[PHP] Brain dead... write image to file
Hey all,
am feeling a bit brain dead, pulled an all nighter and would appreciate some
help as have already wasted over 2hrs on this :(
just not thinking straight.
I got this script off the net, cant even remember where :( its basically to
resize an uploaded image (i have a script that does exactly this that i created
ages ago... but cant find it and if i remember corrrectly that only resizes
jpgs, this one does gifs and pngs as well)
the way the original guy wrote it is that it writes the uploaded image to the
DB, i'm just trying for it to create a resized image *instead of writing to the
db*.
The code is below, needless to say, my little block of code to write an the
image to file is not working... would appreciate some help.
?php
//$conn = mysql_connect(localhost, username, password) or
die(mysql_error());
//mysql_select_db('test', $conn) or die(mysql_error());
if($_FILES['userfile']['tmp_name']){
resize();
//$q = INSERT INTO test VALUES ('', '.$blob_arr['image'].',
'.$blob_arr['thumb'].', '.$blob_arr['type'].');
//$aa = mysql_query($q,$conn) or die(mysql_error());
/## Start my code try
$uploaddir = 'C:\\wamp\\www\\ezee\\funny\\';
$filename = $uploaddir.'test.jpg';
$somecontent = $blob_arr['thumb'];
if (is_writable($filename)) {
if (!$handle = fopen($filename, 'w')) {
echo Cannot open file ($filename);
exit;
}
if (fwrite($handle, $somecontent) === FALSE) {
echo Cannot write to file ($filename);
exit;
}
fclose($handle);
} else {echo The file $filename is not writable;}
header( Content-type: .$blob_arr['type'].);
}
/## End my code try
function resize(){
global $blob_arr;
$temp_name =$_FILES['userfile']['tmp_name'];
$userfile_name =$_FILES['userfile']['name'];
$userfile_size =$_FILES['userfile']['size'];
$userfile_type =$_FILES['userfile']['type'];
$thumb_width=90;
$thumb_height=90;
$image_width=200;
$image_height=200;
if (!($userfile_type ==image/pjpeg OR $userfile_type ==image/jpeg OR
$userfile_type==image/gif OR $userfile_type==image/png OR
$userfile_type==image/x-png)){
die (You can upload just images in .jpg .jpeg .gif and .png format!br
/ );
}
$data = fread(fopen($temp_name, rb), filesize($temp_name));
$src_image = imagecreatefromstring($data);
$width = imagesx($src_image);
$height = imagesy($src_image);
if ($thumb_width ($width $height)) {
$thumb_width = ($thumb_height / $height) * $width;
} else {
$thumb_height = ($thumb_width / $width) * $height;
}
if ($image_width ($width $height)) {
$image_width = ($image_height / $height) * $width;
} else {
$image_height = ($image_width / $width) * $height;
}
$dest_img = imagecreatetruecolor($thumb_width, $thumb_height);
$i_dest_img = imagecreatetruecolor($image_width, $image_height);
imagecopyresized($dest_img, $src_image,0, 0, 0, 0,$thumb_width,
$thumb_height,$width, $height);
imagecopyresized($i_dest_img, $src_image,0, 0, 0, 0,$image_width,
$image_height,$width, $height);
ob_start();
if($userfile_type == image/jpeg OR $userfile_type == image/pjpeg){
imagejpeg($dest_img);
}
if($userfile_type == image/gif){
imagegif($dest_img);
}
if($userfile_type == image/png OR $userfile_type == image/x-png){
imagepng($dest_img);
}
$binaryThumbnail = ob_get_contents();
ob_end_clean();
ob_start();
if($userfile_type == image/jpeg OR $userfile_type == image/pjpeg){
imagejpeg($i_dest_img);
}
if($userfile_type == image/gif){
imagegif($i_dest_img);
}
if($userfile_type == image/png OR $userfile_type == image/x-png){
imagepng($i_dest_img);
}
$binaryImage = ob_get_contents();
ob_end_clean();
if(!get_magic_quotes_gpc()){
$binaryThumbnail=addslashes($binaryThumbnail);
$binaryImage=addslashes($binaryImage);
}
$blob_arr['image']=$binaryImage;
$blob_arr['thumb']=$binaryThumbnail;
$blob_arr['type']=$userfile_type;
}
if($_POST['submit']){
//$Rs = mysql_query(SELECT * FROM funny_test WHERE id =
'.mysql_insert_id().') or die(mysql_error());
//$row_Rs = mysql_fetch_assoc($Rs);
//header( Content-type: .$row_Rs['typ'].);
//echo $row_Rs['thb'];
}else{
?
