I have put together this script with the help of some of you kind members
here. It seems to do what I set out for it to do, but when it finished the
upload it comes up with the following error:
Unexpected character
try for yourself: http://www.alt-design.net/php/new%20admin/add.php
I would be
in input: ' in
/home/altdesign/public_html/php/new admin/do_addauthor.php on line 14
Can you post the section around Line 14 from do_addauthor.php so we can take
a look at it ?
Regards
Girish
--
www.girishnath.co.uk
- Original Message -
From: Will Hives [EMAIL PROTECTED
Hi,
Does anyone know how to change the name of and image and save the new name
to the database?
I know how to do a standard file upload, but can't seem to get the new new
to be saved into the databaseany pointers would be fantastic.
Cheers
Will
--
PHP General Mailing List
That's cool.
How do you then delete and edit that image?
Will
in article [EMAIL PROTECTED], Lerp at
[EMAIL PROTECTED] wrote on 2/1/2002 5:07 PM:
Hi there, here's a bit of code to get you started. It includes a form that
allows uploads, and the code to process the upload on the
:
On Thursday 31 January 2002 05:13, will hives wrote:
I have this version when I add the image on the add article page, and that
works fine...
if ($picture != 'none') {
$path = /home///php/;
$path1 = http://www..net/php/;;
$b_id = mysql_insert_id();
$new_pic_name
:
On Wednesday 30 January 2002 05:32, will hives wrote:
Please do not use HTML mail, thanks!
I have produced an admin area which allows users to add, edit and delete
stories plus images. Everything has gone great, thanks to all of you who
answered previous postings.
Just stuck on the very last point
div align=left/div
/td
/tr
/table
/body
/html
in article [EMAIL PROTECTED], Jason Wong at
[EMAIL PROTECTED] wrote on 30/1/02 9:46 am:
On Wednesday 30 January 2002 17:10, will hives wrote:
The code attached was just the code which ran the html form
2002 19:27, will hives wrote:
THIS IS THE DOING EDIT SCRIPT:
[snip]
if ($picture) {
$path = /home///php/;
$path1 = http://www..net/php/;;
$b_id = mysql_insert_id();
$new_pic_name = $b_id$picture_name;
$imageFile = $path1.$new_pic_name;
copy($picture
Sorry for my past ignorance using this news site forum I can only out this
down to my lack of knowledge.
I have been work on me first script for some time now and have been making
good progress up until now. I have a script that adds a author to the
database, ID, Name, Email, Picture Name.
When
in article [EMAIL PROTECTED], Jason Wong at
[EMAIL PROTECTED] wrote on 25/1/02 3:14 pm:
On Friday 25 January 2002 17:35, Will Hives wrote:
Sorry for my past ignorance using this news site forum I can only out this
down to my lack of knowledge.
I have been work on me first script for some
please help I'm a newb and this is really messing with my head. All I want
to know is what do I need to add to this script to allow it to increment the
ID number. pleasee.
thank you very much for a simple boy
Will
?
$db_name = altone;
$table_name = author;
$connection =
?
Tyler
- Original Message -
From: will hives [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Thursday, January 24, 2002 1:43 PM
Subject: [PHP] Increment help..please
please help I'm a newb and this is really messing with my head. All I want
to know is what do I need to add
I have managed to set the file upload to change the name of the file when
uploading, but when the file is uploaded it takes the original file name not
the renamed one
image upload box is named 'picture' how do I get it to call the new file
name
Thank you for your help so far everyone...
I guess you must all be getting very bored with this type of question..but
away.. I have recently put together a simple news section which allows admin
people to add, edit and delete stories from the site...I would like to add
the function of adding an image to a story...I have managed to use the
in article [EMAIL PROTECTED], Nick Wilson at
[EMAIL PROTECTED] wrote on 1/21/2002 1:36 PM:
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
* On 21-01-02 at 14:32
* Will Hives said
I guess you must all be getting very bored with this type of question..but
away.. I have recently put
Please can someone help, I can't find any answers anywhere
I have this code:
?
$db_name = name;
$table_name = my_contacts;
$connection = @mysql_connect(www.myserver.net, username, password) or
die (couldn't connect.);
$db = @mysql_select_db($db_name, $connection) or die (couldn't select
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