Re: [PHP] function problem
* Richard Lynch <[EMAIL PROTECTED]>:
>
> > '$shippingCountry1')")))
> > {
> >echo "the insertiont cannot be done";
>
> echo mysql_error();
>
> http://php.net/mysql_error
>
> >exit();
> > }
> > header("Location:http://$HTTP_HOST/$DOCROOT/allright.html";);
>
> DON'T DO THAT!!!
>
> Just do:
> include 'allright.html';
>
> The Location header is for a document that has *MOVED* to a new URL.
>
> Your document has not moved.
Actually, untrue. From the W3C HTTP 1.1 specs:
The Location response-header field is used to redirect the recipient
to a location other than the Request-URI for completion of the
request or identification of a new resource.
Yes, one of its primary uses is for 3xx errors, and it is used in that
arena to indicate a change in a page's location. However, it can also be
used in 2xx responses to indicate a page dynamically created for the
request or simply to indicate that a 'pass-thru' was used in the
request.
It's a very common practice in web application programming -- not just
PHP, but the field in general -- after a successful form submission to
redirect to another page. Doing so can help prevent back-button issues
when forms need to be filled out in series -- for example, when you
don't want duplicate records created in the database.
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Re: [PHP] function problem
> '$shippingCountry1')")))
> {
>echo "the insertiont cannot be done";
echo mysql_error();
http://php.net/mysql_error
>exit();
> }
> header("Location:http://$HTTP_HOST/$DOCROOT/allright.html";);
DON'T DO THAT!!!
Just do:
include 'allright.html';
The Location header is for a document that has *MOVED* to a new URL.
Your document has not moved.
> exit();
> }
>
> When i try to insert -> the problem is in the insertion. The message
> "the
> insertiont cannot be done"; appears. Do you know what could be the
> mistake?
We don't know, but any time you get an error like that, the software that
caused that error will have more infomation available if you dig for it.
http://php.net/mysql_error
is just what you use for MySQL.
If it wasn't MySQL you were using, but something else in some other code,
there would still be some function to tell you what went wrong.
Get in the habit of not only checking for errors, but LOGGING them
somewhere and reviewing those logs.
You can use:
http://php.net/error_log to do like this:
error_log(__FILE__ . ': ' __LINE__ . ' ' . @mysql_error() . " $query ");
Or you could even get serious and use http://php.net/error_handler with
http://php.net/trigger_error to catch and log all errors.
By default, your errors will be in the Apache error_log file (usually
/usr/local/apache/logs/error_log) but you can send them any place you
want.
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Re: [PHP] function problem
Hi!
Thank you for the reply!
I have change the function:
function doReg($fname1="",$family1="",$company1="", $MOL1="", $dannum1="",
$bulstat1="", $phone1="", $email1="", $username1="", $password1="",
$payment1="",
$maillist1="", $Addr1="", $City1="", $zipcode1="", $Country1="",
$shippingName1="",
$shippingFamily1="", $shippingphone1="", $shippingAddr1="",
$shippingcity1="",
$shippingzipcode1="", $shippingCountry1="")
{
if(!($link = mysql_pconnect ($DB_SERVER, $DB_LOGIN, $DB_PASSWORD)))
{
echo "the connection failed";
exit();
}
if(!($newresult = mysql_db_query($DB, "insert into
users(name,family,company, MOL,
taxnum, bulstat, phone, email, username, password, payment, maillist, Addr,
City,
zipcode, Country, shippingName, shippingFamily, shippingphone, shippingAddr,
shippingcity, shippingzipcode, shippingCountry)
values('$fname1','$family1','$company1', '$MOL1', '$dannum1', '$bulstat1',
'$phone1',
'$email1', '$username1', '$password1', '$payment1', '$maillist1', '$Addr1',
'$City1',
'$zipcode1', '$Country1','$shippingName1','$shippingFamily1',
'$shippingphone1','$shippingAddr1', '$shippingcity1', '$shippingzipcode1',
'$shippingCountry1')")))
{
echo "the insertiont cannot be done";
exit();
}
header("Location:http://$HTTP_HOST/$DOCROOT/allright.html";);
exit();
}
When i try to insert -> the problem is in the insertion. The message "the
insertiont cannot be done"; appears. Do you know what could be the mistake?
