wow I like it :)
thanks alot Tom you've been a big help and showed me something I never knew
before also at the time
Thanks to all who help me with this , it's very much appreciated
Deadsam
Tom Rogers [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Hi,
I'm trying to get an uploader to work, where you check to make sure it's a
jpeg file that's being uploaded, and no other files allowed. the server is
unix linux using php4.1
http://php.net/getimagesize
--
Like Music? http://l-i-e.com/artists.htm
I'm looking for a PRO QUALITY two-input sound
maybe it would be easier to show you the part of code I'm using,
echo $HTTP_POST_FILES['uploadFile']['size'] . ::.
$HTTP_POST_FILES['uploadFile']['name']['type'] . ::;
echo $HTTP_POST_FILES['uploadFile']['name'] . ::.
$HTTP_POST_FILES['uploadFile']['tmp_name'];
br
*/
?
?php
/** first check that
Hi,
Tuesday, July 30, 2002, 3:52:23 AM, you wrote:
D maybe it would be easier to show you the part of code I'm using,
D }
D /** Check for the type of the image : only allow jpeg's */
D if($_FILES['uploadFile']['name']['type']!=image/jpeg){
D echo You can only upload jpg images.;
D
This looks good but I just have one question about it if you don't mind,
what if someone was uploading a gif or a PNG file? can it be made to only
detect jpgs?
Thanks in advance
Deadsam
Try it this way, it will check if it is a jpeg and it has a horizontal
and vertical size so you can be
On Sat, 27 Jul 2002 01:08:12 -0400, [EMAIL PROTECTED]
(Chris Earle) wrote:
Ahhh, good old UO. I remember GMing my taming,
crazy what they've done to
the game since I've quit (120 skill, insane!).
I'm not completely sure of a few things about your
question and I think that
I could help
Ahhh, good old UO. I remember GMing my taming, crazy what they've done to
the game since I've quit (120 skill, insane!).
I'm not completely sure of a few things about your question and I think that
I could help if you supply the answers to my questions.
Does one tamer's taming of an animal
Hi Gurus,
I am using PHP 4.1.1, Postgresql 7.2 and Perl 5.6.0 on Linux.
I want to execute a perl script throgh a PHP, which will take backup of the
database after submit button on the PHP is clicked.
I am using exec command, but as PHP runs as "nobody" user, it does not have
access to Postgres
John Kelly [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Hi, I have a form using the post method that outputs a dynamic number of
fields each time with the same name but different values - this is
required.
For example ...
filename[]
filename[]
filename[]
John Kelly [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
John Kelly [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Hi, I have a form using the post method that outputs a dynamic number of
fields each time with the same name
1. your while loop should have {} brackets
i can't see where it starts and where it ends.
so does PHP. leaving brackets away tells PHP
that only the next line is part of the while loop.
i don't know if your file has only the three lines
($cust_name, $cust_area, $cust_code) or if
have you included the MAX_FILE_SIZE hidden field?
If yes try increasng the value to above the image size
Walter Enzenhofer [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
hi there,
i´m sorry to post that stupid message, but i really need help in finding
the
this worked for me.. might be useful to someone...
// sort the array..
$current = 0; // we begin at the top
$totals = forum_count_values($r);
while ( $current sizeof($r)) {
$c = $r[$current][FORUM_POST_FIELD_ID];
// count the number of posts that are replies to current.
Php [EMAIL PROTECTED] wrote:
function Authorize() {
global $HTTP_POST_VARS;
$username = $HTTP_POST_VARS['username'];
$password = $HTTP_POST_VARS['password'];
echo T.$username.$password;
}
(insert the line above.)
jim
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Worry not, easier typing:
$search .= ba = '$b';
If above doesn't satify the needs, maybe you are mysql_escape_string();
it?
--
Julio Nobrega
No matter where you go, this.
De Necker Henri [EMAIL PROTECTED] wrote in message
My output is still the same!
