[PHP-DB] vaiable from one page to another
hi, i hav two programs , html form name=f action=p.php mehtod=post input type=text name=fname input type=submit /form /html p.php: ?php echo($fname) ? i tried these programs and i didnt get the value of the variable in another php program. butit is not getting. one our friend told to put like echo $_post(fname) but it is also not working. now getting the error about post array. so what to do. help me regards bis - Do you Yahoo!? Yahoo! Mail Plus - Powerful. Affordable. Sign up now
Re: [PHP-DB] vaiable from one page to another
On Wednesday 05 February 2003 16:36, bismi jomon wrote: hi, i hav two programs , html form name=f action=p.php mehtod=post input type=text name=fname input type=submit /form /html p.php: ?php echo($fname) ? i tried these programs and i didnt get the value of the variable in another php program. butit is not getting. one our friend told to put like echo $_post(fname) but it is also not working. now getting the error about post array. so what to do. manual A simple tutorial manual Variables Variables from outside PHP -- Jason Wong - Gremlins Associates - www.gremlins.biz Open Source Software Systems Integrators * Web Design Hosting * Internet Intranet Applications Development * -- Search the list archives before you post http://marc.theaimsgroup.com/?l=php-db -- /* If you always postpone pleasure you will never have it. Quit work and play for once! */ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: vaiable from one page to another
Hi, you have to write it exactly like this: echo $_POST['fname']; HTH, Bastian echo $_post(fname) -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: vaiable from one page to another
Try p.php: ?php //Not ?php echo($fname); ? Jochem Bismi Jomon wrote: hi, i hav two programs , html form name=f action=p.php mehtod=post input type=text name=fname input type=submit /form /html p.php: ?php echo($fname) ? i tried these programs and i didnt get the value of the variable in another php program. butit is not getting. one our friend told to put like echo $_post(fname) but it is also not working. now getting the error about post array. so what to do. help me regards bis -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Help with select box - multiple...
Hi All,
Sorry to be a pain on this one as it is sorta off topic.. but any help
would be awesome.
I'm creating a report for Product Popularity in which the Admin is
present with a form and they have the ability to select A product or
multiple products from a drop down combo box.
The form is submitted to itself.. and all works.. meaning I can select
multiple products from the drop down box and the results displayed fine.
However, I want to SHOW to the user (in the drop down box) which
products were selected. Things work fine if I am selecting on product -
meaning it show the product name being sleected.. however, if multiple
products are selected it doesn't show any as being selected.
The select box code is below as is the script to iterate through the
products array.
Thanks all.
*** product select box ***
select name=products[] class=selectcontent size=3 multiple
option value= class=selectcontentProduct/option
?php
$productQuery = db_query(SELECT name, SUM(quantity) as mostSold FROM
CartTable WHERE submitted=1 GROUP BY name ORDER BY mostSold DESC);
while ($productResult = db_fetch($productQuery)) {
?
option value=?php echo $productResult['name']; ? ?php if
($productResult['name'] == $str_products) echo selected; ?
class=selectcontent?php echo $productResult['name']; ?/option
?php
}
?
/select
*** array iteration code **
if ($products) {
if (is_array($products)) {
$str_products = implode(,,$products);
}
else {
$str_products = $products;
}
echo $str_products;
}
Aaron
Re: [PHP-DB] Help with select box - multiple...
You can take the selected items posted to your page and use in_array() to
search for the current item as your code creates the list. this is a chunk
of code that i use to do the same thing. I wrote this a long time ago, so
there may be better ways to do thisbut it works. ;-)
snip
if (in_array($value, $array)) {
print option value=\$value\ selected.$value./option\n;
} else {
print option value=\$value\.$value./option\n;
}
snip
hth,
jeff
Aaron Wolski
aaronjw@marte To: [EMAIL PROTECTED]
kbiz.comcc:
Subject: [PHP-DB] Help with select box -
multiple...
02/05/2003
08:42 AM
Hi All,
Sorry to be a pain on this one as it is sorta off topic.. but any help
would be awesome.
I'm creating a report for Product Popularity in which the Admin is
present with a form and they have the ability to select A product or
multiple products from a drop down combo box.
The form is submitted to itself.. and all works.. meaning I can select
multiple products from the drop down box and the results displayed fine.