form name=form1 method=post enctype=multipart/form-data action=?php
echo $PHP_SELF; ?
input id=userfile name=userfile type=file /
input type=submit name=submit value=Submit /
/form
?php
}
?
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[PHP] Brain Dead...
I am trying to get a form to work with pull down menu. But I am so brain dead I forgot how to get the input from the pull down menu. Where do I put input to pass on to php? Chuck Payne -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Brain Dead...
Dropdownmenu is a select name=ddmenuoption/select So the php variable that will hold the user's choice is stored in variable $ddmenu. Thus, no input needed. cheers, --thalis On Sun, 17 Mar 2002, Chuck PUP Payne wrote: I am trying to get a form to work with pull down menu. But I am so brain dead I forgot how to get the input from the pull down menu. Where do I put input to pass on to php? Chuck Payne -- 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] Brain Dead...
Ok, but I have this as an example
select name=abc size=1
option value=a selectedA/option
option value=bB/option
option value=cC/option
option value=dD/option
/select
Are you saying that abc should be $abc? I not clear.
I have for my sql as this...
db_name = xxxinks;
$table_name = ssslinks;
$connection = @mysql_connect(mybox.com,anyusers,!@#$%) or die (Couldn't
connect);
$db = @mysql_select_db($db_name, $connection) or die (Couldn't select
database.);
$sql = INSERT INTO $table_name
(abc, key_word, links, name)
VALUES
('$abc', '$keywords', '$links', '$name');
$result = @mysql_query($sql, $connection) or die(Couldn't execute query.);
but I get that it can't do the query. So I am think it not passing abc...
Chuck
- Original Message -
From: Thalis A. Kalfigopoulos [EMAIL PROTECTED]
To: Chuck PUP Payne [EMAIL PROTECTED]
Cc: PHP General [EMAIL PROTECTED]
Sent: Sunday, March 17, 2002 9:25 PM
Subject: Re: [PHP] Brain Dead...
Dropdownmenu is a select name=ddmenuoption/select
So the php variable that will hold the user's choice is stored in variable
$ddmenu.
Thus, no input needed.
cheers,
--thalis
On Sun, 17 Mar 2002, Chuck PUP Payne wrote:
I am trying to get a form to work with pull down menu. But I am so brain
dead I forgot how to get the input from the pull down menu. Where do I
put
input to pass on to php?
Chuck Payne
--
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To unsubscribe, visit: http://www.php.net/unsub.php
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Re: [PHP] Brain Dead...
What does I get that it can't do the query mean? What error are you getting?
On Sun, 17 Mar 2002, Chuck Payne wrote:
Ok, but I have this as an example
select name=abc size=1
option value=a selectedA/option
option value=bB/option
option value=cC/option
option value=dD/option
/select
Are you saying that abc should be $abc? I not clear.
I have for my sql as this...
db_name = xxxinks;
$table_name = ssslinks;
$connection = @mysql_connect(mybox.com,anyusers,!@#$%) or die (Couldn't
connect);
$db = @mysql_select_db($db_name, $connection) or die (Couldn't select
database.);
$sql = INSERT INTO $table_name
(abc, key_word, links, name)
VALUES
('$abc', '$keywords', '$links', '$name');
$result = @mysql_query($sql, $connection) or die(Couldn't execute query.);
but I get that it can't do the query. So I am think it not passing abc...
