[PHP] problems with permissions or getting access to a file
hi all
having trouble here with a site that I am hosting on go-daddy
I want to keep my php and my images apart
so to do this I have the base directory which contains the php scripts.
On the directory above that I have
my photos stored in /photos
when I run my scripts I get the following:
time -4 hours1265815323 <
*Warning*: filemtime() [function.filemtime
<http://www.patrickrice.net/tomdcam/function.filemtime>]: stat failed
for testPanasonicCamMotion20100201022429140.jpg in
*/home/content/j/a/c/jackcards/html/tomdcam/MotionCamLast1Hours.php* on
line *16*
1265815323
*Warning*: filemtime() [function.filemtime
<http://www.patrickrice.net/tomdcam/function.filemtime>]: stat failed
for testPanasonicCamMotion20100201022429140.jpg in
*/home/content/j/a/c/jackcards/html/tomdcam/MotionCamLast1Hours.php* on
line *28*
0
time -4 hours1265815323 <
*Warning*: filemtime() [function.filemtime
<http://www.patrickrice.net/tomdcam/function.filemtime>]: stat failed
for testPanasonicCamMotion20100201022619570.jpg in
*/home/content/j/a/c/jackcards/html/tomdcam/MotionCamLast1Hours.php* on
line *16
*_The script is the following below _from running this script, I keep
getting this error, now when I copy this script in to the directory
photo where the photos are stored, the script displays the images with
no problems leaving me think that the issue is with the permissions, I
have modified the pre-emptions to give all access as a test and still
problems. Anyone got any Ideas??
Thanks in advance.
Pat
6 8 $d = dir('/home/content/j/a/c/jackcards/html/tomdcam/photo/') or
die($php_errormsg);
9 while (false !== ($f = $d->read())) {
10 // only match alphabetic names
11 if (preg_match('/testPanasonicCamMotion/',$f)) {
12
13 if ($debug == 1) {
14 echo "time -4 hours" . strtotime("- 1 hours" ). " <";
15
16 echo filemtime($f);
17 echo " ";
18 }//end debug
19
20 $FourHoursAgo = intval(strtotime("- 1 hours" ));
21
22 if ($debug == 1) {
23 echo $FourHoursAgo;
24 #
25 echo "";
26 }
27
28 $dateOfFileModify = intval(date((filemtime($f)) ));
29
30 if ($debug == 1) {
31 echo $dateOfFileModify;
32 echo "";
33 }
34
35
36 if ($FourHoursAgo < $dateOfFileModify)
37 # if(intval(strtotime("- 4 hours" )) <
intval(strtotime(filemtime($f))) )
38
39 # if (strcmp(strtotime("- 4 hours" ),strtotime(filemtime($f))
) > 0)
40
41 {
42 #echo ' Images less than 4 hours ';
43
44
45
46
47 print "filename: $f ";
48 print " \n";
49
50 # echo "Last accessed on: " . date("d-m-y H:i:s", strtotime("+
8 hours", fileatime($f)) );
51 # echo "";
52 echo "Last modified on: " . date("d-m-y H:i:s", strtotime("+
8 hours", filemtime($f)) );
53 echo " ";
54 }//end of if statment
55
56
57 # else {
58 # print " None Found ";
59 # }
60 }
61 }
62 $d->close();
63
64
65 ?>
*
*
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[PHP] upgraded apache won't start
Hi; I upgraded to Apache 2.2.6 and it ran fine when I tested it after configuring the httpd.conf file (although instead of getting the PHP configuration page I got a page that said simply "It works.") After I rebooted though, I can't get it to run. The error log refers to a syntax error in the following line in the apache conf/httpd.conf file. LoadModule php5_module c:/php5/php5apache2.dll It says the specified module can't be found. The module certainly is there and it worked with apache 2 before the upgrade. What am I missing here ? Thanks; Pat -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] WWE in Stamford, CT needs a kick ass PHP Developer!