Thank you!
Viktor
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Re: [PHP] function problem
Are you getting an error? If so what does it say. Also in your code you
have nothing to tell you if something is going wrong. Use echo
statements - say something if the connection or the query doesn't go
through. Also noticed that when you made the connection to the DB you
didn't assign it to a variable, but in the query you use $DB as the link.
Ex:
$DB = mysql_pconnect ($DB_SERVER, $DB_LOGIN, $DB_PASSWORD);
if (!$DB)
{
// remove in production or have it sent to a log.
echo 'Failed to connect to the DB';
}
You can do the same thing for the query.
Respectfully,
Ligaya Turmelle
Viktor Popov wrote:
Hi,
I'm trying to do the following but I don't have any success. Could you help
me here...
I have this code in mu page:
if (isset ($_POST['submit'])) {
foreach($_POST as $key=>$value) {
$$key = $value;
}
$valid = $fn = checkLength($fname, 2, 50);
$ln = checkLength($family, 2, 50);
$valid = $valid && $ln;
$cm = checkLength($company,0,50);
$valid = $valid && $cm;
$ml = checkLength($MOL,0,50);
$valid = $valid && $ml;
$dnum = checkLength($dannum,0,12);
$valid = $valid && $dnum;
$bst = checkLength($bulstat,0,12);
$valid = $valid && $bst;
$phn = checkLength($phone,3,20);
$valid = $valid && $phn;
$em = checkEmail($email);
$valid = $valid && $em;
$usr = checkLength($username,4,10);
$valid = $valid && $usr;
$ps = checkLength($password,4,16);
$valid = $valid && $ps;
$ps2 = checkLength($password2,4,16);
$valid = $valid && $ps2;
$ps2 = $password == $password2;
$valid = $valid && $ps2;
$adr = checkLength($Addr,3,70);
$valid = $valid && $adr;
$cty = checkLength($City,2,50);
$valid = $valid && $cty;
$zp = checkLength($zipcode,2,10);
$valid = $valid && $zp;
if ($valid) { //-CHECK
THIS---
doReg($fname,$family,$company, $MOL, $dannum, $bulstat,
$phone, $email, $username, $password, $payment, $maillist, $Addr, $City,
$zipcode,
$Country, $shippingName, $shippingFamily, $shippingphone, $shippingAddr,
$shippingcity, $shippingzipcode, $shippingCountry);
exit;
}
} else {
$fn = $ln = $cm = $ml = $dnum = $bst = $phn = $em = $usr = $ps = $ps2 =
$adr = $cty = $zp = TRUE;
$fname = $family = $company = $MOL = $dannum = $bulstat = $phone =
$email = $username = $password = $password2 = $Addr = $City = $zipcode = '';
}
?>
This is a page with validation. If everything is correct($valid==TRUE), I
would like to call doReg which must do the following:
function doReg($fname1="",$family1="",$company1="", $MOL1="", $dannum1="",
$bulstat1="", $phone1="", $email1="", $username1="", $password1="",
$payment1="", $maillist1="", $Addr1="", $City1="", $zipcode1="",
$Country1="", $shippingName1="", $shippingFamily1="", $shippingphone1="",
$shippingAddr1="", $shippingcity1="", $shippingzipcode1="",
$shippingCountry1="")
{
mysql_pconnect ($DB_SERVER, $DB_LOGIN, $DB_PASSWORD);
mysql_db_query($DB, "insert into users(name,family,company, MOL, taxnum,
bulstat, phone, email, username, password, payment, maillist, Addr, City,
zipcode, Country, shippingName, shippingFamily, shippingphone, shippingAddr,
shippingcity, shippingzipcode, shippingCountry)
values('$fname1','$family1','$company1', '$MOL1', '$dannum1', '$bulstat1',
'$phone1', '$email1', '$username1', '$password1', '$payment1', '$maillist1',
'$Addr1', '$City1', '$zipcode1',
'$Country1','$shippingName1','$shippingFamily1',
'$shippingphone1','$shippingAddr1', '$shippingcity1', '$shippingzipcode1',
'$shippingCountry1')");
}
The problem is that it doesn't work. I have tryed to put the code from the
doReg function in the page and it works. But when I call the function I
can't insert nothing. Why is that? could you tell me?