Chuck
- Original Message -
From: Thalis A. Kalfigopoulos [EMAIL PROTECTED]
To: Chuck PUP Payne [EMAIL PROTECTED]
Cc: PHP General [EMAIL PROTECTED]
Sent: Sunday, March 17, 2002 9:25 PM
Subject: Re: [PHP] Brain Dead...
Dropdownmenu is a select name=ddmenuoption/select
So the php variable that will hold the user's choice is stored in variable
$ddmenu.
Thus, no input needed.
cheers,
--thalis
On Sun, 17 Mar 2002, Chuck PUP Payne wrote:
I am trying to get a form to work with pull down menu. But I am so brain
dead I forgot how to get the input from the pull down menu. Where do I
put
input to pass on to php?
Chuck Payne
--
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To unsubscribe, visit: http://www.php.net/unsub.php
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RE: [PHP] Brain Dead...
do other variables you pass to the script get set?
if not, do you have register_globals (or whatever it is) turned on?
maybe use $_GET or $_POST instead?
-Original Message-
From: Chuck Payne [mailto:[EMAIL PROTECTED]]
Sent: Monday, March 18, 2002 1:36 PM
To: Thalis A. Kalfigopoulos
Cc: PHP General
Subject: Re: [PHP] Brain Dead...
Ok, but I have this as an example
select name=abc size=1
option value=a selectedA/option
option value=bB/option
option value=cC/option
option value=dD/option
/select
Are you saying that abc should be $abc? I not clear.
I have for my sql as this...
db_name = xxxinks;
$table_name = ssslinks;
$connection = @mysql_connect(mybox.com,anyusers,!@#$%) or die (Couldn't
connect);
$db = @mysql_select_db($db_name, $connection) or die (Couldn't select
database.);
$sql = INSERT INTO $table_name
(abc, key_word, links, name)
VALUES
('$abc', '$keywords', '$links', '$name');
$result = @mysql_query($sql, $connection) or die(Couldn't execute query.);
but I get that it can't do the query. So I am think it not passing abc...
Chuck
- Original Message -
From: Thalis A. Kalfigopoulos [EMAIL PROTECTED]
To: Chuck PUP Payne [EMAIL PROTECTED]
Cc: PHP General [EMAIL PROTECTED]
Sent: Sunday, March 17, 2002 9:25 PM
Subject: Re: [PHP] Brain Dead...
Dropdownmenu is a select name=ddmenuoption/select
So the php variable that will hold the user's choice is stored in variable
$ddmenu.
Thus, no input needed.
cheers,
--thalis
On Sun, 17 Mar 2002, Chuck PUP Payne wrote:
I am trying to get a form to work with pull down menu. But I am so brain
dead I forgot how to get the input from the pull down menu. Where do I
put
input to pass on to php?
Chuck Payne
--
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To unsubscribe, visit: http://www.php.net/unsub.php
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To unsubscribe, visit: http://www.php.net/unsub.php
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Re: [PHP] Brain Dead...
When I go to post, I get the following alertCouldn't execute query so
I know I can connect, so it must be my SQL statement. That why I wondering
how to pass the menu information to php so that it can do the SQL statement.
I don't have $abc in the menu but I do have abc. So I am trying to find out
how to do that. Maybe I am going about it the wrong way.
Chuck
- Original Message -
From: Thalis A. Kalfigopoulos [EMAIL PROTECTED]
To: Chuck Payne [EMAIL PROTECTED]
Cc: PHP General [EMAIL PROTECTED]
Sent: Sunday, March 17, 2002 9:40 PM
Subject: Re: [PHP] Brain Dead...
What does I get that it can't do the query mean? What error are you
getting?
On Sun, 17 Mar 2002, Chuck Payne wrote:
Ok, but I have this as an example
select name=abc size=1
option value=a selectedA/option
option value=bB/option
option value=cC/option
option value=dD/option
/select
Are you saying that abc should be $abc? I not clear.
I have for my sql as this...
db_name = xxxinks;
$table_name = ssslinks;
$connection = @mysql_connect(mybox.com,anyusers,!@#$%) or die
(Couldn't
connect);
$db = @mysql_select_db($db_name, $connection) or die (Couldn't select
database.);
$sql = INSERT INTO $table_name
(abc, key_word, links, name)
VALUES
('$abc', '$keywords', '$links', '$name');
$result = @mysql_query($sql, $connection) or die(Couldn't execute
query.);
but I get that it can't do the query. So I am think it not passing
abc...
