RE: [PHP] Re: catch the error
-Original Message- From: Chris [mailto:dmag...@gmail.com] Sent: Thursday, February 26, 2009 6:38 PM To: Boyd, Todd M. Cc: PHP General list Subject: Re: [PHP] Re: catch the error Boyd, Todd M. wrote: -Original Message- From: Chris [mailto:dmag...@gmail.com] Sent: Thursday, February 26, 2009 4:16 PM To: Boyd, Todd M. Cc: PJ; PHP General list Subject: Re: [PHP] Re: catch the error In examples sent to you, people foolishly replaced your $db var with $db_connect ONLY FOR PART OF THE SCRIPT. You've defined your database connection as $db_connect in some versions of the source, but then you reference $db (without _connect) in your mysql_select call in that same source. $db = mysql_connect([option list here]); # -- this code instantiates a connection mysql_select_db([some name], $db); # notice how $db is here? $result = mysql_query([some query], $db); # it's here, too! $db becomes your resource link when you use mysql_connect. That resource link must then be passed to your mysql functions. Otherwise, they have no idea which database connection you are attempting to use. RTFM? If no connection is specified, the last one is used. It is an optional argument (only *really* needed when you have multiple connections in the same script). RTF E-mail I sent? He had used $db_connect instead of $db. $db_connect hadn't been set to anything. He was specifying a connection, but it was null. Unless it falls back to the last connection used in the case of an empty variable, then this was most likely (read: proven to be) the problem. The last two emails I saw (no I haven't read the whole thread) were: $db = mysql_connect($db_host, $db_user, $db_pass); mysql_select_db($db_name,$db); snip $result1 = mysql_query($sql1,$db); and $db = mysql_connect($db_host, $db_user, $db_pass); mysql_select_db($db_name,$db); which have the right variables. Plus I was picking on the you must do this - using the link identifier is an optional thing as I already said. Yes, it was solved before I replied. I just wanted to point out why it worked in the situations that it did and why it did not work in the situations that it did not. As the OP had been staring at it for too long, all of the code started to blur together. :) Anyway, it's a moot point now. I think the information you provided about the link identifier was solid advice... I was just trying to point out that using two different variables when specifying the link identifier in mysql functions was what gave him so much guff. // Todd -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Re: catch the error
Hi PJ,
Could it be that you have //include (lib/db1.php); commented
out? Try uncommenting that line and see what happens. The error
message will always print because the query is never executing
properly if you have the db connections file commented out.
On Thu, Feb 26, 2009 at 12:28 PM, PJ af.gour...@videotron.ca wrote:
What is wrond with this file? same identical insert works from console
but not from this file :-(
html
head
titleUntitled/title
/head
body
?
//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();
}
?
/body
/html
Seems to be good to print out the error message, but that's all. db not
written.
--
Phil Jourdan --- p...@ptahhotep.com
http://www.ptahhotep.com
http://www.chiccantine.com
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[PHP] Re: catch the error
It is commented out because I am using mysql_connect
I don't think it would be good to use both, since the db1 references
another db. But even when I use the db1.php and change the database and
table, I get the same error message.
But what I did miss is my typo in What is wrond with this file? :-)
Hi PJ,
Could it be that you have //include (lib/db1.php); commented
out? Try uncommenting that line and see what happens. The error
message will always print because the query is never executing
properly if you have the db connections file commented out.
On Thu, Feb 26, 2009 at 12:28 PM, PJ af.gour...@videotron.ca wrote:
What is wrond with this file? same identical insert works from console
but not from this file :-(
html
head
titleUntitled/title
/head
body
?
//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();
}
?
/body
/html
Seems to be good to print out the error message, but that's all. db not
written.
--
Phil Jourdan --- p...@ptahhotep.com
http://www.ptahhotep.com
http://www.chiccantine.com
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[PHP] Re: catch the error
what's the error message? On Thu, Feb 26, 2009 at 11:46 AM, PJ af.gour...@videotron.ca wrote: It is commented out because I am using mysql_connect I don't think it would be good to use both, since the db1 references another db. But even when I use the db1.php and change the database and table, I get the same error message. -- Jim Lyons Web developer / Database administrator http://www.weblyons.com
[PHP] Re: catch the error
ok, well if that's the case then do this
$db = mysql_connect('biggie', 'user', 'password', 'test');
That should fix the problem.
On Thu, Feb 26, 2009 at 12:46 PM, PJ af.gour...@videotron.ca wrote:
It is commented out because I am using mysql_connect
I don't think it would be good to use both, since the db1 references
another db. But even when I use the db1.php and change the database and
table, I get the same error message.
But what I did miss is my typo in What is wrond with this file? :-)
Hi PJ,
Could it be that you have //include (lib/db1.php); commented
out? Try uncommenting that line and see what happens. The error
message will always print because the query is never executing
properly if you have the db connections file commented out.
On Thu, Feb 26, 2009 at 12:28 PM, PJ af.gour...@videotron.ca wrote:
What is wrond with this file? same identical insert works from console
but not from this file :-(
html
head
titleUntitled/title
/head
body
?
//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();
}
?
/body
/html
Seems to be good to print out the error message, but that's all. db not
written.
