I want to show a number from a database in the format x,xxx.00 in a
textfield, then if it is changed by the user I want to post the value of the
number to a decimal field. However, once you number_format the number it
becomes a string, and numbers like 3,379.90 give a value of 3 when posted to
could you cast as a float/double before inserting? $number = (double)
$string;
don't know what would happen to the comma, but i assume it would just get
removed??
just a guess/thought
Jeff
On Wed, 5 Nov 2003, Dillon, John wrote:
I want to show a number from a database in the format x,xxx.00 in a
textfield, then if it is changed by the user I want to post the value of the
number to a decimal field. However, once you number_format the number it
becomes a string, and numbers like
Great answer... One question though, how would you convert it back to
X,xxx.00 format??
Thanks
Aleks
-Original Message-
From: Peter Beckman [mailto:[EMAIL PROTECTED]
Sent: Wednesday, November 05, 2003 10:48 AM
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] number_format problem
On Wed,
On Wed, 5 Nov 2003, Aleks @ USA.net wrote:
Great answer... One question though, how would you convert it back to
X,xxx.00 format??
number_format($variable,2);
---
Peter Beckman
From: Dillon, John [EMAIL PROTECTED]
I want to show a number from a database in the format x,xxx.00 in a
textfield, then if it is changed by the user I want to post the value of
the
number to a decimal field. However, once you number_format the number it
becomes a string, and numbers like
From: [EMAIL PROTECTED]
could you cast as a float/double before inserting? $number = (double)
$string;
don't know what would happen to the comma, but i assume it would just get
removed??
Everything after and including the first non-number character would be
dropped.
So $12,000.34 would end
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From: Peter Beckman [EMAIL PROTECTED]
On Wed, 5 Nov 2003, Dillon, John wrote:
I use this:
$x['funds'] = (int)preg_replace(/[\$,]/,,$x['funds']);
where $x['funds'] contains something like $3,249,555.32, and the end
result is an int of 3249555. I drop the cents... you want to keep
sweet. thanks for hte correction. i try, sometimes i fail. :)
CPT John W.
From: [EMAIL PROTECTED]
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No thanks. I like it here.
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sweet. thanks for hte correction. i try, sometimes i fail. :)
But trying is half the battle. GI JOE!
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Hi there,
I need to do a loop where the mysql query starts
at the bottom and goes up, as if the data was in
reverse order. This is for building a category/
sub_category menu.
Data structure: (cat_id, cat_sub, cat_name)
sample routine:
do {
$sql = SELECT * FROM categories
From: Douglas Freake [EMAIL PROTECTED]
I need to do a loop where the mysql query starts
at the bottom and goes up, as if the data was in
reverse order. This is for building a category/
sub_category menu.
Data structure: (cat_id, cat_sub, cat_name)
sample routine:
do {
$sql = SELECT *
This is a basic question but I am all messed up and need to be straightened
out..
Have a select field called customer that works great except when there is a
' in the customer name.
Have tried addslash and stripslashes but I think I might be using them
wrong.
If I addslash to the select
you can do it at sql level by using ORDER BY ... DESC like:
SELECT * FROM categories
WHERE cat_id = '$cat'
ORDER BY cat_id DESC
and then proceed the returned recordset in 'normal' way (order)
--
Douglas Freake wrote:
Hi there,
I need to do a loop where the mysql query starts
at the bottom and
hi!
do not quite understand your problem.. pls post some code?
heres a small snippet that should work well...
$qry = 'SELECT `customer` FROM `customerList` ORDER BY `customer`';
$res = mysql_query($qry);
while($customer = mysql_fetch_object($res)) {
echo stripslashes($res-customer).'br'.\n;
Hi,
Has anybody had any luck using the ODBC libraries in PHP to connect to a SQL
Server 2000 database and passing in a SELECT with a FOR XML clause. I keep
getting
Warning: odbc_execute(): SQL error: [Microsoft][ODBC SQL Server Driver][SQL
Server]The FOR XML clause is not allowed in a CURSOR
Ok..
