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
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 insert the output in a table. Is there a way to ‘echo’ into a
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 $_SERVER['PHP_SELF'].?;
, 2012 2:22 PM
To: php-db@lists.php.net
Subject: Re: [PHP-DB] echo into variable or the like
You echo to the page not to a variable.
php.net is your friend.
:)
GL
On Aug 20, 2012, at 11:01 PM, s@optusnet.com.au s@optusnet.com.au
wrote:
Hi, this is my first post so forgive me if I
thanks to Chris and Dimiter,
I think I am close but still the problem is not solved, when I add
echo img src='/album/img/.$row[photoFileName].' /
to the code as it was sugested by Dimiter there is parse Error :
PHP Parse error: syntax error, unexpected $end in
elk dolk wrote:
thanks to Chris and Dimiter,
I think I am close but still the problem is not solved, when I add
echo img src='/album/img/.$row[photoFileName].' /
to the code as it was sugested by Dimiter there is parse Error :
PHP Parse error: syntax error, unexpected $end in
thank you all,
it did it!
[EMAIL PROTECTED]
You need a trailing
you have echo img src='/album/img/.$row[photoFileName].' /
it needs to be
echo img src='/album/img/.$row[photoFileName].' /;
-
Bored stiff? Loosen up...
Download and play hundreds of games
try
echo img src='/album/img/{$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
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 the images in a
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
I am storing just the name of photos in the database and the photos are in /img
folder ,
and there is no permissions issue. My testing server is IIS And the path would
be
something like this : Inetpub\wwwroot\album\img
as I am running out of time! could someone complete this code just with
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/
--
Hi fellow php late night peoples
Thanks for everyone's help the last couple of days...
I have learned so much. The best way to learn this stuff...is simple trial and
error...
and thanks for good old Ggle!
My problem tonight is very simple. I just don't know what command I want to
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
@lists.php.net
Subject: [PHP-DB] ECHO $variable
Date: Sun, 08 Oct 2006 01:32:13 -0400
In one of my scripts I have
input type=text name=message_title size=40 maxlength=80 value=?echo
$saved_message_title;?
where
$saved_message_title is 1 Peter 5:7 Cast all your cares on Him for He
cares about you
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
In one of my scripts I have
input type=text name=message_title size=40 maxlength=80 value=?echo
$saved_message_title;?
where
$saved_message_title is 1 Peter 5:7 Cast all your cares on Him for He
cares about you
--- note the
When this is displayed on the screen it reads
1 Peter 5:7
I am
Hi
Where's the DB question?
Niel
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
I have the following code in my page as a header.
?
if ($mainarea==Language)
{
header(Location:add_5-14_material.php?mainarea=$mainarea);
}
else
{
}
?
When I echo $mainarea I get \'Language\'. I think it is something to do
with my quotation marks but need some help.
How do I get rid of the '
Hi there everyone,
Thank you all for your responses to my question about the best way to end a
PHP command (Or use Echo to print HTML).
My gut feeling was it should be ok, it's just that *To me* it is MUCH easier
to follow code which doesn't contain echo table width=50 etc . I
much prefer to
Hi I'm trying to print the contents of a cookie (php 4.2.3) the syntax below
is wrong but what should it be?
if ($_cookie[cookiename]== TRUE) {
echopyour cookie is $_COOKIE[cookiename]/p;
}
Regards
Steve Dodkins
IMPORTANT NOTICE The information in this e-mail is confidential and should
only
:
Subject: [PHP-DB] echo printing a
cookie
10/15/2002 09:21 AM
Here ya go..
if ($_cookie[cookiename]) {
echopyour cookie is.$_COOKIE['cookiename']./p;
}
Cheers
Simon
-Original Message-
From: Steve Dodkins [mailto:[EMAIL PROTECTED]]
Sent: 15 October 2002 15:22
To: Php-Db (E-mail)
Subject: [PHP-DB] echo printing a cookie
Hi I'm trying
If (isset($_COOKIE[cookiename]) {
echo 'Your cookie is: '.$_COOKIE[cookiename].' br';
}
-Original Message-
From: Steve Dodkins [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, October 15, 2002 9:22 AM
To: Php-Db (E-mail)
Subject: [PHP-DB] echo printing a cookie
Hi I'm trying
-Original Message-
From: Steve Dodkins [mailto:[EMAIL PROTECTED]]
Sent: 15 October 2002 14:22
To: Php-Db (E-mail)
Hi I'm trying to print the contents of a cookie (php 4.2.3)
the syntax below
is wrong but what should it be?
if ($_cookie[cookiename]== TRUE) {
echopyour cookie
Hi People,
Below I have included some code that just doesn't seem to be functioning all
correctly. I have this running and it appears to stop the query/echo after
the first entry in the database. What is it that I am doing wrong here??
Regards,
Shannon
?
$roundnum = "";
$sql = "select * from
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