Thanks for the insight Tedd. It's good to get real feedback (unlike some of these guys busting my stones here - lol). I can take the blame on putting that Dreamweaver component in there, not our Development team. Gotta remember a lot of times you are dealing with a job spec developed from HR and we don't get the nuances of the space like you folks do. Basically the Dreamweaver is listed there because there is a lot of interdepartmental work that goes on and the less apt employees who are more "Creative" in nature work well in the Dreamweaver application (think Magazine Editors and Photo Departments and such, all of the folks who are more content than anything else). So I need someone who can play both sides, our developers are all excellent and I'd put them up with anyone in terms of ability. So thanks everyone for taking the time to see what World Wrestling Entertainment is all about. I'd really appreciate you keeping us in mind if you hear of anyone who is looking. Thx - Pat -Original Message- From: tedd [mailto:[EMAIL PROTECTED] Bullcrap! If you think that nested tables gets you in the "ballpark" quicker, then you haven't a clue as to what css is used for. My advice, replace your Dreamweaver experience requirement with someone who really knows css and you'll be a lot better off in both reducing time to develop and subsequent compliance issues. IMO, anyone who states that they produce "dynamic and cutting edge web sites" using a WYSIWYG editor is highly suspect and any employer who asks for that experience is doing his business and his clients a disservice. Cheers, tedd -- --- http://sperling.com http://ancientstones.com http://earthstones.com -- 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
[PHP] WWE in Stamford, CT needs a kick ass PHP Developer!
You folks know anyone who fits this one? Description: Your task would be the maintenance and extension of our dynamic and cutting edge Intranet and Internet web sites written in PHP5. PHP/Mysql Experience: 3-4 years, where 1 year must be PHP5. Candidates must be well versed in: * PHP5 * Mysql * JavaScript * Web services * Object-oriented programming Candidates must show: * Experience working through an entire life cycle of a web-based application from concept, implementation, testing to maintenance. * Familiarity with application level network protocols (e.g., HTTP and/or SSL). * Excellent communication and relationship skills along with being a strong team player. * Demonstrable analytical problem-solving skills. * Ability to thrive in a high-pressured, unstructured, customer-oriented environment. Other skills: * AJAX * XHTML/HTML, CSS * Linux * Dreamweaver * Subversion * Unix Shell/Perl The link to apply is as follows. http://www.wwe-careers.com/wwe/jobboard/SearchPositions.asp?ShowJobID=61 1&Keywords <http://www.wwe-careers.com/wwe/jobboard/SearchPositions.asp?ShowJobID=6 11&Keywords> = Thanks, Pat Arbitrio Sr. Staffing Specialist World Wrestling Entertainment, Inc. 1241 East Main Street Stamford, CT 06902
Re: [PHP] Dynamic Form List - how to change values
Thanks to all who helped.
The answer...drumroll arrays for each entry:
load routine (partial):
print "name='imageqty[]' size='5' value='$qty'>";
print "value='".$row[cialbum]."'>";
print "";
print "value='".$row[ciimage]."'>";
print "value='".$row[cisize]."'>";
retrieval routine (partial):
$cartid = $_COOKIE['PHPSESSID'];
$imageqty = $_POST['imageqty'];
$qty = $_POST["imageqty"];
$album = $_POST["hiddenalbum"];
$set = $_POST["hiddenset"];
$image = $_POST["hiddenimage"];
$size = $_POST["hiddensize"];
$n = count($qty);
$index = 0;
while($index < $n)
{
$newqty = $qty[$index];
$queryupdate = "update CartItems set ciqty = '$newqty' where ciid =
'$cartid' and cialbum = '$album[$index]' and ciset = '$set[$index]' and
ciimage = '$image[$index]' and cisize = '$size[$index]'";
querythedatabase($queryupdate);// my routine
$index++;
}
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Re: [PHP] Dynamic Form List - how to change values
tedd wrote:
I have a list from a MYSQL database that I am dumping to a screen. I'd
like the user to be able to change the quantity on the form for any
record on the screen, then post that information back so the user can
review it, and I can then update the database.
