Re: [PHP] update statement
Hi, I got a series of string value in this format ie: 1,2,3, say pretend that these are telephone number, which is one feild in user table. I need to update this telephone number field, I got 3 user ID in series like this too, ie: 3,4,5 and normally it is like update set where userID IN (3,4,5) ( when there is one vaule to update, not series of string to update). I would think about the general layout of the DB. It is not good to store more than one value in one field. Do some normalisation on your DB or read: http://www.devshed.com/Server_Side/MySQL/Normal/ and http://www.devshed.com/Server_Side/MySQL/Normal2/ hope it helps Johannes -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
[PHP] Update statement sent by PHP to MySQL
Hi there,
I am trying to update a row in my MySQL database (I am an newbie)
It does not work, but the DB does not return an error.
Is there an error in the syntax??
// Create the user record
//--
--
if (!($newresult = mysql_db_query($DB,
update $T1
set
caption = $caption
where name = $name
)))
{
HEADER(Location:admin.php?content=edit_dataentryerrormsg=baduserrecord);
}
//--
--
Thanx for your help
Cheers
Andy
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Re: [PHP] Update statement sent by PHP to MySQL
Hi,
update $T1 set caption = $caption where name = $name
put single quotes aroud the variables: caption='$caption'.
You can do this aswell:
$result = mysql_query(UPDATE $T1 SET caption='$caption WHERE name='$name',
$DB)
or die(mysql_error());
if (! $result) {
send the haeder...
}
hope it helps
Johannes
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[PHP] update statement
I got a series of string value in this format ie: 1,2,3, say pretend that these are telephone number, which is one feild in user table. I need to update this telephone number field, I got 3 user ID in series like this too, ie: 3,4,5 and normally it is like update set where userID IN (3,4,5) ( when there is one vaule to update, not series of string to update). How do I do that with this case then? cheers Jack [EMAIL PROTECTED] There is nothing more rewarding than reaching the goal you set for yourself
[PHP] Update statement?
hey i'm working on a shopping cart..but have hit a snag whit this statement..
function modify_quantity($table, $session, $itemid, $quantity)
{
$sql = UPDATE $table SET quantity='$quantity' WHERE
session='$session'
AND itemid='$itemid';
mysql_query($sql);
}
if there is a value in the db field quantity then after the statement is
executed it is set to 0
here is my calling code.
if ($action == uq || $action == Update ){
modify_quantity($table,$session,$itemid,$quanity);
}
and the requested url.
http://www.harborfresh.com/retail/cart.php?itemid=21591quantity=5action=Update
any ideas?
~kurth
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Re: [PHP] Update statement?
Kurth Bemis [EMAIL PROTECTED] wrote:
function modify_quantity($table, $session, $itemid, $quantity)
{
$sql = UPDATE $table SET quantity='$quantity' WHERE session='$session'
AND itemid='$itemid';
mysql_query($sql);
Add 'echo $sql;' here (no quotes) and look at the output. If the RHS of
quantity= is zero, then $quantity passed to modify_quantity is zero and
that's the problem.
if there is a value in the db field quantity then after the statement is
executed it is set to 0
I think that $quantity is the problem.
--
Steve Werby
President, Befriend Internet Services LLC
http://www.befriend.com/
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Re: [PHP] Update statement?:SOLVED
At 01:03 PM 4/30/2001, Steve Werby wrote:
I hate the letter i :-)
it was missing when i passed quantity to the function
~kurth
Kurth Bemis [EMAIL PROTECTED] wrote:
function modify_quantity($table, $session, $itemid, $quantity)
{
$sql = UPDATE $table SET quantity='$quantity' WHERE session='$session'
AND itemid='$itemid';
mysql_query($sql);
Add 'echo $sql;' here (no quotes) and look at the output. If the RHS of
quantity= is zero, then $quantity passed to modify_quantity is zero and
that's the problem.
if there is a value in the db field quantity then after the statement is
executed it is set to 0
I think that $quantity is the problem.
--
Steve Werby
President, Befriend Internet Services LLC
http://www.befriend.com/
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