Thank you in advance!!
Viktor
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Re: [PHP] Function Problem
> I'm having a problem with a php application;
>
> I have two files: one is ccadduser wich adds users to a controlcenter
> that I am currently designing for a website.
>
> In that ccaduserfile I call for a function checkpermission(); this
> function is defined in another file called ccfunctions
>
> When a user does not have access to the script it should abort the
> script, this is done using a header("location: ccnopermission.php");
> statement
>
> But now it seems that while executing the function checkpermission()
> the code in ccadduser just keeps running and the database query that
> inserts the new user is executed before the user can be redirected to
> ccnopermission.
>
> Is there a way to make php wait until checkpermission is completely
executed?
>
> I know it is not a simple question, but I really need a solution to
> ensure the safety of my system.
>
> grtz & thanks
>
> DragonEye
>
After calling the header redirect call exit();
example:
header('Location: somewhere.php');
exit();
Jim Grill
Web-1 Hosting, LP
http://www.web-1hosting.net
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RE: [PHP] Function Problem (Long-ish)
I think this a lesson to me and anyone else to make more use of
mysql_error() in sql statements.
I soon as I saw the Engishized explanation via mysql_error() it was obvious
and easy to fix.
Note to self: Use mysql_error() as standard ;-)
I humbly thank you all for you help.
Dave C
-Original Message-
From: Vincent Jansen [mailto:[EMAIL PROTECTED]
Sent: 13 January 2004 13:05
To: 'Dave Carrera'; 'Richard Davey'
Cc: [EMAIL PROTECTED]
Subject: RE: [PHP] Function Problem (Long-ish)
I doubt that...
MySQL says:
Mixing of GROUP columns (MIN(),MAX(),COUNT()...) with no GROUP columns is
illegal if there is no GROUP BY clause
---
Vincent Jansen
-Original Message-
From: Dave Carrera [mailto:[EMAIL PROTECTED]
Sent: dinsdag 13 januari 2004 13:58
To: 'Richard Davey'
Cc: [EMAIL PROTECTED]
Subject: RE: [PHP] Function Problem (Long-ish)
Thanks for the reply Richard.
If I use this sql in my func:
$sql = mysql_query("select count(fieldtocount) as cnt, sum(fieldforsum) as
total from $tab3 where fieldtocompare =\"$varcomparedwith\"");
Then all is fine and works as expected.
If I then add:
$sql = mysql_query("select fieldtoselect, count(fieldtocount) as cnt,
sum(fieldforsum) as total from $tab3 where fieldtocompare
=\"$varcomparedwith\"");
So adding the extra field to select, This dose not work :-(
Although in other apps not sql-ing within a func this kind of sql query
works.
So I am puzzled why and if you or the list can help I would appreciate it.
Thank you
Dave C
-Original Message-
From: Richard Davey [mailto:[EMAIL PROTECTED]
Sent: 13 January 2004 12:48
To: Dave Carrera
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP] Function Problem (Long-ish)
Hello Dave,
Tuesday, January 13, 2004, 12:34:33 PM, you wrote:
DC> sql = mysql_query("select *, count(id) as cnt from table where
DC> somefield=\"somevar\""){
It's nothing to do with your function, simply that your SQL is invalid.
--
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Richardmailto:[EMAIL PROTECTED]
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RE: [PHP] Function Problem (Long-ish)
I doubt that...
