Re: [PHP] Re: catch the error
Additionally regarding the error handling , add this to the op of your script.
ini_set(display_errors,true);
error_reporting(E_STRICT|E_ALL);
and post the output of your error message.
On Thu, Feb 26, 2009 at 1:40 PM, Ashley Sheridan
a...@ashleysheridan.co.uk wrote:
On Thu, 2009-02-26 at 13:34 -0500, Darryle Steplight wrote:
Hi PJ,
$db_host = 'biggie';
$db_user = 'root';
$db_pass = 'gu...@#$';
$db_name = 'biblane';
Everyone here is trying to help you and that's cool, but EVERYONE on
this list may not be so nice. The above credentials is definitely the
type of information you want to keep private, unless you don't mind
people potentially accessing your database tables and doing whatever
they like with them.
I suggest doing something like
$db_host = 'localhost;
$db_user = 'foo';
$db_pass= ''bar;
$db_name =''xx;
if you are going to post it on the list.
On Thu, Feb 26, 2009 at 1:22 PM, PJ af.gour...@videotron.ca wrote:
Ricardo Dias Marques wrote:
Hi PJ,
On Thu, Feb 26, 2009 at 17:28, PJ af.gour...@videotron.ca wrote:
What is wrond with this file? same identical insert works from console
but not from this file :-(
[snip]
?
//include (lib/db1.php); // Connect to database
mysql_connect('biggie', 'user', 'password', 'test');
$sql1 = INSERT INTO example (name, age) VALUES ('Joe Blow', '69');
$result1 = mysql_query($sql1,$db);
if (!$result1) {
echo(PError performing 1st query: .
mysql_error() . /P);
exit();
}
?
I haven't coded in PHP for a long time, but I think that your problem
is in this line:
$result1 = mysql_query($sql1,$db);
Up to that point, $db (that should point to a database link
identifier) is not defined. You probably want to assign the
mysql_connect result to that $db variable.
So, I think that you will solve your problem by changing your
mysql_connect line FROM the current form:
mysql_connect('biggie', 'user', 'password', 'test');
.. TO this one:
$db = mysql_connect('biggie', 'user', 'password', 'test');
Am I right?
Partly. I had an error in the location of the include. Ashley corrected
the rest but it only works with the include. Not as whown below
?
//include (../lib/db1.php); // Connect to database
$db_host = 'biggie';
$db_user = 'root';
$db_pass = 'gu...@#$';
$db_name = 'biblane';
$db_connect = mysql_connect($db_host, $db_user, $db_pass);
$db_select = mysql_select_db($db_name, $db_connect);
$sql1 = INSERT INTO test (name, age) VALUES ('Arnie Shwartz', '75');
$result1 = mysql_query($sql1,$db);
if (!$result1) {
echo(PError performing 1st query: .
mysql_error() . /P);
exit();
}
?
--
Phil Jourdan --- p...@ptahhotep.com
http://www.ptahhotep.com
http://www.chiccantine.com
--
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I agree. I wouldn't trust me at all! ;)
Ash
www.ashleysheridan.co.uk
--
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Re: [PHP] Re: catch the error
Ashley Sheridan wrote:
On Thu, 2009-02-26 at 13:34 -0500, Darryle Steplight wrote:
Hi PJ,
$db_host = 'biggie';
$db_user = 'root';
$db_pass = 'gu...@#$';
$db_name = 'biblane';
Everyone here is trying to help you and that's cool, but EVERYONE on
this list may not be so nice. The above credentials is definitely the
type of information you want to keep private, unless you don't mind
people potentially accessing your database tables and doing whatever
they like with them.
I suggest doing something like
$db_host = 'localhost;
$db_user = 'foo';
$db_pass= ''bar;
$db_name =''xx;
if you are going to post it on the list.
On Thu, Feb 26, 2009 at 1:22 PM, PJ af.gour...@videotron.ca wrote:
Ricardo Dias Marques wrote:
Hi PJ,
On Thu, Feb 26, 2009 at 17:28, PJ af.gour...@videotron.ca wrote:
What is wrond with this file? same identical insert works from console
but not from this file :-(
[snip]
?