It looks like this : ba = \'DA\'
:)
-Original Message-
From: Julio Nobrega Trabalhando
[mailto:[EMAIL PROTECTED]]
Sent: 29 November 2001 13:54
To: [EMAIL PROTECTED]
Subject: [PHP] Re: Need help with a stupid problem!
Worry not, easier typing:
$search
name the checkbox as:
input type='checkbox' value='?=$field['id']?' name='selectedplayers'
where ID is used to identify this current field!
then read in the posted page as:
?
while (list(, $selected) = each($selectedplayers))
{
// $selected holds the ID of the field that was previously
: [PHP] Re: Need help on putting variable into form
No problem.
With the space in there, it considered Company as something separate...
-Andy
Hugh Danaher [EMAIL PROTECTED] wrote in message
000801c12f74$62b93020$3907f4d8@win95">news:000801c12f74$62b93020$3907f4d8@win95...
Andy,
T
On Tue, 28 Aug 2001, Hugh Danaher wrote:
I think at one time I tried using double quotes but didn't get good results.
If you have time to answer, why the backslash and what the hell is foo?
Using the backslash escapes the double quote ... tells php to not use the
double quote (as it does to
Wow, I'll try it! I don't think someone will put extra info into the cell,
but you never know.
Thanks,
Hugh
- Original Message -
From: James Holloway [EMAIL PROTECTED]
To: Hugh Danaher [EMAIL PROTECTED]
Sent: Tuesday, August 28, 2001 7:11 AM
Subject: Re: Need help on putting variable
Well,think about it like this:
If you did replace $comp name with ABC Company, this is what you'd get:
input type=text size=30 name=comp_name value=ABC Company
Not quite right, eh? This will solve your problem:
print input type=text size=30 name=comp_name value=\$comp_name\;
=)
-Andy
Hugh
Andy,
Thank you. It Works!
Hugh
- Original Message -
From: Andy Ladouceur [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Monday, August 27, 2001 8:35 PM
Subject: [PHP] Re: Need help on putting variable into form
Well,think about it like this:
If you did replace $comp name with ABC
--
From: Andy Ladouceur [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Monday, August 27, 2001 8:35 PM
Subject: [PHP] Re: Need help on putting variable into form
Well,think about it like this:
If you did replace $comp name with ABC Company, this is what you'd get:
input type=text size
Try this:
printf(TABLE);
while($row=...fetch()) {
printf(TRTD%s/TDTD%s/TD/TR,$string_variable1,
$string_variable2);
}
- Original Message -
From: John Bass [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Thursday, August 23, 2001 6:29 PM
Subject: Re: [PHP] Need help to create HTML
John Bass [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Hi All,
I am trying to generate a table with two columns. Each column should
print
2 fields named image_link and web_url. Have spent over week looking
documentations and trying different coding but
Ah! But never forget...the code in some books have bugs too.
Van
At 12:47 AM 7/23/01, you wrote:
it was my error reporting along i copied that script dtraight from the
text book so i knew it wasnt the variables themselves
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that error is because the php error checking had been changed from the
default...but still some do have problems... :D
- Original Message -
From: Van Tate Jr. [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Monday, July 23, 2001 5:01 PM
Subject: Re: [PHP] Re: need help w/ variables
Ah
it was my error reporting along i copied that script dtraight from the
text book so i knew it wasnt the variables themselves
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For additional commands, e-mail: [EMAIL PROTECTED]
To contact the list
yep, thought so :D
- Original Message -
From: Virgil Claritt [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Monday, July 23, 2001 3:47 PM
Subject: [PHP] Re: need help w/ variables
it was my error reporting along i copied that script dtraight from the
text book so i knew it wasnt
$result = mysql_query($db, SELECT count(*) AS song_count FROM table_name);
if ($array = mysql_fetch_array($result)) $song_count = $array(song_count);
Tim Ward
Senior Systems Engineer
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