Ohlson Stefan wrote:
Hi!
I have a db where we stores images in the database but I haven't figured out how I display
that image after I've selected it from the table.
I've tried with IMG SRC='$data' and IMG SRC=$data but $data just return the Select-query:
htmlheadtitle/title/head
body
Dear all
Does php have any script which can check if a directory exist
in specific
folder?
Thx a lot
Jack
if (is_dir (/path/to/your/folder) )
echo Dir exists;
else
echo Dir doesn't exist;
Regards
Joakim
-
This message
Hi,
Since i dont know any mysql group, i will try here - it is
anyway related to
php.
When i want to make php/mysql application i have mysql on the
server and i
connect to it with mysql_pconnect(localhost,root,) and it works.
I assume that any user or hacker could connect to mysql
-Ursprungligt meddelande-
Från: Snijders, Mark [mailto:Mark.Snijders;atosorigin.com]
Skickat: måndag den 11 november 2002 09:03
Till: '[EMAIL PROTECTED]'; [EMAIL PROTECTED]
Ämne: RE: [PHP-DB] Find out a pic size?
check this out:
http://www.php.net/manual/en/ref.image.php
Hi,
I'm working with a lot of picture in a website and the size is not
always the same.
I need some help on how to find out the size (width and height) of a
picture. Is there any way to do this (especially with PHP)?
I need it so I can calculate the width and height to be
$sql = select quarter($qdate) or die(not work #3);
change to
$sql = select quarter($qdate) as my_quarter; //Added an alias to
quarter($qdate). Easier to access that way...
And there's no need for an 'or die()' here. You're just assigning a variable
and most likely this will allways work!
$yyy
printf(tda href=\%s?id=%sdelete=yes\Delete/a/td,
$PHP_SELF,
$myrow[id]);
printf(tda
href=\%s?id=%ssubmit=yes\Update/tdtd%s/tdtd
%s/tdtd %s/td/a/tr,
update-inv.php, $myrow[id], $myrow[name],
$myrow[details], $yyy);
My bad.
Before this printf statement you need
$my_var =
On Monday 04 November 2002 19:58, [EMAIL PROTECTED] wrote:
$sql = select quarter($qdate) or die(not work #3);
change to
$sql = select quarter($qdate) as my_quarter; //Added an alias to
quarter($qdate). Easier to access that way...
And there's no need for an 'or die()' here. You're
How can I get mySQL to group stuff by the day? my date
coloumn is a UNIX
timestamp.
SELECT whatever FROM my_table GROUP BY FROM_UNIXTIME(timestamp_col,
'%Y-%m-%d')
Regards
Joakim
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Hi Guys,
I did try that but I will try it again.
Also, I have echo'd the data and it does appear as I expect.
Any other thoughts while I try this?
Don't know if this has anything to do with your problem, but
$headers = Return-Path: $support_email\r\n;
should be
$headers .=
Hi,
I think you need to remove echo $src_img; from your non working code as this
is not what you want to do.
And do you have support for GIF compiled in?
Regards
Joakim
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Try instead:
if( $_POST['action'] ==test ){ echo Test;}
HTH
Ignatius
This should really be
if( $_GET['action'] ==test ){ echo Test;}
since it's a get-request we're dealing with...
Regards
Joakim
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OK,
Looks like there are two options
I can either convert the unix timestamp colomn to a mysql
time one *OR*
someone can tell me how to use unix timestamps in this query...
Try something like this:
SELECT
CONCAT(extract(year FROM FROM_UNIXTIME(time)),
extract(month FROM
Hello PHP people. I was wondering if I could get a
question answered?
How would I print a Database Table into a table when
the DB Table
has 4 columns, AND 4 rows.
See I only know how to print one column.
?php
// Your dB-stuff here...
$separator = ;
echo
I do not think php is the problem here.
Take a look here
http://www.mysql.com/documentation/mysql/bychapter/manual_Problems.html#Gone
_away
There are some good pointers on what to do...
Regards
Joakim
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VALUES ($insert_id, $max_second_id + 1);
$result = mysql_query($sql);
?
Might be some errors in it, but you get the idea.
Regards
Joakim Andersson
-Original Message-
From: Graeme McLaren [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, September 03, 2002 5:42 PM
To: [EMAIL PROTECTED
'%$search_string%'
and let your MySQL-server do the counting. No need to transfer all data to
PHP just to count the rows.
You still have to run two queries, but I don't think you need to worry too
much about performance issues, unless you have a really, really huge table.
Regards
Joakim Andersson
to know if you have made this function and if
so we probably need to see the function aswell.
Regards
Joakim Andersson
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From: Clive Bruton [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, July 16, 2002 10:24 PM
Joakim, thanks, that sorted it. Just one note, numrows in
the sql query
should be num_rows? That's how I got it to work anyway.
Yes, that's totally correct. Just a typo.
Joakim
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PHP Database Mailing
$rows_to_list ?/a
This oughta do it...
Regards
Joakim Andersson
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the data for output to the form, then just don't use
nl2br().
Regards
Joakim Andersson
Thanx for any help,
Andy
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mytable WHERE LEFT(my_datetime, 10) = ' .
substr($form_date, 3, 2) . . substr($form_date, 0, 2) . .
substr($form_date, 6, 4) . ';
Regards
Joakim Andersson
-Original Message-
From: Gabor Niederlaender [mailto:[EMAIL PROTECTED]]
Sent: Thursday, July 11, 2002 8:36 AM
To: [EMAIL
?php
do_html_footer();
?
Regards
Joakim Andersson
-Original Message-
From: JJ Harrison [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, July 10, 2002 8:30 AM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Re: Getting Percentage of coloumn value
Here is my script:
?
include includes
This doesn't look like a database question to me...
Anyway, try $_POST['variable_name'] if the variable comes from a form with
the POST-method
$_GET['variable_name'] - Form using GET-method or passing variables in the
URI (mypage.php?variable_name=42)
Regards
Joakim Andersson
-Original
Does the user 'gn' exist?
Are you using the correct password?
Does that user have permissions on the database?
Hint: Login failed for user 'gn' suggests that there is something wrong
with that account...
Regards
Joakim Andersson
-Original Message-
From: Gabor Niederlaender [mailto
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