Simple checkout routine. Or so I thought.
I have these fields displayed on the screen before the user changes
the quantity:
$imageqty = $row["cialbum"]."-".$row["ciset"]."-".$row["ciimage"];
$qty = $row["ciqty"];
print "name='$imageqty' size='5' value='$qty'>";
I've set the $imageqty variable to hold the initial value from the
database. The initial screen works fine.
However, when I reprocess the file, how can I retrieve the original
value of the field that is named the same as $imageqty?
The form outputs something like this:
http://.../chkout.php?NATURE-1.jpg=6&continue=Continue+Checkout, which
does show the $imageqty of "NATURE-1.jpg" value with the new qty of "6".
How can I get the script to pull the value?
$imageqty = $row["cialbum"]."-".$row["ciset"]."-".$row["ciimage"];
$newqty = $_GET["$imageqty"];
Retrieves nothing.
Thanks for your help.
Pat
Pat:
I would suggest changing to:
http://.../chkout.php?imagename=NATURE-1.jpg&count=6&continue=Continue+Checkout,
$newqty = $_GET['count'];
$imagename = $_GET['imagename'];
Also, to see what you are actually "getting" try:
echo ("");
echo ("");
echo ("GET info:\n");
print_r($_GET);
echo("");
You might also want to consider taking care of undefined variables --
here's a reference:
http://www.weberdev.com/get_example-3823.html
HTH's
tedd
Ok, this works for one item, but my shopping cart may have (hopefully)
dozens of items in it. How do I refer to each new quantity changes for
each new row in the table? I was hoping that by making the qty field a
name of the image, I would be able to match the new qty up with the
right image quickly, but that won't work, obviously.
How can I change multiple items in a list (or all) when the list is
dynamically created? My approach may be wrong -- suggestions?
Pat
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[PHP] Dynamic Form List - how to change values
I have a list from a MYSQL database that I am dumping to a screen. I'd like the user to be able to change the quantity on the form for any record on the screen, then post that information back so the user can review it, and I can then update the database. Simple checkout routine. Or so I thought. I have these fields displayed on the screen before the user changes the quantity: $imageqty = $row["cialbum"]."-".$row["ciset"]."-".$row["ciimage"]; $qty = $row["ciqty"]; print "name='$imageqty' size='5' value='$qty'>"; I've set the $imageqty variable to hold the initial value from the database. The initial screen works fine. However, when I reprocess the file, how can I retrieve the original value of the field that is named the same as $imageqty? The form outputs something like this: http://.../chkout.php?NATURE-1.jpg=6&continue=Continue+Checkout, which does show the $imageqty of "NATURE-1.jpg" value with the new qty of "6". How can I get the script to pull the value? $imageqty = $row["cialbum"]."-".$row["ciset"]."-".$row["ciimage"]; $newqty = $_GET["$imageqty"]; Retrieves nothing. Thanks for your help. Pat -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] empty variable
So, you have to know, first, what is your meaning of your empty variabe:
null, 0, "", array()?
For example, "$var = 0;", empty($var) says true... but you?
"Jason Davidson" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
> as suggeted the isset and empty functions are usefull,
> or if you know it may be set, but could be empty, how about just
> if($variable) {
>
> }
>
> that is the same as
> if($var == "")
>
> "Chris Mach" <[EMAIL PROTECTED]> wrote:
> >
> > What is the best way to determine if a variable is empty?
> >
> > I have a feeling there is a better way than the way I'm doing it now...
> >
> > if ($variable == "")
> >
> >
> > ---
> > Outgoing mail is certified Virus Free.
> > Checked by AVG anti-virus system (http://www.grisoft.com).