MySQL says:
Mixing of GROUP columns (MIN(),MAX(),COUNT()...) with no GROUP columns
is illegal if there is no GROUP BY clause
---
Vincent Jansen
-Original Message-
From: Dave Carrera [mailto:[EMAIL PROTECTED]
Sent: dinsdag 13 januari 2004 13:58
To: 'Richard Davey'
Cc: [EMAIL PROTECTED]
Subject: RE: [PHP] Function Problem (Long-ish)
Thanks for the reply Richard.
If I use this sql in my func:
$sql = mysql_query("select count(fieldtocount) as cnt, sum(fieldforsum)
as total from $tab3 where fieldtocompare =\"$varcomparedwith\"");
Then all is fine and works as expected.
If I then add:
$sql = mysql_query("select fieldtoselect, count(fieldtocount) as cnt,
sum(fieldforsum) as total from $tab3 where fieldtocompare
=\"$varcomparedwith\"");
So adding the extra field to select, This dose not work :-(
Although in other apps not sql-ing within a func this kind of sql query
works.
So I am puzzled why and if you or the list can help I would appreciate
it.
Thank you
Dave C
-Original Message-
From: Richard Davey [mailto:[EMAIL PROTECTED]
Sent: 13 January 2004 12:48
To: Dave Carrera
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP] Function Problem (Long-ish)
Hello Dave,
Tuesday, January 13, 2004, 12:34:33 PM, you wrote:
DC> sql = mysql_query("select *, count(id) as cnt from table where
DC> somefield=\"somevar\""){
It's nothing to do with your function, simply that your SQL is invalid.
--
Best regards,
Richardmailto:[EMAIL PROTECTED]
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RE: [PHP] Function Problem (Long-ish)
Thanks for the reply Richard.
If I use this sql in my func:
$sql = mysql_query("select count(fieldtocount) as cnt, sum(fieldforsum) as
total from $tab3 where fieldtocompare =\"$varcomparedwith\"");
Then all is fine and works as expected.
If I then add:
$sql = mysql_query("select fieldtoselect, count(fieldtocount) as cnt,
sum(fieldforsum) as total from $tab3 where fieldtocompare
=\"$varcomparedwith\"");
So adding the extra field to select, This dose not work :-(
Although in other apps not sql-ing within a func this kind of sql query
works.
So I am puzzled why and if you or the list can help I would appreciate it.
Thank you
Dave C
-Original Message-
From: Richard Davey [mailto:[EMAIL PROTECTED]
Sent: 13 January 2004 12:48
To: Dave Carrera
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP] Function Problem (Long-ish)
Hello Dave,
Tuesday, January 13, 2004, 12:34:33 PM, you wrote:
DC> sql = mysql_query("select *, count(id) as cnt from table where
DC> somefield=\"somevar\""){
It's nothing to do with your function, simply that your SQL is invalid.
--
Best regards,
Richardmailto:[EMAIL PROTECTED]
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Re: [PHP] Function Problem (Long-ish)
On Tuesday 13 January 2004 20:34, Dave Carrera wrote: [snip] > My Question is Why ? > > Any help is a appreciated and I thank you fully in advance. You're not checking for errors. Incorporate error checking code and make use of mysql_error(). -- 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-general -- /* The goal of science is to build better mousetraps. The goal of nature is to build better mice. */ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Function Problem (Long-ish)
From: "Dave Carrera" <[EMAIL PROTECTED]> > I get a Warning: > mysql_fetch_array(): supplied argument is not a valid MySQL result resource Whenever you get this warning it's because your query failed for some reason and you're trying to use a result that's not valid. Use mysql_error() to see what the error is. ---John Holmes... -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Function Problem (Long-ish)
Hello Dave,
Tuesday, January 13, 2004, 12:34:33 PM, you wrote:
DC> sql = mysql_query("select *, count(id) as cnt from table where
DC> somefield=\"somevar\""){
It's nothing to do with your function, simply that your SQL is invalid.
--
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Richardmailto:[EMAIL PROTECTED]
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RE: [PHP] function problem?
[snip] Not sure why the last section won't work... ...so much code it made my head hurt [/snip] Not sure either. Did you have a question? -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] function problem (simple redirect)
--- Frank Tudor <[EMAIL PROTECTED]> wrote:
> function $payment{
You probably mean payment, not $payment.
