Hi,
I have a MySQL varchar column which I wish to display. Most of the
entries are numbers eg 1, 2, 3 etc etc. But some are numbers and letters
eg 34A, 34B, 35A.
When I display the column I am able to sort by number using the order
by column_name+0 command, as in:
$result = mysql_query(SELECT
Thanks Ramil,
I knew it had something to do with the \, but I still can't figure out
how to get the variable into the database without the extra '\'
It is entered into the database via a form using the input below:
input name=ident type=hidden value=?php echo
$_SERVER[AUTH_USER];?
Where would I
No, still no idea :)
I did read the manual but I can't figure out how to use it in my script.
I'll paste the page below:
FORM ACTION=?php echo($PHP_SELF); ? METHOD=POST TARGET=_self
table border=1 width=80% bgcolor=#0D9BA4
tr
td width=29%bAEC Submission Title:/td
td width=81%
I made the change but the domain\username now appears in my database as:
domainusername
-Original Message-
From: Ramil Sagum [mailto:[EMAIL PROTECTED]
Sent: Friday, 1 October 2004 11:40 AM
To: Baiocchi, Justin (CSIRO IT, Armidale)
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] The
Hello, I wonder if someone could point me in the right direction here.
I have a table that is displayed that is also a form, and allowed a
person to select a record to update using a radio button. With one of
the fields of the form/table however, I would like it to display the
value in the db (if
Thanks Eric,
I have changed it somewhat but am just getting a parse error,
unexpected T_STRING on that line.
My revised code is below:
print bCurrent Staff Working Alone/b;
print /td/tr;
print /table\n;
print table border=\1\ cellpadding=\3\ cellspacing=\0\\n;
print
Still no joy. The parse error actually refers to the -- if
($row[timeout] IS NULL); --- line.
-Original Message-
From: Eric Schwartz [mailto:[EMAIL PROTECTED]
Sent: Friday, 6 August 2004 11:32 AM
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Php if statement in a form
Sorry. Forgot to
Well, this works: so you were all on the right track
if($row['timeout']==)
{
print input name='timeout' type='text' value='echo
date('H:i')';
}else{
print $row[timeout];
}
But now I get a parse error on the input line
-Original Message-
From: Ng Hwee Hwee
Not to worry - manage to solve the problem using the line below
$date=date('H:i');
echo input name=\timeout\ type=\text\ value=\$date\;
Thanks for all the input
-Original Message-
From: Baiocchi, Justin (CSIRO IT, Armidale)
Sent: Friday, 6 August 2004 12:50 PM
To: [EMAIL PROTECTED]
Hi Neil,
I have tried it like that with no luck, what is below just happens to be
my latest attempts to get it working.
What is happening is that it is saying that $C and $G are zero, when in
fact there is data in there.
How do I get it to 'refresh' the variables? That is what I am trying to
do
Hi there, I hoped someone might be able to tell me why this is not
working.
I have a form that updates the table 'media' with the 3 values below,
that and some other variables are the only data in the table. That works
nicely.
I then want to update one of the other columns ($E) based on the
Hi,
I am trying to do the following:
- retrieve data from a table based on the $id field (easy enough)
- add a few more fields to the data set, then
- enter the data (and new fields) into a new table (based on the
selected $id field)
I'm just not sure how to structure this, do I use a temporary
Hello,
I have a page with a button that when clicked loads a pre-determined
text file into my database. The code is posted below.
However, when clicked the button does nothing, just opens up the same
page again. The data does not get loaded into the database. any ideas?
Thanks
Justin
The
Actually, I have just figured it out. It had to do with the 'slashes'
around the query. The original $sql code was created using phpmyadmin,
but I messed around with it until it worked, now using the following
query:
mysql_query(LOAD DATA INFILE 'C:abc.txt' INTO TABLE abc FIELDS
TERMINATED BY
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