> > Version: 6.0.766 / Virus Database: 513 - Release Date: 18-Sep-2004
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[PHP] Re: DROP TABLE IF EXISTS
Maybe something like this: http://dev.mysql.com/doc/mysql/en/TRUNCATE.html "John Taylor-Johnston" <[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED] > MySQL question: > > DROP TABLE IF EXISTS > > Instead of dropping the table, I want to truncate the table if it contains something. ANyone know of a way? I couldn't find anything under truncate in the manual. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] imagejpeg
I'm trying to create thumbnails for my images and everything seems to
work correctly except for trying to save them. I've tried outputting
directly with imagejpeg($im) and that works, but when trying
imagejpeg($im, "new.jpeg") it does not create a new image in the current
directory. I even tried changing the ownership of the script to root to
see if my permissions were screwed up but that didn't work. Here's my
function:
$fileName is the full path name to the image
$newName is just the file name
$thumb_path is the full path to my thumbnail directory
function createThumb($fileName, $newName, $thumb_path, $MAX_H, $MAX_W){
$new_h = 0;
$new_w = 0;
$divisor = 1;
list($width, $height, $type, $attr) = getimagesize($fileName);
if ($width > $MAX_W || $height > $MAX_H){
if ($width > $height)
$divisor = $width/$MAX_W;
else
$divisor = $height/$MAX_H;
}
switch($type){
case 1: $src_img = imagecreatefromgif($fileName);
break;
case 2: $src_img = imagecreatefromjpeg($fileName);
break;
default: DisplayErrMsg("Unknown image format, please
alert webmaster!");
break;
}
if (isset($src_img)){
$new_h = $height/$divisor;
$new_w = $width/$divisor;
$dest_img = imagecreatetruecolor($new_w,$new_h);
imagecopyresampled($dest_img,$src_img,0,0,0,0,$new_w,$new_h,$width,$heig
ht);
switch($type){
case 1: imagegif($dest_img,
$thumb_path.$newName);
break;
case 2: imagejpeg($dest_img,
$thumb_path.$newName);
break;
default: DisplayErrMsg("Unknown image format,
please alert webmaster!");
break;
}
imagedestroy($dest_img);
imagedestroy($src_img);
}
}
Thanks for any help!
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RE: [PHP] Strange coding bug
I think I've fixed it but in an odd way. Apparently I was getting error
1046: no database seleted. So I ran mysql_db_query and that worked. But
on the other pages I have, mysql_query always works. I have a function
called connect() that I run on every page:
function connect()
{
global $DB_SERVER, $DB_LOGIN, $DB_PASSWORD;
if (@mysql_pconnect($DB_SERVER, $DB_LOGIN, $DB_PASSWORD))
{
return 1;
}
else
header("Location: blahblahblah");
}
This should establish a permanent connection to the database server. Is
there a way to permanatly select a certain database?
Pat
-Original Message-
From: Jay Blanchard [mailto:[EMAIL PROTECTED]
Sent: Monday, January 05, 2004 3:20 PM
To: Pat Hanna; php help
Subject: RE: [PHP] Strange coding bug
[snip]
$query="SELECT * FROM products WHERE stock_number = '$item_number'";
$result=mysql_query($query); $row=mysql_fetch_array($result);
When I do 'echo $result;' on the first page, I get the result number.
[/snip]
If you
echo $result;
with this code all you should get back is 'array' ($result is an array)
Do this
print_r($row) and see what happens.
P.S. Attachements are stripped on the way to the list usually, so post
the relevant small bits of code in your e-mail.