//include (lib/db1.php);// Connect to database
mysql_connect('biggie', 'user', 'password', 'test');
$sql1 = INSERT INTO example (name, age) VALUES ('Joe Blow', '69');
$result1 = mysql_query($sql1,$db);
if (!$result1) {
echo(PError performing 1st query: .
mysql_error() . /P);
exit();
}
?
I haven't coded in PHP for a long time, but I think that your problem
is in this line:
$result1 = mysql_query($sql1,$db);
Up to that point, $db (that should point to a database link
identifier) is not defined. You probably want to assign the
mysql_connect result to that $db variable.
So, I think that you will solve your problem by changing your
mysql_connect line FROM the current form:
mysql_connect('biggie', 'user', 'password', 'test');
.. TO this one:
$db = mysql_connect('biggie', 'user', 'password', 'test');
Am I right?
Partly. I had an error in the location of the include. Ashley corrected
the rest but it only works with the include. Not as whown below
?
//include (../lib/db1.php);// Connect to database
$db_host = 'biggie';
$db_user = 'root';
$db_pass = 'gu...@#$';
$db_name = 'biblane';
$db_connect = mysql_connect($db_host, $db_user, $db_pass);
$db_select = mysql_select_db($db_name, $db_connect);
$sql1 = INSERT INTO test (name, age) VALUES ('Arnie Shwartz', '75');
$result1 = mysql_query($sql1,$db);
if (!$result1) {
echo(PError performing 1st query: .
mysql_error() . /P);
exit();
}
?
--
Phil Jourdan --- p...@ptahhotep.com
http://www.ptahhotep.com
http://www.chiccantine.com
--
MySQL General Mailing List
For list archives: http://lists.mysql.com/mysql
To unsubscribe:http://lists.mysql.com/mysql?unsub=dstepli...@gmail.com
I agree. I wouldn't trust me at all! ;)
Ash
www.ashleysheridan.co.uk
Yeah very stupid of me...but I found the error: see if you can catch it:
?
//include (../lib/db1.php);// Connect to database
$db_host = 'xxx';
$db_user = 'xxx;
$db_pass = 'xxx';
$db_name = 'xxx';
$db = mysql_connect($db_host, $db_user, $db_pass);
mysql_select_db($db_name,$db);
$sql1 = INSERT INTO test (name, age) VALUES ('Arnie Shwartz', '75');
$result1 = mysql_query($sql1,$db);
if (!$result1) {
echo(PError performing 3st query: .
mysql_error() . /P);
}
echo $sql1;
echo br /;
echo $db_select;
exit();
?
--
Phil Jourdan --- p...@ptahhotep.com
http://www.ptahhotep.com
http://www.chiccantine.com
--
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Re: [PHP] Re: catch the error
Here's the working code...
?
//include (../lib/db1.php);// Connect to database
$db_host = '';
$db_user = '';
$db_pass = '';
$db_name = '';
$db = mysql_connect($db_host, $db_user, $db_pass);
mysql_select_db($db_name,$db);
$sql1 = INSERT INTO test (name, age) VALUES ('Arnie Shwartz', '75');
$result1 = mysql_query($sql1,$db);
if (!$result1) {
echo(PError performing 3st query: .
mysql_error() . /P);
}
echo $sql1;
echo br /;
echo $db_select;
exit();
?
This works fine either as is or using the include... :-)
9el wrote:
But the question is PJ, have you got it out of errors yet? :)
www.twitter.com/nine_L http://www.twitter.com/nine_L
www.lenin9l.wordpress.com http://www.lenin9l.wordpress.com
---
Use FreeOpenSourceSoftwares, Stop piracy, Let the developers live. Get
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--
2009/2/27 Ashley Sheridan a...@ashleysheridan.co.uk
mailto:a...@ashleysheridan.co.uk
On Thu, 2009-02-26 at 13:34 -0500, Darryle Steplight wrote:
Hi PJ,
$db_host = 'biggie';
$db_user = 'root';
$db_pass = 'gu...@#$';
$db_name = 'biblane';
Everyone here is trying to help you and that's cool, but EVERYONE on
this list may not be so nice. The above credentials is
definitely the
type of information you want to keep private, unless you don't mind
people potentially accessing your database tables and doing whatever
they like with them.
