not 
defined.

-Steve

On Tuesday, December 18, 2001, at 07:26  AM, J.F.Kishor wrote:

> hi all,
>
>   I have got a problem, when I execute the following script it gives
> a Fatal error, could any one tell me why is it ?, If this is a silly
> problem please execuse me but, plz do reply me.
>
> The script is
> -
> 
> 
>  if(!$Phone)
> {
>   ?>
>  
> 
> Telephone Number : 
> 
> 
> 
> }
> else
> {
>
>   function is_allnumbers ($text)
> {
>   if( (gettype($text)) == "integer")
>   {
>   print "the value is an integer";
> return true;
>   }
>
>   $Bad = $this->strip_numbers($text);
>
>   if(empty($Bad))
>   {
>   print "the value is  empty";
> return true;
>   }
>   return false;
> }
>
>   function clear_error ()
> {
>   $this->ERROR = "this is an error";
> }
>
>
>   function is_phone ($Phone ="")
> {
>   if($this->CLEAR)
>   {
> $this->clear_error();
>   }
>
>   if(empty($Phone))
>   {
> $this->ERROR = "is_phone: No Phone number
> submitted";
> return false;
>   }
>
>   $Num = $Phone;
>   $Num = $this->strip_space($Num);
>   $Num = eregi_replace("(\(|\)|\-|\+)","",$Num);
>   if(!$this->is_allnumbers($Num))
>   {
> $this->ERROR = "is_phone: bad data in phone
> number";
> return false;
>   }
>
>   if ( (strlen($Num)) < 7)
>   {
>   print "the number is less then 7";
> $this->ERROR = "is_phone: number is too short
> [$Num][$Phone]";
> return false;
>   }
>
>   if( (strlen($Num)) > 13)
>   {
>   print "the number is > then 13";
> $this->ERROR = "is_phone: number is too long
> [$Num][$Phone]";
> return false;
>   }
>
>   return true;
> }
> }
> $result = is_phone($Phone);
> if($result == "true")
> {
>  echo "success";
> }
> else
> {
>   echo "failure";
> }
>
> 
> 
> 
>
> The error is
> 
> Fatal error: Call to undefined function: is_phone() in
> /home/kuruvi1/kishor/public_html/IMS/ADMIN/test/is_phone.php on line 87
>
>
> Thanks for your tolerance,
>
>   - JFK
>  kishor
>
>
>
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[PHP] Querying an object's functions

2001-09-24 Thread Steve Cayford

Is there any way in PHP 4.0.6 to query an object for its member 
functions? For example I have a couple classes that have an htmlString() 
function to display themselves in some special way as html. I'd like to 
go through a list of various objects, find out if each can run 
htmlString() and if so use that, otherwise use some generic 
object-to-html function.

I could just put a member variable in the object and check for that, but 
it would be more elegant to be able to check for the function directly.

Thanks for any suggestions.

-Steve

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Re: [PHP] Querying an object's functions

2001-09-24 Thread Steve Cayford

Thanks. Sorry for missing that.

-Steve

On Monday, September 24, 2001, at 05:54  PM, Rasmus Lerdorf wrote:

> http://www.php.net/manual/en/function.get-class-methods.php
>
> On Mon, 24 Sep 2001, Steve Cayford wrote:
>
>> Is there any way in PHP 4.0.6 to query an object for its member
>> functions? For example I have a couple classes that have an 
>> htmlString()
>> function to display themselves in some special way as html. I'd like to
>> go through a list of various objects, find out if each can run
>> htmlString() and if so use that, otherwise use some generic
>> object-to-html function.
>>
>> I could just put a member variable in the object and check for that, 
>> but
>> it would be more elegant to be able to check for the function directly.
>>
>> Thanks for any suggestions.
>>
>> -Steve
>>
>>
>

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Re: [PHP] Feature?

2001-09-25 Thread Steve Cayford

Maybe this? A bit verbose, but functional.