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RE: [PHP] Strange coding bug
mailto:[EMAIL PROTECTED] Sent: Monday, January 05, 2004 3:32 PM To: Pat Hanna; Brad Pauly Cc: php help Subject: RE: [PHP] Strange coding bug [snip] I have tried printing out the query and got the exact same thing on both pages. The only relevent code is the code I included. The $item_number gets passed to both pages exactly the same. The query comes up no different on both pages. It is accessing the exact same product and yet it will not get a hit. [/snip] Please post more of the relevant code so that we can help you. -- 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] Strange coding bug
I have tried printing out the query and got the exact same thing on both pages. The only relevent code is the code I included. The $item_number gets passed to both pages exactly the same. The query comes up no different on both pages. It is accessing the exact same product and yet it will not get a hit. Pat -Original Message- From: Brad Pauly [mailto:[EMAIL PROTECTED] Sent: Monday, January 05, 2004 3:28 PM To: Pat Hanna Cc: 'php help' Subject: Re: [PHP] Strange coding bug On Mon, 2004-01-05 at 12:59, Pat Hanna wrote: > I cannot seem to find the problem in my code. I have two pages. Both > need to select a product from a database. The first page works fine, > the product is found and I use the data in the page. The second one > does not fetch the product. The code is exactly the same: > > $query="SELECT * FROM products WHERE stock_number = '$item_number'"; > $result=mysql_query($query); $row=mysql_fetch_array($result); > > When I do 'echo $result;' on the first page, I get the result number. > But if I do it on the second page I get nothing. Both pages are > connected to the database and I get no errors in code. I have included > the full code of each page in the attachment. Item_details.php3 is the > working page and addtocart.php3 is the one that won't seem to connect. I think the attachments got removed. Sorting through the source of two pages is a lot to look through. Maybe you could paste the relevant code from each of the pages. Specifically, the select statements and where $item_number comes from or is manipulated. That should help find the problem. You might also try echoing $query on both pages to make sure they same query actually is being executed. - Brad -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Strange coding bug
I cannot seem to find the problem in my code. I have two pages. Both need to select a product from a database. The first page works fine, the product is found and I use the data in the page. The second one does not fetch the product. The code is exactly the same: $query="SELECT * FROM products WHERE stock_number = '$item_number'"; $result=mysql_query($query); $row=mysql_fetch_array($result); When I do 'echo $result;' on the first page, I get the result number. But if I do it on the second page I get nothing. Both pages are connected to the database and I get no errors in code. I have included the full code of each page in the attachment. Item_details.php3 is the working page and addtocart.php3 is the one that won't seem to connect. Thanks, Patrick -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] or return problem
On Tue, 7 Oct 2003, Curt Zirzow wrote:
>> function istrue() {
>> return true;
>> }
>> function retor_test() {
>> istrue() or return( "False" );
>> return "True";
>> }
> return (istrue()? 'True': 'False');
>
>hmm.. less typing, easier to understand and logically readable.
This doesn't answer the problem because it does not follow the same
logic as the orignial code example. In your example you want to return a
value regardless of what istrue() returns. In my example I only wanted to
return a value if istrue() failed, otherwise I wanted to continue in the
scope of the function. That may not have been obvious because the example
was a little contrived.
Pat Carmody
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Re: [PHP] or return problem
So far everyone is telling me that it won't work, but no one is telling me
why. (btw I did search extensively for the answer to this question but so
far have found nothing). Robert, could you be more specific in your
reference to the http://www.php.net documentation? I see nothing on the
basic syntax page that addresses this.
Pat Carmody
On Tue, 7 Oct 2003, Robert Cummings wrote:
>On Tue, 2003-10-07 at 13:02, Pat Carmody wrote:
>>
>>
>> Calling the following retor_test() function causes a "Parse error: parse
>> error, unexpected T_RETURN" message when the script is run:
>>
>> function istrue() {
>> return true;
>> }
>> function retor_test() {
>> istrue() or return( "False" );
>> return "True";
>> }
>>
>> The problem is with the "or return" part. Any ideas why? I realize that
>> I could use an if statement instead, but I'm a lazy, lazy man and I don't
>> want to.
>
>Your laziness is causing you problems. Please check out ALL of the
>documenation located at http://www.php.net since this is covered under
>basic syntax and we don't cater to lazy people (at the very least *I*
>don't cater to lazy people).