Chuck
- Original Message -
From: Thalis A. Kalfigopoulos [EMAIL PROTECTED]
To: Chuck PUP Payne [EMAIL PROTECTED]
Cc: PHP General [EMAIL PROTECTED]
Sent: Sunday, March 17, 2002 9:25 PM
Subject: Re: [PHP] Brain Dead...
Dropdownmenu is a select name=ddmenuoption/select
So the php variable that will hold the user's choice is stored in
variable
$ddmenu.
Thus, no input needed.
cheers,
--thalis
On Sun, 17 Mar 2002, Chuck PUP Payne wrote:
I am trying to get a form to work with pull down menu. But I am so
brain
dead I forgot how to get the input from the pull down menu. Where do
I
put
input to pass on to php?
Chuck Payne
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
--
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To unsubscribe, visit: http://www.php.net/unsub.php
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RE: [PHP] Brain Dead...
Another thing you might want to try doing is throwing in phpinfo() somewhere
to see what variables you are actually getting
-Original Message-
From: Chuck Payne [mailto:[EMAIL PROTECTED]]
Sent: Monday, March 18, 2002 1:44 PM
To: Thalis A. Kalfigopoulos
Cc: PHP General
Subject: Re: [PHP] Brain Dead...
When I go to post, I get the following alertCouldn't execute query so
I know I can connect, so it must be my SQL statement. That why I wondering
how to pass the menu information to php so that it can do the SQL statement.
I don't have $abc in the menu but I do have abc. So I am trying to find out
how to do that. Maybe I am going about it the wrong way.
Chuck
- Original Message -
From: Thalis A. Kalfigopoulos [EMAIL PROTECTED]
To: Chuck Payne [EMAIL PROTECTED]
Cc: PHP General [EMAIL PROTECTED]
Sent: Sunday, March 17, 2002 9:40 PM
Subject: Re: [PHP] Brain Dead...
What does I get that it can't do the query mean? What error are you
getting?
On Sun, 17 Mar 2002, Chuck Payne wrote:
Ok, but I have this as an example
select name=abc size=1
option value=a selectedA/option
option value=bB/option
option value=cC/option
option value=dD/option
/select
Are you saying that abc should be $abc? I not clear.
I have for my sql as this...
db_name = xxxinks;
$table_name = ssslinks;
$connection = @mysql_connect(mybox.com,anyusers,!@#$%) or die
(Couldn't
connect);
$db = @mysql_select_db($db_name, $connection) or die (Couldn't select
database.);
$sql = INSERT INTO $table_name
(abc, key_word, links, name)
VALUES
('$abc', '$keywords', '$links', '$name');
$result = @mysql_query($sql, $connection) or die(Couldn't execute
query.);
but I get that it can't do the query. So I am think it not passing
abc...
Chuck
- Original Message -
From: Thalis A. Kalfigopoulos [EMAIL PROTECTED]
To: Chuck PUP Payne [EMAIL PROTECTED]
Cc: PHP General [EMAIL PROTECTED]
Sent: Sunday, March 17, 2002 9:25 PM
Subject: Re: [PHP] Brain Dead...
Dropdownmenu is a select name=ddmenuoption/select
So the php variable that will hold the user's choice is stored in
variable
$ddmenu.
Thus, no input needed.
cheers,
--thalis
On Sun, 17 Mar 2002, Chuck PUP Payne wrote:
I am trying to get a form to work with pull down menu. But I am so
brain
dead I forgot how to get the input from the pull down menu. Where do
I
put
input to pass on to php?
Chuck Payne
--
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To unsubscribe, visit: http://www.php.net/unsub.php
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To unsubscribe, visit: http://www.php.net/unsub.php
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Re: [PHP] Brain Dead...
Is it possible to check the Mysql log and tell us its point of view?