I suggest doing something like
$db_host = 'localhost;
$db_user = 'foo';
$db_pass= ''bar;
$db_name =''xx;
if you are going to post it on the list.
On Thu, Feb 26, 2009 at 1:22 PM, PJ af.gour...@videotron.ca
mailto:af.gour...@videotron.ca wrote:
Ricardo Dias Marques wrote:
Hi PJ,
On Thu, Feb 26, 2009 at 17:28, PJ af.gour...@videotron.ca
mailto:af.gour...@videotron.ca wrote:
What is wrond with this file? same identical insert works
from console
but not from this file :-(
[snip]
?
//include (lib/db1.php);// Connect to database
mysql_connect('biggie', 'user', 'password', 'test') or die(Error
connecting DB.mysql_error());
$sql1 = INSERT INTO example (name, age) VALUES ('Joe Blow', '69');
$result1 = mysql_query($sql1,$db) or die(PError performing 1st
query: .mysql_error() . /P);
?
I haven't coded in PHP for a long time, but I think that your problem
is in this line:
$result1 = mysql_query($sql1,$db);
Up to that point, $db (that should point to a database link
identifier) is not defined. You probably want to assign the
mysql_connect result to that $db variable.
So, I think that you will solve your problem by changing your
mysql_connect line FROM the current form:
mysql_connect('biggie', 'user', 'password', 'test');
.. TO this one:
$db = mysql_connect('biggie', 'user', 'password', 'test');
Am I right?
Partly. I had an error in the location of the include. Ashley
corrected
the rest but it only works with the include. Not as whown below
?
//include (../lib/db1.php);// Connect to database
$db_host = 'biggie';
$db_user = 'root';
$db_pass = 'gu...@#$';
$db_name = 'biblane';
$db_connect = mysql_connect($db_host, $db_user, $db_pass);
$db_select = mysql_select_db($db_name, $db_connect);
$sql1 = INSERT INTO test (name, age) VALUES ('Arnie Shwartz', '75');
$result1 = mysql_query($sql1,$db);
if (!$result1) {
echo(PError performing 1st query: .
mysql_error() . /P);
exit();
}
?
--
Phil Jourdan --- p...@ptahhotep.com mailto:p...@ptahhotep.com
http://www.ptahhotep.com
http://www.chiccantine.com
--
MySQL General Mailing List
For list archives: http://lists.mysql.com/mysql
To unsubscribe:
http://lists.mysql.com/mysql?unsub=dstepli...@gmail.com
I agree. I wouldn't trust me at all! ;)
Ash
www.ashleysheridan.co.uk http://www.ashleysheridan.co.uk
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
--
Phil Jourdan --- p...@ptahhotep.com
http://www.ptahhotep.com
http://www.chiccantine.com
--
MySQL General Mailing List
For list archives: http://lists.mysql.com/mysql
To unsubscribe:http://lists.mysql.com/mysql?unsub=arch...@jab.org
Re: [PHP] Re: catch the error
Ashley Sheridan wrote:
On Thu, 2009-02-26 at 13:56 -0500, PJ wrote:
Ashley Sheridan wrote:
On Thu, 2009-02-26 at 13:34 -0500, Darryle Steplight wrote:
Hi PJ,
$db_host = 'biggie';
$db_user = 'root';
$db_pass = 'gu...@#$';
$db_name = 'biblane';
Everyone here is trying to help you and that's cool, but EVERYONE on
this list may not be so nice. The above credentials is definitely the
type of information you want to keep private, unless you don't mind
people potentially accessing your database tables and doing whatever
they like with them.
I suggest doing something like
$db_host = 'localhost;
$db_user = 'foo';
$db_pass= ''bar;
$db_name =''xx;
if you are going to post it on the list.
On Thu, Feb 26, 2009 at 1:22 PM, PJ af.gour...@videotron.ca wrote:
Ricardo Dias Marques wrote:
Hi PJ,
On Thu, Feb 26, 2009 at 17:28, PJ af.gour...@videotron.ca wrote:
What is wrond with this file? same identical insert works from console
but not from this file :-(
[snip]
?