>
>Cheers,
>Rob.
>--
>..
>| InterJinn Application Framework - http://www.interjinn.com |
>::
>| An application and templating framework for PHP. Boasting |
>| a powerful, scalable system for accessing system services |
>| such as forms, properties, sessions, and caches. InterJinn |
>| also provides an extremely flexible architecture for |
>| creating re-usable components quickly and easily. |
>`'
>
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>
>
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[PHP] or return problem
Calling the following retor_test() function causes a "Parse error: parse
error, unexpected T_RETURN" message when the script is run:
function istrue() {
return true;
}
function retor_test() {
istrue() or return( "False" );
return "True";
}
The problem is with the "or return" part. Any ideas why? I realize that
I could use an if statement instead, but I'm a lazy, lazy man and I don't
want to.
Pat Carmody
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[PHP] Re: Register globals on and off
I've read that an include file in each of your pages with the lines below should do the trick for you with register_globals OFF.. Not sure if this is a valid way to go though... Regards, Pat "Davy Obdam" <[EMAIL PROTECTED]> wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... > Hello people, > > On my development machine (win XP/Apache 2.0.44/PHP 4.3.0/MySQL 3.23.55) > i have several websites that i made some time ago that require register > globals to be On in the php.ini. Ofcourse i know thats not a good idea > at all for security, but rewriting all this code is not an option. > However in my php.ini i have set register globals to Off because that > better. Is it possible to configure my webserver/php so that only those > sites that require register globals to be On have that setting, for > instance in a .htacces file?? Any help is appreciated:-) > > Best regards, > > Davy Obdam > mailto:[EMAIL PROTECTED] > > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] help!
This may or may not have anything to do with PHP but my goal deals with it. I have a box that I'm trying to turn into a linux box for php purposes. I have reformatted the hard drive so I could install it. The problem is that when I try to partition the disk during the installation, my hard drive is not recognized. I am new at this and I would appreciate any help I could get. Please help me!! -- Patrick Hanna Database integration--E-commerce solutions The Wentworth Company Get paid cash every time you receive email! Sign up FREE at: http://www.MintMail.com/?m=1080349
Re: [PHP] addslashes, stripslashes
--On Monday, November 26, 2001 04:47:35 PM -0800 Scott Aikin <[EMAIL PROTECTED]> wrote: > I've come across a strange problem working backwards with stripslashes > after running addslashes. I take a string like: > > "\t\4" > > and run it through addslashes, the result is: > > \\t\\4 > > After grabbing this data from the database and running 'stripslashes', > the data comes out as: > > "\t" > > without the \4, for some reason stripslashes always removes any > combination of \ and a number. Does anybody know a way around this or > can maybe provide some insight about why this is happening? It probably isn't removing it, it's converting it into an EOT character (0x04). That's pretty standard for most environments that do backslash substitution. It should also convert \48 and \060 into a '0' character. (The first being decimal, the second octal due to the leading zero.) -Pat -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
[PHP] PHP versus all other languages
I'm doing my senior exit project on database languages online. I'm asking for help from anyone who can provide any information on the comparison between the different languages. I'm comparing languages such as PHP, ASP, ColdFussion, perl and any others that I might not know about that you guys might know. Thank you to anyone who helps me out in the least. Patrick -- Database integration--E-commerce solutions The Wentworth Company Get paid cash every time you receive email! Sign up FREE at: http://www.MintMail.com/?m=1080349
[PHP] Warning: Page has Expired
When using sessions with forms, a user presses the back button and gets the message: Page has Expired. How can I get around this happening? I have to use sessions because I store many different variables at a time for a shopping cart. Forms are submitted depending on what item the user is shopping for. What can I do to get around the expired page? Thanks! Patrick -- Patrick Hanna Database integration--E-commerce solutions The Wentworth Company