However, I want to SHOW to the user (in the drop down box) which
products were selected. Things work fine if I am selecting on product -
meaning it show the product name being sleected.. however, if multiple
products are selected it doesn't show any as being selected.
The select box code is below as is the script to iterate through the
products array.
Thanks all.
*** product select box ***
select name=products[] class=selectcontent size=3 multiple
option value= class=selectcontentProduct/option
?php
$productQuery = db_query(SELECT name, SUM(quantity) as mostSold FROM
CartTable WHERE submitted=1 GROUP BY name ORDER BY mostSold DESC);
while ($productResult = db_fetch($productQuery)) {
?
option value=?php echo $productResult['name']; ? ?php if
($productResult['name'] == $str_products) echo selected; ?
class=selectcontent?php echo $productResult['name']; ?/option
?php
}
?
/select
*** array iteration code **
if ($products) {
if (is_array($products)) {
$str_products = implode(,,$products);
}
else {
$str_products = $products;
}
echo $str_products;
}
Aaron
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Re: [PHP-DB] UPDATE doesn't work
Thank Jason! It counts now.
So if i want to make something like
Most view tutorial is $row[title] How can i do this?
Regards,
- Original Message -
From: Jason Wong [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Wednesday, February 05, 2003 1:44 PM
Subject: Re: [PHP-DB] UPDATE doesn't work
On Wednesday 05 February 2003 12:50, Le Hoang wrote:
// Add 1 view to the view column
$v = $row[view];
$vplus = $v+1;
$view = mysql_query(update photoshop_tutorial where id=$id set
view=$vplus); // Problem here!
You should ALWAYS check the result of a call to mysql_query():
if ($view === false) { echo Fatal error: . mysql_error(); }
Your problem is that the SQL for update is incorrect and it should be:
UPDATE photoshop_tutorial SET view = $vplus WHERE id = $id
You can also do the increment inside the SQL so:
UPDATE photoshop_tutorial SET view = view + 1 WHERE id = $id
--
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Open Source Software Systems Integrators
* Web Design Hosting * Internet Intranet Applications Development *
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[PHP-DB] Argh ! array help required :(
Hi All,
Again.. a little off topic - sorry (again!).
Have a form that allows me to select a bunch of product names from a
drop down combo box.
I'm trying to create code like in ('Product name1','Product Name2) to
be inserted into an SQL statement.
I figured that since $products was ALREADY in an array format the code
below would work.. but it doesn't:
$prod_search = in(;
for ($i=0;$icount($products);$i++) {
$prod_search .= escapeValue($products[$i]);
if ($i != (count($products) - 1)) {
$prod_search .= ,;
}
}
$prod_search .= );
Can anyone see where I am going wrong?
Thanks!
Aaron
RE: [PHP-DB] odbc query with single quote in string
The 1st one didn't work, but the 2nd one did work. Fantastic! Thanks alot,
you saved me from frustrations.
Quoting John W. Holmes [EMAIL PROTECTED]:
I'm having problem querying a table with a field value containing an
appostropy, please help. Using ODBC to connect MSAcess database.
$mQuery = CustomerID='$mCust';
$mCur2 = odbc_do( $mCnx, select Login from Emails where $mQuery
);
When it hits O'Donald, James, I get error in odbc_exec(). I tried
variations of $mQuery, including:
$mQuery = addslashes( $mQuery );
You don't want to use addslashes() on $mQuery, you want to use it on the
data you're inserting between the quotes, i.e. $mCust.
$mQuery = CustomerID=' . addslashes($mCust) . ';
If that still causes an error, your database may require quotes to be
escaped with another quote, instead of a backslash. In that case, you
can create a function like this
function addslashes2($data)
{ return str_replace(','',$data); }
Hope that helps.
---John W. Holmes...
PHP Architect - A monthly magazine for PHP Professionals. Get your copy
today. http://www.phparch.com/
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Re: [PHP-DB] Argh ! array help required :(
On Thursday 06 February 2003 01:13, Aaron Wolski wrote:
Hi All,
Again.. a little off topic - sorry (again!).
Have a form that allows me to select a bunch of product names from a
drop down combo box.