On Sun, 17 Mar 2002, Chuck Payne wrote:
When I go to post, I get the following alertCouldn't execute query so
I know I can connect, so it must be my SQL statement. That why I wondering
how to pass the menu information to php so that it can do the SQL statement.
I don't have $abc in the menu but I do have abc. So I am trying to find out
how to do that. Maybe I am going about it the wrong way.
Chuck
- Original Message -
From: Thalis A. Kalfigopoulos [EMAIL PROTECTED]
To: Chuck Payne [EMAIL PROTECTED]
Cc: PHP General [EMAIL PROTECTED]
Sent: Sunday, March 17, 2002 9:40 PM
Subject: Re: [PHP] Brain Dead...
What does I get that it can't do the query mean? What error are you
getting?
On Sun, 17 Mar 2002, Chuck Payne wrote:
Ok, but I have this as an example
select name=abc size=1
option value=a selectedA/option
option value=bB/option
option value=cC/option
option value=dD/option
/select
Are you saying that abc should be $abc? I not clear.
I have for my sql as this...
db_name = xxxinks;
$table_name = ssslinks;
$connection = @mysql_connect(mybox.com,anyusers,!@#$%) or die
(Couldn't
connect);
$db = @mysql_select_db($db_name, $connection) or die (Couldn't select
database.);
$sql = INSERT INTO $table_name
(abc, key_word, links, name)
VALUES
('$abc', '$keywords', '$links', '$name');
$result = @mysql_query($sql, $connection) or die(Couldn't execute
query.);
but I get that it can't do the query. So I am think it not passing
abc...
Chuck
- Original Message -
From: Thalis A. Kalfigopoulos [EMAIL PROTECTED]
To: Chuck PUP Payne [EMAIL PROTECTED]
Cc: PHP General [EMAIL PROTECTED]
Sent: Sunday, March 17, 2002 9:25 PM
Subject: Re: [PHP] Brain Dead...
Dropdownmenu is a select name=ddmenuoption/select
So the php variable that will hold the user's choice is stored in
variable
$ddmenu.
Thus, no input needed.
cheers,
--thalis
On Sun, 17 Mar 2002, Chuck PUP Payne wrote:
I am trying to get a form to work with pull down menu. But I am so
brain
dead I forgot how to get the input from the pull down menu. Where do
I
put
input to pass on to php?
Chuck Payne
--
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To unsubscribe, visit: http://www.php.net/unsub.php
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Re: [PHP] Brain Dead...
I think it I am the right path now, it has to do with $_GET or $_POST not
passing the information on to mysql.
But I will look at the log shortly.
Chuck
- Original Message -
From: Thalis A. Kalfigopoulos [EMAIL PROTECTED]
To: Chuck Payne [EMAIL PROTECTED]
Cc: PHP General [EMAIL PROTECTED]
Sent: Sunday, March 17, 2002 10:00 PM
Subject: Re: [PHP] Brain Dead...
Is it possible to check the Mysql log and tell us its point of view?
On Sun, 17 Mar 2002, Chuck Payne wrote:
When I go to post, I get the following alertCouldn't execute query
so
I know I can connect, so it must be my SQL statement. That why I
wondering
how to pass the menu information to php so that it can do the SQL
statement.
I don't have $abc in the menu but I do have abc. So I am trying to find
out
how to do that. Maybe I am going about it the wrong way.
Chuck
- Original Message -
From: Thalis A. Kalfigopoulos [EMAIL PROTECTED]
To: Chuck Payne [EMAIL PROTECTED]
Cc: PHP General [EMAIL PROTECTED]
Sent: Sunday, March 17, 2002 9:40 PM
Subject: Re: [PHP] Brain Dead...
What does I get that it can't do the query mean? What error are you
getting?
On Sun, 17 Mar 2002, Chuck Payne wrote:
Ok, but I have this as an example
select name=abc size=1
option value=a selectedA/option
option value=bB/option
option value=cC/option
option value=dD/option
/select
Are you saying that abc should be $abc? I not clear.