[PHP] Tip for cookies
I found out today after suffering from minor headaches and hair loss that if you have functions defined in and outside file name 'foo.php', and in those functions you use cookies, you must pass those cookies to the function called. I tried using a value from a cookie in a function and it never could access it unless I passed it. Funny -- Patrick Hanna Database integration--E-commerce solutions The Wentworth Company
Re: [PHP] cookies
I've tried that but they still don't go away. - Original Message - From: "Brian Clark" <[EMAIL PROTECTED]> To: "PHP is not a drug." <[EMAIL PROTECTED]> Sent: Monday, April 16, 2001 6:23 PM Subject: Re: [PHP] cookies > Hi Pat, > > @ 6:13:20 PM on 4/16/2001, Pat Hanna wrote: > > > I'm setting cookies for a shopping cart and they are setting fine > > but they will not delete. I've tried every way of deleting them and > > they just won't go away. Someone please help. > > http://www.php.net/manual/en/function.setcookie.php > > Read Example 2. > > -Brian > -- > PGP is spoken here: 0xE4D0C7C8 > Please, DO NOT carbon copy me on list replies. > > > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, e-mail: [EMAIL PROTECTED] > For additional commands, e-mail: [EMAIL PROTECTED] > To contact the list administrators, e-mail: [EMAIL PROTECTED] > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
[PHP] cookies
I'm setting cookies for a shopping cart and they are setting fine but they will not delete. I've tried every way of deleting them and they just won't go away. Someone please help. -- Patrick Hanna Database integration--E-commerce solutions The Wentworth Company
[PHP] database
Hi I need some scripts written in PHP to interface with a mySQL database. There will be a need for a high level of security. Anyone interested in making some money ? Call me Regards Pat -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
[PHP] edit plus
Anyone who uses Edit Plus. Can you send me the php function files and auto complete file. The edit plus web page has the links but none of them work. Thanks Pat
[PHP] directory reading
I'm trying to read the contents of a directory and it's subdirectories. I
use w while loop with readdir($handle) and then an if statement with
filetype($file)=="dir" and that works that far. The contents of the
subdirectory is listed. After that breaks out of the if statement though,
the script goes back through the while loop and reads the next entry. The
problem is that the next file is a dir but the filetype function doesn't
return "dir", it returns nothing. I can only get through one subdirectory,
then all the other files in the main directory are listed even though they
are directories. Their contents are not displayed. I need to use the names
of the subdirectories for viewing and their locations. What is going wrong?
//sub directory function
function subdir($path, $root)
{
chdir($path);
$handle2=opendir(".");
$file2=readdir($handle2);
$file2=readdir($handle2);
while($file2=readdir($handle2))
{
?>
">
http://www.wentworthco.com/shoponline/customerimages/">[VIEW]
[DELETE]
">
http://www.wentworthco.com/shoponline/customerimages//">[VIEW]
[DELETE]
[PHP] picture quality question
I have these pictures that I'm using for my webpage. Their background changes from blue to white. When I loaded them onto the server, they looked very nice and they always looked good on netscape browsers. Every once and a while though, the background turns bad and all the pictures look terrible on AOL and Internet explorer. The colors don't fade evenly between each other and it's rather ugly. I have to reload them or redraw them in order for them to look the same on all the browsers. How can I fix this? Very frustrated, Patrick
[PHP] is_dir
I have a snipet of code i'm using for a file maintanance program. I have no
idea why but when the main loop repeats, the is_dir function doesn't return
true even though it prints out a directory in the server.
chdir("/certain/place/for/files/");
$handle=opendir(".");
$file=readdir($handle); //to get rid of "." and ".."
$file=readdir($handle);
while($file=readdir($handle))
{
if (is_dir($file)){
chdir("/certain/place/for/files/".$file."/");
$handle2=opendir(".");
$file2=readdir($handle2);
$file2=readdir($handle2);
while($file2=readdir($handle2))
{
?>
">
[view]
[delete]
">
[view]
[delete]