I'm trying to create code like in ('Product name1','Product Name2) to
be inserted into an SQL statement.
I figured that since $products was ALREADY in an array format the code
below would work.. but it doesn't:
$prod_search = in(;
for ($i=0;$icount($products);$i++) {
$prod_search .= escapeValue($products[$i]);
if ($i != (count($products) - 1)) {
$prod_search .= ,;
}
}
$prod_search .= );
Can anyone see where I am going wrong?
$doo = array('dah', 'dee', 'dib');
$doo = in (' . implode(', ', $doo) . ');
echo $doo;
--
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Open Source Software Systems Integrators
* Web Design Hosting * Internet Intranet Applications Development *
--
Search the list archives before you post
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[PHP-DB] checking for empty array from a form field? grrrrrrrrrrr!
Argh
HOW does one check for an array being empty from a form field??
Tried a billion different things and NOTHING works
I've tried:
if ($products == \n) {
echo hello;
}
else {
echo bye;
}
if ($products == ) {
echo hello;
}
else {
echo bye;
}
if (empty($products)) {
echo hello;
}
else {
echo bye;
}
if (!isset($products)) {
echo hello;
}
else {
echo bye;
}
if (count($products) == 0) {
echo hello;
}
else {
echo bye;
}
NOTHING works!!!
Any help.. puuulease?
Aaron
Re: [PHP-DB] checking for empty array from a form field? grrrrrrrrrrr!
First try isset($product) to see if it's being POSTed or GETted
properly. You might need to use $_POST['products'] or
$_GET['products'].
--- Aaron Wolski [EMAIL PROTECTED] wrote:
Argh
HOW does one check for an array being empty from a form
field??
Tried a billion different things and NOTHING works
I've tried:
if ($products == \n) {
echo hello;
}
else {
echo bye;
}
if ($products == ) {
echo hello;
}
else {
echo bye;
}
if (empty($products)) {
echo hello;
}
else {
echo bye;
}
if (!isset($products)) {
echo hello;
}
else {
echo bye;
}
if (count($products) == 0) {
echo hello;
}
else {
echo bye;
}
NOTHING works!!!
Any help.. puuulease?
Aaron
=
Mark Weinstock
[EMAIL PROTECTED]
***
You can't demand something as a right unless you are willing to fight to death to
defend everyone else's right to the same thing.
-Stolen from the now-defunct Randy's Random mailing list.
***
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[PHP-DB] calendário
Eu tenho dois SELECT um para os meses e outros para os anos, como faço para garantir que o utilizador não possa escolher um ano já passado, e durante o presente ano apenas os meses seguintes e o actual. Desde já obrigado __ Rui Palma ICQ#: 171381429 Current ICQ status: + More ways to contact me __
RE: [PHP-DB] checking for empty array from a form field? grrrrrrrrrrr!
What about:
count($arrayname)
Should tell you how many items are in an array.
-Original Message-
From: Aaron Wolski [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, February 05, 2003 4:04 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] checking for empty array from a form field?
grrr!
Argh
HOW does one check for an array being empty from a form
field??
Tried a billion different things and NOTHING works
I've tried:
if ($products == \n) {
echo hello;
}
else {
echo bye;
}
if ($products == ) {
echo hello;
}
else {
echo bye;
}
if (empty($products)) {
echo hello;
}
else {
echo bye;
}
if (!isset($products)) {
echo hello;
}
else {
echo bye;
}
if (count($products) == 0) {
echo hello;
}
else {
echo bye;
}
NOTHING works!!!
Any help.. puuulease?
Aaron
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Re: [PHP-DB] checking for empty array from a form field?grrrrrrrrrrr!
I usually use:
while(list($key,$value) = each($products)) {
echo $key:$value;
}
On Wed, 2003-02-05 at 15:03, Aaron Wolski wrote:
Argh
HOW does one check for an array being empty from a form field??
Tried a billion different things and NOTHING works
I've tried:
if ($products == \n) {
echo hello;
}
else {
echo bye;
}
if ($products == ) {
echo hello;
}
else {
echo bye;
}
if (empty($products)) {
echo hello;
}
else {
echo bye;
}
if (!isset($products)) {
echo hello;
}
else {
echo bye;
}
if (count($products) == 0) {
echo hello;
}
else {
echo bye;
}
NOTHING works!!!