> header ("location:test_page.html");
The Location header has an uppercase L, a space after the colon, and an
absolute URL after the space. Your example violates all three.
Hope that helps.
Chris
=
My Blog
http://shiflett.org/
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RE: [PHP] function problem (simple redirect)
Make that:
function payment{
global $payment;
-Gregory
-Original Message-
From: Gregory Kornblum [mailto:[EMAIL PROTECTED]
Sent: Monday, October 27, 2003 1:30 PM
To: 'Frank Tudor'; 1PHP
Subject: RE: [PHP] function problem (simple redirect)
>function $payment{
Change that to:
function payment{
$global $payment;
Regards,
-Gregory
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RE: [PHP] function problem (simple redirect)
>function $payment{
Change that to:
function payment{
$global $payment;
Regards,
-Gregory
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Re: [PHP] function problem
On 2003.02.15 00:11 Peter Gumbrell wrote: [...] $option_block .= " [...] Are you sure it's a scope problem ? You haven't instantiated any $option_block variable when the loop starts, so you're concatenating a string to a bunch of uninitialized memory :) I can't run your script at the moment, so it's just a supposition... Hope it helps, Nicholas -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] function problem
Here is the string that is being used in this function example. Everything
print except the option_block
// Vendor string
$vendor_string .= <<
Select the vendor of your choice:
$option_block;
EOVS;
-Original Message-
From: Kevin Stone [mailto:[EMAIL PROTECTED]]
Sent: Friday, February 14, 2003 6:41 PM
To: Php-General
Subject: Re: [PHP] function problem
Exactly where does $option_block get put into $string?
-Kevin
- Original Message -
From: "Peter Gumbrell" <[EMAIL PROTECTED]>
To: "Php-General" <[EMAIL PROTECTED]>
Sent: Friday, February 14, 2003 4:11 PM
Subject: [PHP] function problem
> I have a function below which populates a select list, based on the query
> string and the field name. It then prints a string:
>
> function populate_selects($query_name, $db_field, $link, $string)
> {
> $result = mysql_query($query_name, $link) or die ("Could not execute
> query.");
> while ($columns = mysql_fetch_array($result))
> {
> $column = $columns[$db_field];
> global $option_block;
> $option_block .= " value=\"$column\">$column\n";
> }
> print $string;
> }
>
> Part of the string that is printed in $string is the $option_block which
is
> in the function. There seems to be some problem with scope here. I made
the
> $option_block variable global but it still won't print. I have tested that
> the $column field is being populated and it is, so I believe that it must
be
> the $option_block part that isn't working. Does anyone have any
suggestions?
>
> Peter Gumbrell
> [EMAIL PROTECTED]
>
>
>
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Re: [PHP] function problem
Exactly where does $option_block get put into $string?
-Kevin
- Original Message -
From: "Peter Gumbrell" <[EMAIL PROTECTED]>
To: "Php-General" <[EMAIL PROTECTED]>
Sent: Friday, February 14, 2003 4:11 PM
Subject: [PHP] function problem
> I have a function below which populates a select list, based on the query
> string and the field name. It then prints a string:
>
> function populate_selects($query_name, $db_field, $link, $string)
> {
> $result = mysql_query($query_name, $link) or die ("Could not execute
> query.");
> while ($columns = mysql_fetch_array($result))
> {
> $column = $columns[$db_field];
> global $option_block;
> $option_block .= " value=\"$column\">$column\n";
> }
> print $string;
> }
>
> Part of the string that is printed in $string is the $option_block which
is
> in the function. There seems to be some problem with scope here. I made
the
> $option_block variable global but it still won't print. I have tested that
> the $column field is being populated and it is, so I believe that it must
be
> the $option_block part that isn't working. Does anyone have any
suggestions?
>
> Peter Gumbrell
> [EMAIL PROTECTED]
>
>
>
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Re: [PHP] Function Problem
Hey,
Of course they do, they work if I put the echo in the function its self, but
I dont need the echo in the function.
