Message- From: Karl DeSaulniers
Sent: Tuesday, August 21, 2012 2:22 PM
To: php-db@lists.php.net
Subject: Re: [PHP-DB] echo into variable or the like
You echo to the page not to a variable.
php.net is your friend.
:)
GL
On Aug 20, 2012, at 11:01 PM, s@optusnet.com.au s
On Tue, Aug 21, 2012 at 12:01 AM, s@optusnet.com.au wrote:
Hi, this is my first post so forgive me if I missed a rule and do something
wrong.
I have this code,
echo $_SERVER['PHP_SELF'].?;
foreach ($_GET as $urlvar=$urlval)
echo $urlvar.=.$urlval.;
It works by it’s self.
I want to
You echo to the page not to a variable.
php.net is your friend.
:)
GL
On Aug 20, 2012, at 11:01 PM, s@optusnet.com.au s@optusnet.com.au
wrote:
Hi, this is my first post so forgive me if I missed a rule and do
something wrong.
I have this code,
echo $_SERVER['PHP_SELF'].?;
, 2012 2:22 PM
To: php-db@lists.php.net
Subject: Re: [PHP-DB] echo into variable or the like
You echo to the page not to a variable.
php.net is your friend.
:)
GL
On Aug 20, 2012, at 11:01 PM, s@optusnet.com.au s@optusnet.com.au
wrote:
Hi, this is my first post so forgive me if I
elk dolk wrote:
thanks to Chris and Dimiter,
I think I am close but still the problem is not solved, when I add
echo img src='/album/img/.$row[photoFileName].' /
to the code as it was sugested by Dimiter there is parse Error :
PHP Parse error: syntax error, unexpected $end in
try
echo img src='/album/img/{$row[photoFileName]}' / ;
warpping the array element in braces allows for proper evaluation
bastien
From: elk dolk [EMAIL PROTECTED]
To: php-db@lists.php.net
Subject: [PHP-DB] echo
Date: Thu, 29 Mar 2007 05:08:36 -0700 (PDT)
thanks to Chris and Dimiter,
I
something like this : Inetpub\wwwroot\album\img
as I am running out of time! could someone complete this code just with one
echo and img src so that I can retrive my photos ?
MySQL columns : photoID=seq number
photoFileName=name of my photo like
3sw.jpg
if you view the source of the generated page, is the image name correct? is
the path to the image correct?
bastien
From: elk dolk [EMAIL PROTECTED]
To: php-db@lists.php.net
Subject: [PHP-DB] echo Date: Tue, 27 Mar 2007 22:07:37 -0700 (PDT)
Hi all,
I am new to web programming.
I have code
You might be missing a quote so here and there unless you have the
quotes stored in the database too.
Since your photos are stored on disk, make sure the webserver has
access to them.
Then make sure that your string is something like
img src=pathphotoFileName
in php: printf(img src=\%s%s\,
elk dolk wrote:
Hi all,
I am new to web programming.
I have code to add pictures to a MYSQL database. Now I can't seem to figure out how to get them back out ! so we can see them.
The MySQL doesn't seem to be a problem (yet), also I'm trying to learn PHP.
What I usually do is to load
elk dolk wrote:
I am storing just the name of photos in the database and the photos are in /img
folder ,
and there is no permissions issue.
So it's a path issue.
You need to reference the image as:
img src=/img/image_name_here
--
Postgresql php tutorials
http://www.designmagick.com/
--
I would suggest placing all the data into divs and using some js to show
those divs when the time is right
bastien
From: Matthew Ferry [EMAIL PROTECTED]
To: php-db@lists.php.net
Subject: [PHP-DB] echo delay...
Date: Fri, 16 Feb 2007 03:37:17 -0500
Hi fellow php late night peoples
easist way is to wrap the entire value in single quotes not double
quotes...kinda breaks the rules but it will work..the other option is to
search your value and do a replace on the double quotes
bastien
From: Ron Piggott (PHP) [EMAIL PROTECTED]
Reply-To: [EMAIL PROTECTED]
To: PHP DB
You can as well add a backslash BEFORE the
eg. echo text \more text\ ;
So that will return this:
text more text
- Original Message -
From: Bastien Koert [EMAIL PROTECTED]
To: [EMAIL PROTECTED]; php-db@lists.php.net
Sent: Sunday, October 08, 2006 3:35 PM
Subject: RE: [PHP-DB] ECHO
Hi
Where's the DB question?
Niel
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
I belive $_cookie is different then $_COOKIE
Jeff
Steve Dodkins
Here ya go..
if ($_cookie[cookiename]) {
echopyour cookie is.$_COOKIE['cookiename']./p;
}
Cheers
Simon
-Original Message-
From: Steve Dodkins [mailto:[EMAIL PROTECTED]]
Sent: 15 October 2002 15:22
To: Php-Db (E-mail)
Subject: [PHP-DB] echo printing a cookie
Hi I'm trying to
If (isset($_COOKIE[cookiename]) {
echo 'Your cookie is: '.$_COOKIE[cookiename].' br';
}
-Original Message-
From: Steve Dodkins [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, October 15, 2002 9:22 AM
To: Php-Db (E-mail)
Subject: [PHP-DB] echo printing a cookie
Hi I'm trying to
-Original Message-
From: Steve Dodkins [mailto:[EMAIL PROTECTED]]
Sent: 15 October 2002 14:22
To: Php-Db (E-mail)
Hi I'm trying to print the contents of a cookie (php 4.2.3)
the syntax below
is wrong but what should it be?
if ($_cookie[cookiename]== TRUE) {
echopyour cookie
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