Any help.. puuulease?
Aaron
--
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Web Developer
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713.466.0277 x 3461
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RE: [PHP-DB] checking for empty array from a form field? grrrrrrrrrrr!
Ok... well
When I select the select product which has a option value= it tells
me the count is one (1). If I select an ACTUAL product it tells me
one(1) was select.. if I select two ACTUAL products it tells me two (2)
products were selected.
For SOME reason it doesn't seem to be recognizing that the option
value= is empty!
Any more ideas??
*sigh*
Aaron
-Original Message-
From: Hutchins, Richard [mailto:[EMAIL PROTECTED]]
Sent: February 5, 2003 4:03 PM
To: 'Aaron Wolski'; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] checking for empty array from a form field?
grrr!
What about:
count($arrayname)
Should tell you how many items are in an array.
-Original Message-
From: Aaron Wolski [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, February 05, 2003 4:04 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] checking for empty array from a form field?
grrr!
Argh
HOW does one check for an array being empty from a form
field??
Tried a billion different things and NOTHING works
I've tried:
if ($products == \n) {
echo hello;
}
else {
echo bye;
}
if ($products == ) {
echo hello;
}
else {
echo bye;
}
if (empty($products)) {
echo hello;
}
else {
echo bye;
}
if (!isset($products)) {
echo hello;
}
else {
echo bye;
}
if (count($products) == 0) {
echo hello;
}
else {
echo bye;
}
NOTHING works!!!
Any help.. puuulease?
Aaron
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RE: [PHP-DB] checking for empty array from a form field? grrrrrrrrrrr!
to find out if an array is empty use:
empty()
it must be unset
NULL returns false
0 returns false
it must be truly empty to return true...
Ok... well
When I select the select product which has a option value= it tells
me the count is one (1). If I select an ACTUAL product it tells me
one(1) was select.. if I select two ACTUAL products it tells me two (2)
products were selected.
For SOME reason it doesn't seem to be recognizing that the option
value= is empty!
Any more ideas??
*sigh*
Aaron
-Original Message-
From: Hutchins, Richard [mailto:[EMAIL PROTECTED]]
Sent: February 5, 2003 4:03 PM
To: 'Aaron Wolski'; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] checking for empty array from a form field?
grrr!
What about:
count($arrayname)
Should tell you how many items are in an array.
-Original Message-
From: Aaron Wolski [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, February 05, 2003 4:04 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] checking for empty array from a form field?
grrr!
Argh
HOW does one check for an array being empty from a form
field??
Tried a billion different things and NOTHING works
I've tried:
if ($products == \n) {
echo hello;
}
else {
echo bye;
}
if ($products == ) {
echo hello;
}
else {
echo bye;
}
if (empty($products)) {
echo hello;
}
else {
echo bye;
}
if (!isset($products)) {
echo hello;
}
else {
echo bye;
}
if (count($products) == 0) {
echo hello;
}
else {
echo bye;
}
NOTHING works!!!
Any help.. puuulease?
Aaron
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RE: [PHP-DB] checking for empty array from a form field? grrrrrrrrrrr!
Somebody else replied with isset($varname). Try that. The count() function
is working as expected (see
http://www.php.net/manual/en/function.count.php). So maybe it's not the best
choice for trying to find out what you want.
-Original Message-
From: Aaron Wolski [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, February 05, 2003 4:53 PM
To: 'Hutchins, Richard'; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] checking for empty array from a form field?
grrr!
Ok... well
When I select the select product which has a option
value= it tells
me the count is one (1). If I select an ACTUAL product it tells me
one(1) was select.. if I select two ACTUAL products it tells
me two (2)
products were selected.
For SOME reason it doesn't seem to be recognizing that the option
value= is empty!
Any more ideas??
*sigh*
Aaron
-Original Message-
From: Hutchins, Richard [mailto:[EMAIL PROTECTED]]
Sent: February 5, 2003 4:03 PM
To: 'Aaron Wolski'; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] checking for empty array from a form field?
grrr!
What about:
count($arrayname)
Should tell you how many items are in an array.
-Original Message-
From: Aaron Wolski [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, February 05, 2003 4:04 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] checking for empty array from a form field?
grrr!