//include (lib/db1.php);// Connect to database
mysql_connect('biggie', 'user', 'password', 'test');
$sql1 = INSERT INTO example (name, age) VALUES ('Joe Blow', '69');
$result1 = mysql_query($sql1,$db);
if (!$result1) {
echo(PError performing 1st query: .
mysql_error() . /P);
exit();
}
?
I haven't coded in PHP for a long time, but I think that your problem
is in this line:
$result1 = mysql_query($sql1,$db);
Up to that point, $db (that should point to a database link
identifier) is not defined. You probably want to assign the
mysql_connect result to that $db variable.
So, I think that you will solve your problem by changing your
mysql_connect line FROM the current form:
mysql_connect('biggie', 'user', 'password', 'test');
.. TO this one:
$db = mysql_connect('biggie', 'user', 'password', 'test');
Am I right?
Partly. I had an error in the location of the include. Ashley corrected
the rest but it only works with the include. Not as whown below
?
//include (../lib/db1.php);// Connect to database
$db_host = 'biggie';
$db_user = 'root';
$db_pass = 'gu...@#$';
$db_name = 'biblane';
$db_connect = mysql_connect($db_host, $db_user, $db_pass);
$db_select = mysql_select_db($db_name, $db_connect);
$sql1 = INSERT INTO test (name, age) VALUES ('Arnie Shwartz', '75');
$result1 = mysql_query($sql1,$db);
if (!$result1) {
echo(PError performing 1st query: .
mysql_error() . /P);
exit();
}
?
--
Phil Jourdan --- p...@ptahhotep.com
http://www.ptahhotep.com
http://www.chiccantine.com
--
MySQL General Mailing List
For list archives: http://lists.mysql.com/mysql
To unsubscribe:http://lists.mysql.com/mysql?unsub=dstepli...@gmail.com
I agree. I wouldn't trust me at all! ;)
Ash
www.ashleysheridan.co.uk
Yeah very stupid of me...but I found the error: see if you can catch it:
?
//include (../lib/db1.php);// Connect to database
$db_host = 'xxx';
$db_user = 'xxx;
$db_pass = 'xxx';
$db_name = 'xxx';
$db = mysql_connect($db_host, $db_user, $db_pass);
mysql_select_db($db_name,$db);
$sql1 = INSERT INTO test (name, age) VALUES ('Arnie Shwartz', '75');
$result1 = mysql_query($sql1,$db);
if (!$result1) {
echo(PError performing 3st query: .
mysql_error() . /P);
}
echo $sql1;
echo br /;
echo $db_select;
exit();
?
--
Phil Jourdan --- p...@ptahhotep.com
http://www.ptahhotep.com
http://www.chiccantine.com
$db_user has not had the string terminated. pray tell was that the
answer you were looking for?!
Ash
www.ashleysheridan.co.uk
No. Damn those typos!
What seems to have made it work is just
$db = mysql_connect($db_host, $db_user, $db_pass);
mysql_select_db($db_name,$db);
not using mysql_select in a string
but would you use it in a string? how why?
--
Phil Jourdan --- p...@ptahhotep.com
http://www.ptahhotep.com
http://www.chiccantine.com
--
MySQL General Mailing List
For list archives: http://lists.mysql.com/mysql
To unsubscribe:http://lists.mysql.com/mysql?unsub=arch...@jab.org
Re: [PHP] Re: catch the error
Ashley Sheridan wrote:
On Thu, 2009-02-26 at 14:15 -0500, PJ wrote:
Ashley Sheridan wrote:
On Thu, 2009-02-26 at 13:56 -0500, PJ wrote:
Ashley Sheridan wrote:
On Thu, 2009-02-26 at 13:34 -0500, Darryle Steplight wrote:
Hi PJ,
$db_host = 'biggie';
$db_user = 'root';
$db_pass = 'gu...@#$';
$db_name = 'biblane';
Everyone here is trying to help you and that's cool, but EVERYONE on
this list may not be so nice. The above credentials is definitely the
type of information you want to keep private, unless you don't mind
people potentially accessing your database tables and doing whatever
they like with them.