I have for my sql as this...
db_name = xxxinks;
$table_name = ssslinks;
$connection = @mysql_connect(mybox.com,anyusers,!@#$%) or die
(Couldn't
connect);
$db = @mysql_select_db($db_name, $connection) or die (Couldn't
select
database.);
$sql = INSERT INTO $table_name
(abc, key_word, links, name)
VALUES
('$abc', '$keywords', '$links', '$name');
$result = @mysql_query($sql, $connection) or die(Couldn't execute
query.);
but I get that it can't do the query. So I am think it not passing
abc...
Chuck
- Original Message -
From: Thalis A. Kalfigopoulos [EMAIL PROTECTED]
To: Chuck PUP Payne [EMAIL PROTECTED]
Cc: PHP General [EMAIL PROTECTED]
Sent: Sunday, March 17, 2002 9:25 PM
Subject: Re: [PHP] Brain Dead...
Dropdownmenu is a select name=ddmenuoption/select
So the php variable that will hold the user's choice is stored in
variable
$ddmenu.
Thus, no input needed.
cheers,
--thalis
On Sun, 17 Mar 2002, Chuck PUP Payne wrote:
I am trying to get a form to work with pull down menu. But I am
so
brain
dead I forgot how to get the input from the pull down menu.
Where do
I
put
input to pass on to php?
Chuck Payne
--
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To unsubscribe, visit: http://www.php.net/unsub.php
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Re: [PHP] Brain dead - need help!
use a phpinfo() in one of your pages...that'll show you various places to get your data... btw, php does have the $HTTP_GET_VARS that'll have what you want... jack Gaylen Fraley wrote: I know that I know the answer - it's just buried in the recesses of my mind .. PHP has the various $HTTP_*_VARS arrays. Is there one to get the formliteral pairs from the hyperlinks, as in a href=somepage.php?var1=test? I want to extract the $var1 value pair. I know I can do this through javascript, but I want somepage.php to parse and use the variable/value. If there is not an array, then what is the simplest way? Thanks! -- Gaylen [EMAIL PROTECTED] Home http://www.gaylenandmargie.com/ PHP KISGB v2.6 Guest Book http://www.gaylenandmargie.com/phpwebsite/ -- 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]
Re: [PHP] Brain dead - need help!
DUH! I had tried that and it didn't work, so I panicked and posted! Come to find out, I had a typo. I am so red thanks! BTW, to the others who answered privately that it's automatically there - Not if register_globals is set to OFF, which is what I am using. But thanks for responding! -- Gaylen [EMAIL PROTECTED] Home http://www.gaylenandmargie.com/ PHP KISGB v2.6 Guest Book http://www.gaylenandmargie.com/phpwebsite/ Jack Dempsey [EMAIL PROTECTED] wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... use a phpinfo() in one of your pages...that'll show you various places to get your data... btw, php does have the $HTTP_GET_VARS that'll have what you want... jack Gaylen Fraley wrote: I know that I know the answer - it's just buried in the recesses of my mind .. PHP has the various $HTTP_*_VARS arrays. Is there one to get the formliteral pairs from the hyperlinks, as in a href=somepage.php?var1=test? I want to extract the $var1 value pair. I know I can do this through javascript, but I want somepage.php to parse and use the variable/value. If there is not an array, then what is the simplest way? Thanks! -- Gaylen [EMAIL PROTECTED] Home http://www.gaylenandmargie.com/ PHP KISGB v2.6 Guest Book http://www.gaylenandmargie.com/phpwebsite/ -- 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] Brain dead - need help!
I know that I know the answer - it's just buried in the recesses of my mind .. PHP has the various $HTTP_*_VARS arrays. Is there one to get the formliteral pairs from the hyperlinks, as in a href=somepage.php?var1=test? I want to extract the $var1 value pair. I know I can do this through javascript, but I want somepage.php to parse and use the variable/value. If there is not an array, then what is the simplest way? Thanks! -- Gaylen [EMAIL PROTECTED] Home http://www.gaylenandmargie.com/ PHP KISGB v2.6 Guest Book http://www.gaylenandmargie.com/phpwebsite/ -- 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]