Argh
HOW does one check for an array being empty from a form
field??
Tried a billion different things and NOTHING works
I've tried:
if ($products == \n) {
echo hello;
}
else {
echo bye;
}
if ($products == ) {
echo hello;
}
else {
echo bye;
}
if (empty($products)) {
echo hello;
}
else {
echo bye;
}
if (!isset($products)) {
echo hello;
}
else {
echo bye;
}
if (count($products) == 0) {
echo hello;
}
else {
echo bye;
}
NOTHING works!!!
Any help.. puuulease?
Aaron
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AW: [PHP-DB] checking for empty array from a form field? grrrrrrrrrrr!
hallo,
look like here?
http://www.php.net/manual/en/function.sizeof.php
echo sizeof($HTTP_POST_VARS[myArrayObject]);
a.v.l
-Ursprüngliche Nachricht-
Von: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]
Gesendet: Mittwoch, 5. Februar 2003 02:54
An: Aaron Wolski
Cc: [EMAIL PROTECTED]
Betreff: RE: [PHP-DB] checking for empty array from a form field?
grrr!
to find out if an array is empty use:
empty()
it must be unset
NULL returns false
0 returns false
it must be truly empty to return true...
Ok... well
When I select the select product which has a option value= it tells
me the count is one (1). If I select an ACTUAL product it tells me
one(1) was select.. if I select two ACTUAL products it tells me two (2)
products were selected.
For SOME reason it doesn't seem to be recognizing that the option
value= is empty!
Any more ideas??
*sigh*
Aaron
-Original Message-
From: Hutchins, Richard [mailto:[EMAIL PROTECTED]]
Sent: February 5, 2003 4:03 PM
To: 'Aaron Wolski'; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] checking for empty array from a form field?
grrr!
What about:
count($arrayname)
Should tell you how many items are in an array.
-Original Message-
From: Aaron Wolski [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, February 05, 2003 4:04 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] checking for empty array from a form field?
grrr!
Argh
HOW does one check for an array being empty from a form
field??
Tried a billion different things and NOTHING works
I've tried:
if ($products == \n) {
echo hello;
}
else {
echo bye;
}
if ($products == ) {
echo hello;
}
else {
echo bye;
}
if (empty($products)) {
echo hello;
}
else {
echo bye;
}
if (!isset($products)) {
echo hello;
}
else {
echo bye;
}
if (count($products) == 0) {
echo hello;
}
else {
echo bye;
}
NOTHING works!!!
Any help.. puuulease?
Aaron
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RE: [PHP-DB] checking for empty array from a form field? grrrrrrrrrrr!
Well well well.. with the help of someone else pointing me in a
different direction altogether here's what we came up with that totally
works!
$PRODUCT_SELECTED = TRUE;
foreach($products as $value) {
if(empty($value)) {
$PRODUCT_SELECTED = FALSE;
}
last;
}
if(!$PRODUCT_SELECTED) {
$productQuery = db_query(SELECT name FROM CartTable
WHERE submitted=1 GROUP BY name);
while ($productResult = db_fetch($productQuery)) {
$products[] = $productResult[name];
}
$prod_search = in(;
for ($i=0;$icount($products);$i++) {
$prod_search .=
escapeQuote($products[$i]);
if ($i !=
(count($products) - 1)) {
$prod_search .=
,;
}
}
$prod_search .= );
$product = AND name $prod_search;
}
else {
$prod_search = in(;
for ($i=0;$icount($products);$i++) {
$prod_search .= escapeQuote($products[$i]);
if ($i != (count($products) - 1)) {
$prod_search .= ,;
}
}
$prod_search .= );
$product = AND name $prod_search;
}
Thanks to all for their support and help on this. Longest 4 hours of my
life :(
Aaron
-Original Message-
From: Hutchins, Richard [mailto:[EMAIL PROTECTED]]
Sent: February 5, 2003 4:53 PM
To: [EMAIL PROTECTED]
Subject: RE: [PHP-DB] checking for empty array from a form field?
grrr!
Somebody else replied with isset($varname). Try that. The count()
function
is working as expected (see
http://www.php.net/manual/en/function.count.php). So maybe it's not the
best
choice for trying to find out what you want.