First I build my select list:
SELECT NAME=Cid size=1
OPTION Selected VALUE=All Customers/OPTION
?
While ($Site = mysql_fetch_array($S))
{
$Sid = $Site[CID];
$SName = htmlspecialchars($Site[Customer]);
echo(option value='$SName'$SName/options\n);
}
?
/select
The
From: Aleks @ USA.net [EMAIL PROTECTED]
First I build my select list:
SELECT NAME=Cid size=1
OPTION Selected VALUE=All Customers/OPTION
?
While ($Site = mysql_fetch_array($S))
{
$Sid = $Site[CID];
$SName = htmlspecialchars($Site[Customer]);
echo(option
Thanks John for the answer But...
Now my select statement on the Result.php page errors out when
The value has the [']in it. What the select statement looks like now
Is
Select *
From customer
Where customer.customer LIKE 'St Mary's Hospital'
Error message is
Warning
hi
think you should use ' when you create the query and in the SQL-statement
for comparison:
$qry = 'SELECT *
FROM customer
WHERE customer.customer LIKE '.$FF.'';
_ma
# life would be easier if i knew the source code...
Von: Aleks @ USA.net [EMAIL PROTECTED]
Datum: Wed, 5 Nov 2003 13:52:51
Sorry I left out the exact form
It is
$info = mysql_query( Select * From customer Where customer.customer LIKE 'St
Mary's Hospital');
-Original Message-
From: ma [mailto:[EMAIL PROTECTED]
Sent: Wednesday, November 05, 2003 1:56 PM
To: PHP-DB
Subject: Re: [PHP-DB] Select Value
hi
ok - than make it this way:
$info = mysql_query( Select * From customer Where customer.customer LIKE St
Mary's Hospital);
anyways - shouldn't it be like this?:
$FF = St Mary's Hospital;
$info = mysql_query('Select * From customer Where customer.customer LIKE
'.$FF.'');
_ma
# life would be
Tried both... Still no joy...
The statement becomes
$info = mysql_query('Select * From customer Where customer.customer LIKE St
Mary's Hospital');
Maybe I need to be a little clearer... Seem that the sql statement is now
getting the correct value
But the extra ['] is confusing it
hi
hm - it would help if you'd send us the code where you generate the query
_ma
# life would be easier if i knew the source code...
Von: Aleks @ USA.net [EMAIL PROTECTED]
Datum: Wed, 5 Nov 2003 14:23:06 -0500
An: 'ma' [EMAIL PROTECTED], 'PHP-DB' [EMAIL PROTECTED]
Betreff: RE: [PHP-DB]
Using your variables and query, the following, based on one of my own
functional pages, the following should work:
$FF = addslashes($_POST[Cid]);
$info = mysql_query(Select * From customer Where customer.customer LIKE
'$FF' );
Give it a shot. Hope this helps.
Rich
-Original Message-
BINGO Thanks Rich... I just realized were I was going wrong with my
attempt
Of addslashes I forgot to remove the ['s] in the $_POST statement.
I had $FF = addslashes($_POST ['Cid']);
Thanks for MA and John for your help also Hope to return the favors..
Aleks
-Original
You are most welcome, Aleks. Glad it helped.