-Steve

 'hello', 'banana' => 'world'));

?>

On Tuesday, September 25, 2001, at 06:25  AM, Andrey Hristov wrote:

> Yeap, I know that. I thought about this bit of hack but this breaks the 
> conception on giving parameters. Also this trick cannot
> solve the problem with default parameters. If I have function with 5 
> params and all of them have default values and I want to pass
> value only to the second parameter what I have to do?
>
> Andrey Hristov
> IcyGEN Corporation
> http://www.icygen.com
> BALANCED SOLUTIONS
>
>
> - Original Message -
> From: <[EMAIL PROTECTED]>
> To: "Andrey Hristov" <[EMAIL PROTECTED]>
> Sent: Tuesday, September 25, 2001 1:14 PM
> Subject: Re: [PHP] Feature?
>
>
>> On Tue, 25 Sep 2001 11:36:25 +0300, you wrote:
>>
>>> In the case when I've few parameters I've to remember their order, so 
>>> why not
>>> $bar=foo('par2'=>10);
>>> I want to pass value to only one or more but not to all params.
>>> Also this will make the code clearer I think.
>>> Comments are welcome!
>>
>> >
>> function foo($p)
>> {
>>   echo  $p['fred'] . $p['banana'];
>> }
>>
>> foo(array('fred' => 'hello', 'banana' => 'world'));
>>
>> ?>
>>
>
>
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Re: [PHP] Getting my head around nulls

2001-09-25 Thread Steve Cayford

On Tuesday, September 25, 2001, at 10:30  AM, [EMAIL PROTECTED] 
wrote:

>> Thanks for correcting my misspelling, Andrey, "IS NULL" not "IS_NULL".
>> Another thing to look at, Ben, is IFNULL(). I would give you an 
>> example,
> but
>> I never got it to work like I thought it should ;) I think it is 
>> supposed
> to
>> return an alternate value for NULLs.
>
> Just had a look at manual, ifnull() seems close to useless.  However if
> they had NVL(), as Oracle douse
>
> NVL( column|litrel, '*null*' )
>
> This takes a column or literal as its first arg and if arg is null 
> returns
> 2nd arg otherwise returns arg.  In other words if the column is null it 
> is
> replaces by arg2 otherwise it is simply returned.
>

Um, that's what ifnull() does. At least it works for me.

-Steve


>
> Next week I will sing the prases of the oracle decode statement -;)  The
> week after that I will get all exited about MySQL freetext indexes, yum,
> yum.
>
> Neb
>
>
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Re: [PHP] Getting my head around nulls

2001-09-26 Thread Steve Cayford


On Wednesday, September 26, 2001, at 02:48  AM, [EMAIL PROTECTED] 
wrote:

>
>
>
>
>
>
> Steve Cayford <[EMAIL PROTECTED]> on 25/09/2001 17:28:57
>
>
>
> To:   [EMAIL PROTECTED]
> cc:   [EMAIL PROTECTED], [EMAIL PROTECTED]
> Subject:  Re: [PHP] Getting my head around nulls
>
>
> On Tuesday, September 25, 2001, at 10:30  AM, [EMAIL PROTECTED]
> wrote:
>
>>>
>>> Just had a look at manual, ifnull() seems close to useless.  However 
>>> if
>>> they had NVL(), as Oracle douse
>>>
>>> NVL( column|litrel, '*null*' )
>>>
>>> This takes a column or literal as its first arg and if arg is null
>>> returns
>>> 2nd arg otherwise returns arg.  In other words if the column is null 
>>> it
>>> is
>>> replaces by arg2 otherwise it is simply returned.
>>>
>>
>> Um, that's what ifnull() does. At least it works for me.
>>
>> -Steve
>
> Verry true, sorry, was looking at nullif.  Also noticed the if function,
> can this have more than three parameters.  i.e
>
> if ( col, 'A', 'Accept', 'C', 'Complete', 'Unknown' )
>
> like Oracle decode. and if elseif else structure.
>
> Neb
>

Looks like you want the CASE statement. I haven't used it but it looks 
pretty much like what you're asking.

CASE value WHEN [compare-value] THEN result [WHEN [compare-value] THEN 
result ...] [ELSE result] END

Guess this discussion should really be on php-db :)

-Steve

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[PHP] Extract. Was Re: Feature?

2001-09-26 Thread Steve Cayford

Just looking up that extract function, the manual says it extracts into 
the current symbol table. I assume that means that within foo() the 
array would be extracted into local variables. Is that right?