-Original Message-
From: Aaron Wolski [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, February 05, 2003 4:53 PM
To: 'Hutchins, Richard'; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] checking for empty array from a form field?
grrr!
Ok... well
When I select the select product which has a option
value= it tells
me the count is one (1). If I select an ACTUAL product it tells me
one(1) was select.. if I select two ACTUAL products it tells
me two (2)
products were selected.
For SOME reason it doesn't seem to be recognizing that the option
value= is empty!
Any more ideas??
*sigh*
Aaron
-Original Message-
From: Hutchins, Richard [mailto:[EMAIL PROTECTED]]
Sent: February 5, 2003 4:03 PM
To: 'Aaron Wolski'; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] checking for empty array from a form field?
grrr!
What about:
count($arrayname)
Should tell you how many items are in an array.
-Original Message-
From: Aaron Wolski [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, February 05, 2003 4:04 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] checking for empty array from a form field?
grrr!
Argh
HOW does one check for an array being empty from a form
field??
Tried a billion different things and NOTHING works
I've tried:
if ($products == \n) {
echo hello;
}
else {
echo bye;
}
if ($products == ) {
echo hello;
}
else {
echo bye;
}
if (empty($products)) {
echo hello;
}
else {
echo bye;
}
if (!isset($products)) {
echo hello;
}
else {
echo bye;
}
if (count($products) == 0) {
echo hello;
}
else {
echo bye;
}
NOTHING works!!!
Any help.. puuulease?
Aaron
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[PHP-DB] Oracle + PHP + Linux Red Hat {again}
Please, any idea to resolve my problem!!!
I have the output:
Connected to DataBase Oracle!
Warning: OCIStmtExecute: ORA-03106: fatal two-task communication protocol
error in /var/www/html/ora2.php on line 21
Warning: OCIFetchStatement: ORA-24374: define not done before fetch or
execute and fetch in /var/www/html/ora2.php on line 22
Found: 0 results
I'm running redhat 7.3 with oracle 9i client connecting to oracle 8.0.5
Server on NT. I put the putEnv to $ORACLE_HOME,$ORACLE_SID,I use
OCIDefineByName, but the result is the error. When I'm using the oci
functions I can't receive the error above, but if I use the old oracle
functions of php, the results are good.
Do someone have any idea to resolve my problem??
Thanks in advance.
P.D. I probe to set up the User=oracle and Group dba in apache web server,
but either, it's not working.
The source code:
?php
PutEnv(ORACLE_SID=upbback);
PutEnv(ORACLE_HOME=/opt/oracle/product/9.2.0);
$db_conn = ocilogon(generador,sinok,upbback);
if (!$db_conn)
{
echo Connection failed;
echo Error Message: [ . OCIError($db_conn) . ];
exit;
}
else
{
echo Connected to DataBase Oracle!;
}
$cmdstr = select cod_aula,nombre from aulas;
$parsed = ociparse($db_conn, $cmdstr);
OCIDefineByName($parsed,COD_AULA,$codaula);
OCIDefineByName($parsed,NOMBRE,$nombre);
OCIExecute($parsed);
$nrows = ocifetchstatement($parsed, $results);
echo Found: $nrows results\n;
echo table border=1 cellspacing='0' width='50%' align=center\n;
echo tr\n;
echo tdbCodigo/b/td\n;
echo tdbNombre/b/td\n;
echo /tr\n;
for ($i = 0; $i $nrows; $i++ ) {
echo tr\n;
echo td . $codaula. /td;
echo td . $nombre. /td;
echo /tr\n;
}
echo /table\n;
OCIFreeStatement($parsed);
OCILogoff($db_conn);
?
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Re: [PHP-DB] checking for empty array from a form field? grrrrrrrrrrr!
try :
if (strlen($products) == 0) {
echo nothing entered in field;
}
http://www.php.net/manual/en/function.strlen.php
Jeff
- Original Message -
From: Aaron Wolski [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Wednesday, February 05, 2003 3:03 PM
Subject: [PHP-DB] checking for empty array from a form field? grrr!
Argh
HOW does one check for an array being empty from a form field??