I suggest doing something like
$db_host = 'localhost;
$db_user = 'foo';
$db_pass= ''bar;
$db_name =''xx;
if you are going to post it on the list.
On Thu, Feb 26, 2009 at 1:22 PM, PJ af.gour...@videotron.ca wrote:
Ricardo Dias Marques wrote:
Hi PJ,
On Thu, Feb 26, 2009 at 17:28, PJ af.gour...@videotron.ca wrote:
What is wrond with this file? same identical insert works from console
but not from this file :-(
[snip]
?
//include (lib/db1.php);// Connect to database
mysql_connect('biggie', 'user', 'password', 'test');
$sql1 = INSERT INTO example (name, age) VALUES ('Joe Blow', '69');
$result1 = mysql_query($sql1,$db);
if (!$result1) {
echo(PError performing 1st query: .
mysql_error() . /P);
exit();
}
?
I haven't coded in PHP for a long time, but I think that your problem
is in this line:
$result1 = mysql_query($sql1,$db);
Up to that point, $db (that should point to a database link
identifier) is not defined. You probably want to assign the
mysql_connect result to that $db variable.
So, I think that you will solve your problem by changing your
mysql_connect line FROM the current form:
mysql_connect('biggie', 'user', 'password', 'test');
.. TO this one:
$db = mysql_connect('biggie', 'user', 'password', 'test');
Am I right?
Partly. I had an error in the location of the include. Ashley corrected
the rest but it only works with the include. Not as whown below
?
//include (../lib/db1.php);// Connect to database
$db_host = 'biggie';
$db_user = 'root';
$db_pass = 'gu...@#$';
$db_name = 'biblane';
$db_connect = mysql_connect($db_host, $db_user, $db_pass);
$db_select = mysql_select_db($db_name, $db_connect);
$sql1 = INSERT INTO test (name, age) VALUES ('Arnie Shwartz', '75');
$result1 = mysql_query($sql1,$db);
if (!$result1) {
echo(PError performing 1st query: .
mysql_error() . /P);
exit();
}
?
--
Phil Jourdan --- p...@ptahhotep.com
http://www.ptahhotep.com
http://www.chiccantine.com
--
MySQL General Mailing List
For list archives: http://lists.mysql.com/mysql
To unsubscribe:
http://lists.mysql.com/mysql?unsub=dstepli...@gmail.com
I agree. I wouldn't trust me at all! ;)
Ash
www.ashleysheridan.co.uk
Yeah very stupid of me...but I found the error: see if you can catch
it:
?
//include (../lib/db1.php);// Connect to database
$db_host = 'xxx';
$db_user = 'xxx;
$db_pass = 'xxx';
$db_name = 'xxx';
$db = mysql_connect($db_host, $db_user, $db_pass);
mysql_select_db($db_name,$db);
$sql1 = INSERT INTO test (name, age) VALUES ('Arnie Shwartz', '75');
$result1 = mysql_query($sql1,$db);
if (!$result1) {
echo(PError performing 3st query: .
mysql_error() . /P);
}
echo $sql1;
echo br /;
echo $db_select;
exit();
?
--
Phil Jourdan --- p...@ptahhotep.com
http://www.ptahhotep.com
http://www.chiccantine.com
$db_user has not had the string terminated. pray tell was that the
answer you were looking for?!
Ash
www.ashleysheridan.co.uk
No. Damn those typos!
What seems to have made it work is just
$db = mysql_connect($db_host, $db_user, $db_pass);
mysql_select_db($db_name,$db);
not using mysql_select in a string
but would you use it in a string? how why?
--
Phil Jourdan --- p...@ptahhotep.com
http://www.ptahhotep.com
http://www.chiccantine.com
Yeah, you'd typo'd on the variable name. Also, the $ sign doesn't
actually denote a string, but a scaler variable, which can be any type,
complex or simple.
I type too fast and am too speedy... :-)
I'll have to look up about the variables.
Thanks good night. 'Til the morrow.
--
Phil Jourdan --- p...@ptahhotep.com
http://www.ptahhotep.com
http://www.chiccantine.com
--
MySQL General Mailing List
For list archives: http://lists.mysql.com/mysql
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