-Original Message-
From: Aleks @ USA.net [mailto:[EMAIL PROTECTED]
Sent: Wednesday, November 05, 2003 3:01 PM
To: 'Hutchins, Richard'; 'PHP-DB'
Subject: RE: [PHP-DB] Select Value with 's
BINGO Thanks Rich... I just realized were I
Hello,
I'm having problems surfing to different search engines (e.g. yahoo,
google, and the like) since after putting the address to the address bar,
it redirects me to another site which is called searchalot.com and makes a
prefix of a certain IP address and making the address I input look
I cannot seem to get a SELECT COUNT for a query from fields in two different tables
and a WHERE clause. Does anyone know if this is not possible with php/mysql or am I
doing something wrong? I have tried a number of variations on the following code:
$sql = SELECT COUNT(*), bandid, bandname,
Mark Gordon wrote:
I cannot seem to get a SELECT COUNT for a query from fields in two different tables and a WHERE clause. Does anyone know if this is not possible with php/mysql or am I doing something wrong? I have tried a number of variations on the following code:
$sql = SELECT COUNT(*),
[EMAIL PROTECTED] wrote:
Hello,
I'm having problems surfing to different search engines (e.g. yahoo,
google, and the like) since after putting the address to the address bar,
it redirects me to another site which is called searchalot.com and makes a
prefix of a certain IP address and making
Yes, query is definitely working without COUNT(*). Even in the most stripped down
form, the query fails:
$sql = SELECT COUNT(bandid), genre
FROM bands, genre;
$result=mysql_query($sql);
while ($gen=mysql_fetch_row($result)) {
echo $gen[1];
}
John W. Holmes [EMAIL PROTECTED] wrote:
Mark Gordon
Mark Gordon wrote:
Yes, query is definitely working without COUNT(*). Even in the most stripped down form, the query fails:
$sql = SELECT COUNT(bandid), genre
FROM bands, genre;
$result=mysql_query($sql);
while ($gen=mysql_fetch_row($result)) {
echo $gen[1];
}
Fails how? If it echos zero, it's
maybe mysql cannot COUNT the result from more than 1 table, hence the mysql_num_rows
function - but isn't it good programming practice to get the SQL to do as much work up
front?
-
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Putting more than one table in the FROM clause means tables are joined,
then at least following problems could arise:
- using WHERE clause you can have empty recordset returned and then
COUNT conflicts with it because there is actually no any data to be
returned;
- joining two (or more)
A bit difficult to debug this without the file included (config.php);
providing the error message would also be helpful.
At first glance, I'm just wondering what does the dot mean in the table
name used in the FROM clause:
FROM school.physics_chris_rockets
It shouldn't generate a php syntacs
if tables are joined correctly it shouldn't be any problem to get count
of a column, and yes - delegating that task to the database should be
more efficient concerning the execution time
boyan
--
[EMAIL PROTECTED] wrote:
maybe mysql cannot COUNT the result from more than 1 table, hence the
On Thu, 06 Nov 2003 05:46:03 +0100
Boyan Nedkov [EMAIL PROTECTED] wrote:
A bit difficult to debug this without the file included (config.php);
providing the error message would also be helpful.
At first glance, I'm just wondering what does the dot mean in the
table name used in the FROM
Thanks for the debug advice - I will start using my_sql_error
First I got this error:
Mixing of GROUP columns (MIN(),MAX(),COUNT()...) with no GROUP columns is illegal if
there is no GROUP BY clause
So the correct code ended up:
$sql = SELECT COUNT(bandid), genre
FROM bands, genre
GROUP BY
Thanks for your help, I was able to go through line by line and fix what
was wrong, I guess a simple rewriting of some code solved it because now its
better and faster then before. Thank You all.
Jacob
I am seeing some errors with a program I wrote and I need to write
everything to a log file that the program is doing.
The following syntax I KNOW is wrong, but not sure they are important to
put here correctly yet.
//script addtocart
$Addcart (info1, info2)
Mysqlquey($addcart)
I am seeing
$Addcart() -- is Addcart() a function (in which case you should remove
the dollar sign) or are you specifically trying to do some magic there
by running a function whose name is stored in that variable?
Bogdan
Robert Sossomon wrote:
I am seeing some errors with a program I wrote and I need to
Like I said, syntactically bogus, I was trying to write it out fast
while I was away from the actual scripts... Like I said, for the MOST
part the pieces are working, however on some items it is acting
correctly yet the items are not showing up in the shopping cart or in
the database anywhere.
if you want actual errors that PHP generates and not logic based errors,
then writing your own error handling function and calling it with
set_error_handler() at the top of your scripts may work for you. in your
function you'd just need to use $errorstr, $errorfile and $errline and log
it using
From: Becoming Digital [EMAIL PROTECTED]
php|cruise is coming this March.
Final word on this, I promise! :)
I'll be on the cruise, so I'm looking forward to meeting anyone else that'll
be there. Contact me offline if you want.
I wanted to say think you to all of those that contributed to the
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