Tried a billion different things and NOTHING works
I've tried:
if ($products == \n) {
echo hello;
}
else {
echo bye;
}
if ($products == ) {
echo hello;
}
else {
echo bye;
}
if (empty($products)) {
echo hello;
}
else {
echo bye;
}
if (!isset($products)) {
echo hello;
}
else {
echo bye;
}
if (count($products) == 0) {
echo hello;
}
else {
echo bye;
}
NOTHING works!!!
Any help.. puuulease?
Aaron
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[PHP-DB] Use php with sequel server2000?
I have a situation where I have a network at work with only Sequel Server 2000 on a Microsoft IIs Server. Is php a viable option for accessing a sequel database? Thanks -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: php-db Digest 5 Feb 2003 17:13:51 -0000 Issue 1661
Hi, I would like to know how to get error messages to appear if something goes wrong. Like if the mysql finds the database account but fails to update the table, finds the table but a unique entry is inputted so can't add. I'd like to be able, if possible, to customize the error message. Could someone please show me how to achieve this? At pressent it only produces an error if the database is not found. Nothing else! Jerry http://movies.yahoo.com.au - Yahoo! Movies - What's on at your local cinema? -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Remnant data from previous queries when using mysql_fetch_array...
Hello,
I am a novice PHP dev having a strange issue with mysql_fetch_array.
For some reason, it is keeping data from previous queries in the array, and
displaying it in the web page.
My development platform is Win2k SP2, Apache 1.3.23, PHP 4.3.0,
MySQL-MAX 3.23.47-nt.
Here is the code that I am using:
?php
// open database connection
$connection = mysql_connect(localhost, username, password) or die (Unable
to
connect!);
// select database
mysql_select_db(vendorNameChange) or die (Unable to select database!);
// generate and execute query
$query = select * from changelog where changedBy = '$changedBy' or
vendorNumber = '$vendorNumber'
or oldName like '$oldName' or newName like '$newName' or newVendorNumber =
'$newVendorNumber';
$result = mysql_query($query) or die (Error in query: $query. .
mysql_error());
if ($row = mysql_fetch_array($result)) {
do {
print buOld vendor number/u: /b;
print $row[vendorNumber];
print br;
snip
/snip
} while($row = mysql_fetch_array($result));
} else {print Sorry, no records were found!;}
// close database connection
mysql_close($connection);
?
I am looking for a way to initialize the array, and display only the
data returned from the current query on the page. Any help with this would
be greatly appreciated.
Thanks,
Mike Hilty
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[PHP-DB] ``List-Unsubscribe''
``List-Unsubscribe''
RE: [PHP-DB] Use php with sequel server2000?
I have a situation where I have a network at work with only Sequel Server 2000 on a Microsoft IIs Server. Is php a viable option for accessing a sequel database? Sure... ---John W. Holmes... PHP Architect - A monthly magazine for PHP Professionals. Get your copy today. http://www.phparch.com/ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Remnant data from previous queries when using mysql_fetch_array...
I am a novice PHP dev having a strange issue with
mysql_fetch_array.
For some reason, it is keeping data from previous queries in the
array,
and
displaying it in the web page.
My development platform is Win2k SP2, Apache 1.3.23, PHP 4.3.0,
MySQL-MAX 3.23.47-nt.
Here is the code that I am using:
?php
// open database connection
$connection = mysql_connect(localhost, username, password) or die
(Unable
to
connect!);
// select database
mysql_select_db(vendorNameChange) or die (Unable to select
database!);
// generate and execute query
$query = select * from changelog where changedBy = '$changedBy' or
vendorNumber = '$vendorNumber'
or oldName like '$oldName' or newName like '$newName' or
newVendorNumber =
'$newVendorNumber';
$result = mysql_query($query) or die (Error in query: $query. .
mysql_error());
if ($row = mysql_fetch_array($result)) {
do {
print buOld vendor number/u: /b;
print $row[vendorNumber];
print br;
snip
/snip
} while($row = mysql_fetch_array($result));
} else {print Sorry, no records were found!;}
// close database connection
mysql_close($connection);
?
I am looking for a way to initialize the array, and display only
the
data returned from the current query on the page. Any help with this
would
be greatly appreciated.
$row = array();
will initialize the $row variable to an empty array. Or you can use
unset($row) to just get rid of it entirely.