- Lee
"Marco Tabini" <[EMAIL PROTECTED]> wrote in message
news:1036911387.23753.1054.camel@;localhost.localdomain...
> Either add
>
> global $db;
>
> as the first line of the function or change your function call to:
>
> $db = db_conn("$host","$user","$pass","$dab");
>
> I assume that $host, $user, $pass and $dab actually contain some correct
> values.
>
>
> Marco
> --
>
> php|architect - The magazine for PHP Professionals
> The first monthly worldwide magazine dedicated to PHP programmers
> Check us out on the web at http://www.phparch.com
>
>
>
> On Sun, 2002-11-10 at 01:58, conbud wrote:
> > Hey
> > Im trying to create a fucntion in an include file
> > function db_conn($host,$user,$pass,$dab)
> > {
> > $db = mysql_connect("$host",
"$user","$pass")mysql_select_db("$dab",$db);
> > }
> >
> >
> > In the page ide use:
> > require('func.inc.php');
> > db_conn("$host","$user","$pass","$dab");
> > $result = mysql_query("select * FROM $page",$db);
> > $myrow = mysql_fetch_array($result);
> >
> > echo $myrow['general_info'];
> >
> > ...
> > ...
> > however when I use the db_conn on the webpage I just get a mysql error
> > saying not a valid resource, so how do I get the db_conn to actually
return
> > the data, ive tried using
> >
> > function db_conn($host,$user,$pass,$db)
> > {
> > $db = mysql_connect("$host", "$user","$pass")mysql_select_db("$db",$db);
> >
> > return $db
> > }
> >
> > and various other thing but still nothing, I have to actually put the
echo
> > statements in the function too to get it to work properlly, any ideas?
> >
> >
> >
> > --
> > PHP General Mailing List (http://www.php.net/)
> > To unsubscribe, visit: http://www.php.net/unsub.php
> >
>
>
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Re: [PHP] Function Problem
Either add
global $db;
as the first line of the function or change your function call to:
$db = db_conn("$host","$user","$pass","$dab");
I assume that $host, $user, $pass and $dab actually contain some correct
values.
Marco
--
php|architect - The magazine for PHP Professionals
The first monthly worldwide magazine dedicated to PHP programmers
Check us out on the web at http://www.phparch.com
On Sun, 2002-11-10 at 01:58, conbud wrote:
> Hey
> Im trying to create a fucntion in an include file
> function db_conn($host,$user,$pass,$dab)
> {
> $db = mysql_connect("$host", "$user","$pass")mysql_select_db("$dab",$db);
> }
>
>
> In the page ide use:
> require('func.inc.php');
> db_conn("$host","$user","$pass","$dab");
> $result = mysql_query("select * FROM $page",$db);
> $myrow = mysql_fetch_array($result);
>
> echo $myrow['general_info'];
>
> ...
> ...
> however when I use the db_conn on the webpage I just get a mysql error
> saying not a valid resource, so how do I get the db_conn to actually return
> the data, ive tried using
>
> function db_conn($host,$user,$pass,$db)
> {
> $db = mysql_connect("$host", "$user","$pass")mysql_select_db("$db",$db);
>
> return $db
> }
>
> and various other thing but still nothing, I have to actually put the echo
> statements in the function too to get it to work properlly, any ideas?
>
>
>
> --
> PHP General Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
>
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To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] function problem ...
Try require_once($your_file) or verify in $your_file use the function function_exists() []'s Frederico Augusto Costa [EMAIL PROTECTED] On Wed, 7 Feb 2001, Miguel Loureiro wrote: > Hello > I have a script that use other script file(php) twice, I call a function with >certain parameters and call it again with other parameters and when runnig the >script, the first function works well, but on second time I got a error message: >"Fatal error: Cannot redeclare getndays() in Unknown on line 5". > Anyone kows what is my problem ??? > T.Y. > Miguel Loureiro <[EMAIL PROTECTED] > > -- 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]