-Steve

On Wednesday, September 26, 2001, at 06:50  AM, Alister wrote:

> Follow up to my own message:
>
> If you want to also skip even needing the empty 'array()' (new fianl
> example) - check if it is an array, and only do the first extract of
> the parameters if there's something there.
>
>  function foo($p='')
> {
>   $foodefault = array(
>   'foo_fred'=> 'default fred',
>   'foo_banana' => 'default bananananana',
>   );
>   if (is_array($p))
>   extract ($p, EXTR_PREFIX_ALL, 'foo');   // get values
>   extract ($foodefault, EXTR_SKIP, "foo");// get defaults
>   echo  "$foo_fred / $foo_banana";
> }
>
> echo "Both in place: ";
> foo(array('fred' => 'hello', 'banana' => 'world'));
> echo " Now with a missing param: ";
> foo(array('fred' => 'hello'));
> echo " Now both missing params: ";
> foo(array());
> echo " and with no array: ";
> foo();
> ?>

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Re: [PHP] Building Dynamic Value list using ohp

2001-09-26 Thread Steve Cayford

Also look for missing semicolons and unclosed braces in the lines above 
22.

-Steve

On Wednesday, September 26, 2001, at 09:19  AM, Derek Mailer wrote:

> oops...I was going to add...
>
> 2) -
>
> $result = mysql_query($query, $mysql_link);
>
> doesn't require the $mysql_link argument, try
>
> $result = mysql_query($query);
>
> Another general comment is that you have copied a bit of your code that
> seems to include the root password to mysql.  Not a good idea - you 
> should
> check and edit your code if necessary before posting.
>
> Hope this helps - good luck
>
> Derek
>
> - Original Message -
> From: "George Pitcher" <[EMAIL PROTECTED]>
> To: <[EMAIL PROTECTED]>
> Cc: <[EMAIL PROTECTED]>
> Sent: Wednesday, September 26, 2001 2:54 PM
> Subject: Re: [PHP] Building Dynamic Value list using ohp
>
>
>> Thanks Gottfried, but that doesn't change the response I get.
>>
>> Any more suggestions.
>>
>> George P in Edinburgh
>>
>> PS I am hoping to come over to Frankfurt for the conference in November
>> (boss might pay)
>>
>> - Original Message -
>> From: <[EMAIL PROTECTED]>
>> To: "George Pitcher" <[EMAIL PROTECTED]>
>> Sent: Wednesday, September 26, 2001 2:46 PM
>> Subject: Re: [PHP] Building Dynamic Value list using ohp
>>
>>
>>> hi!
>>>
>>> try:
>>>
>>> $query = "select HEI from heronuser"; // line 22
>>>
>>> greetinx!
>>> gottfried
>>>
 Hi all,

 I'm having a problem with dynamically building a value list.

 My code:

 >>> mysql_connect ('localhost', 'root', 'monty');
 mysql_select_db ('Heronsql');
$query = ("select HEI from heronuser"); // LINE 22 on original
> script
$result = mysql_query($query, $mysql_link);
  if(mysql_num_rows($result)) {
// show all HEIs as options in select form
while($row = mysql_fetch_row($result))
{
   print("$row[0]");
}
  }
 ?>

 But this generates:

 "Parse error:  parse error in c:\program files\apache
 group\apache\htdocs\bizflyers\login.php on line 22"

 Any suggestions?

 Regards

 George


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Re: [PHP] Array Elements & While Loops

2001-10-01 Thread Steve Cayford

Where is the closing brace for your while loop?

-Steve

On Monday, October 1, 2001, at 07:04  AM, Tom Churm wrote:

> hi,
>
> my problem is this:  i'm using a while loop to check elements in an
> Array for valid email syntax.  if $User[0] is a valid email address but
> $User[1] is not, the code for $User[0] is still executed before the die
> statement.  i need my loop to finish checking ALL array elements for
> validity, and to then die BEFORE any further code is executed.  here's
> what i have now (it doesn't work):
>
> //loop to check for bad email addresses:
> $j = 0;
> $flag = 0;
> while ($j < count($User)){
> if
> (($User[$j]!="")&&!eregi("^[_a-z0-9-]+(\.[_a-z0-9-]+)*@[a-
> z0-9-]+(\.[a-z0-9-]+)*$",
> $User[$j]))
> {
>   $flag = 1;
>   $errorNo = $j + 1;
> }
> //die if flag is 1
> if ($flag = 1) {
> die ("Email #$errorNo is not a valid e-mail
> address! href='javascript:window.history.back();'>Please return and correct
> this.");
>   }
> else {
> continue...
> }
>
> any suggestions would be great!
>
> thanks much,
>
> tom
>
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