The problem isn't with mysql_query(), it's just that $row is retaining
data from a previous loop in your script.
---John W. Holmes...
PHP Architect - A monthly magazine for PHP Professionals. Get your copy
today. http://www.phparch.com/
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[PHP-DB] PHP Database Abstraction Layer
I once read a great article in the first or second issue of http://www.phparch.com/ on database abstraction layers. At which point I used the tutorial as a starting point for creating a very similar structure I named dbWave. There are only minor differences and a postgresql driver is now included for the most common pg_* functions. I was just wondering if anyone has developed a database abstraction layer that allows you to separate your SQL queries from your application logic like dbWave does? I'm looking for a more advanced way of doing this? Attached is dbWave for anyone to look at/use. To run it you need to use the following tags in your file: // DBWave include files include( [attached_filename].php ); To instantiate the dbWave object you use the following code in a file name connect.php ?php /* This file instantiates dbWave using our chosen API */ /* It is automatically generated by the database setup program */ // Instantiate dbWave using the MySQL API $dbWave = new Mysql(); // Connect to the database $dbWave-connect( 'yourhost', 'yourport', 'yourdbname', 'yourdbuser', 'yourdbpass' ); ? Thanks, Luke Woollard Programmer / Analyst TABORVISION.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] subdate
hello, thank you for your time. please note the time intervals that are working. what i want is to select only data that is two weeks old or younger. i did a test with one day and it worked great. i inserted the two-week/14days and neither of them worked. 1 month worked. i could not find anything on php.net that directed me to this. thank you again, addison this works:(*note end of syntax interval 1 day) $obj = mysql_db_query($dbname,select a.*,s.name as subcategory_name,c.name as category_name from ads a,subcategory s, category c where a.subcategory=s.id and s.category=c.id and a.subcategory=$id and a.que='checked' and createdate =subdate(now(), interval 1 day)); this does not: (*note end of syntax interval 14 days) $obj = mysql_db_query($dbname,select a.*,s.name as subcategory_name,c.name as category_name from ads a,subcategory s, category c where a.subcategory=s.id and s.category=c.id and a.subcategory=$id and a.que='checked' and createdate =subdate(now(), interval 14 days)); this does not: (*note end of syntax interval 2 weeks) $obj = mysql_db_query($dbname,select a.*,s.name as subcategory_name,c.name as category_name from ads a,subcategory s, category c where a.subcategory=s.id and s.category=c.id and a.subcategory=$id and a.que='checked' and createdate =subdate(now(), interval 2 weeks)); this works: (*note end of syntax interval 1 month) $obj = mysql_db_query($dbname,select a.*,s.name as subcategory_name,c.name as category_name from ads a,subcategory s, category c where a.subcategory=s.id and s.category=c.id and a.subcategory=$id and a.que='checked' and createdate =subdate(now(), interval 1 month)); -- Addison Ellis small independent publishing co. 114 B 29th Avenue North Nashville, TN 37203 (615) 321-1791 [EMAIL PROTECTED] [EMAIL PROTECTED] subsidiaries of small independent publishing co. [EMAIL PROTECTED] [EMAIL PROTECTED] -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] subdate
please note the time intervals that are working. what i want is to select only data that is two weeks old or younger. i did a test with one day and it worked great. i inserted the two-week/14days and neither of them worked. 1 month worked. i could not find anything on php.net that directed me to this. thank you again, addison this works:(*note end of syntax interval 1 day) createdate =subdate(now(), interval 1 day)); this does not: (*note end of syntax interval 14 days) createdate =subdate(now(), interval 14 days)); this does not: (*note end of syntax interval 2 weeks) createdate =subdate(now(), interval 2 weeks)); this works: (*note end of syntax interval 1 month) createdate =subdate(now(), interval 1 month)); Take a look at http://www.mysql.com/documentation/mysql/bychapter/manual_Reference.html #Date_and_time_functions and see why DAYS and WEEKS is causing you trouble. You're still not checking mysql_error() when you issue a query, are you? Otherwise you may have figured it out already. And, you're still using mysql_db_query(), which is depreciated. :( ---John W. Holmes... PHP Architect - A monthly magazine for PHP Professionals. Get your copy today. http://www.phparch.com/ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