--
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
--
Phil Jourdan --- p...@ptahhotep.com
http://www.ptahhotep.com
http://www.chiccantine.com
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[PHP] Re: catch the error
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?
Cheers,
Ricardo Dias Marques
lists AT ricmarques DOT net
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[PHP] Re: catch the error
Jim Lyons wrote:
what's the error message?
it's in the script/// Error performing query: of Error performing 1st
query: - whatever I input.
But I had an error in the include location... that's fixed and it works,
but not the rest as corrected:
?
//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();
}
?
On Thu, Feb 26, 2009 at 11:46 AM, PJ af.gour...@videotron.ca
mailto:af.gour...@videotron.ca wrote:
It is commented out because I am using mysql_connect
I don't think it would be good to use both, since the db1 references
another db. But even when I use the db1.php and change the
database and
table, I get the same error message.
--
Jim Lyons
Web developer / Database administrator
http://www.weblyons.com
--
Phil Jourdan --- p...@ptahhotep.com
http://www.ptahhotep.com
http://www.chiccantine.com
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[PHP] Re: catch the error
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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[PHP] Re: catch the error
Darryle Steplight wrote:
ok, well if that's the case then do this
$db = mysql_connect('biggie', 'user', 'password', 'test');
Ashley pointed out that the 4th parameter is not right - belongs in
mysql_select_db. Here it is corrected: (but it still does not work)
?
//include (../lib/db1.php);// Connect to database
$db_host = 'biggie';
$db_user = 'my_user';
$db_pass = 'my_pass';
$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();
}
?
That should fix the problem.
On Thu, Feb 26, 2009 at 12:46 PM, PJ af.gour...@videotron.ca wrote:
It is commented out because I am using mysql_connect
I don't think it would be good to use both, since the db1 references
another db. But even when I use the db1.php and change the database and
table, I get the same error message.
But what I did miss is my typo in What is wrond with this file? :-)
Hi PJ,
Could it be that you have //include (lib/db1.php); commented
out? Try uncommenting that line and see what happens. The error
message will always print because the query is never executing
properly if you have the db connections file commented out.
On Thu, Feb 26, 2009 at 12:28 PM, PJ af.gour...@videotron.ca wrote:
What is wrond with this file? same identical insert works from console
but not from this file :-(
html
head
titleUntitled/title
/head
body
?
//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();
}
?
/body
/html
Seems to be good to print out the error message, but that's all. db not
written.
--
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
--
Phil Jourdan --- p...@ptahhotep.com
http://www.ptahhotep.com
http://www.chiccantine.com
--
Phil Jourdan --- p...@ptahhotep.com
http://www.ptahhotep.com
http://www.chiccantine.com
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Re: catch the error
What does this output?
?
//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, $db_pass);
if (!is_resource($db_connect)) {
echo 'p$db_connect is not a valid resource: ' . mysql_error() . '/p';
exit();
}
if (!mysql_select_db($db_name, $db_connect)) {
echo pUnable to select $db_name: . mysql_error() . '/p';
exit();
}
$sql1 = INSERT INTO test (name, age) VALUES ('Arnie Shwartz', '75');
$result1 = mysql_query($sql1,$db);
if (!is_resource($result1)) {
echo pError performing 1st query: . mysql_error() . /p;
exit();
} else {
echo 'p' . mysql_affected_rows() . 'row(s) affected by last
statement/p';
exit();
}
?
Andrew
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[PHP] Re: catch the error
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
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Re: [PHP] Re: catch the error
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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[PHP] RE: catch the error
-Original Message-
From: PJ [mailto:af.gour...@videotron.ca]
Sent: Thursday, February 26, 2009 12:28 PM
To: php-general@lists.php.net; MySql
Subject: catch the error
What is wrond with this file? same identical insert works from console
but not from this file :-(
html
head
titleUntitled/title
/head
body
?
//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();
}
?
/body
/html
Seems to be good to print out the error message, but that's all. db not
written.
[JS] You need
$db = mysql_connect('biggie', 'user', 'password', 'test');
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infoshop.com
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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
--
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
--
PHP General Mailing List (http://www.php.net/)
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Re: [PHP] Re: catch the error
But the question is PJ, have you got it out of errors yet? :)
www.twitter.com/nine_L
www.lenin9l.wordpress.com
---
Use FreeOpenSourceSoftwares, Stop piracy, Let the developers live. Get
a Free CD of Ubuntu mailed to your door without any cost. Visit :
www.ubuntu.com
--
2009/2/27 Ashley Sheridan 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 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') 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
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
--
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[PHP] Re: catch the error
Jerry Schwartz wrote:
-Original Message-
From: PJ [mailto:af.gour...@videotron.ca]
Sent: Thursday, February 26, 2009 12:28 PM
To: php-general@lists.php.net; MySql
Subject: catch the error
What is wrond with this file? same identical insert works from console
but not from this file :-(
html
head
titleUntitled/title
/head
body
?
//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();
}
?
/body
/html
Seems to be good to print out the error message, but that's all. db not
written.
[JS] You need
$db = mysql_connect('biggie', 'user', 'password', 'test');
--
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http://www.ptahhotep.com
http://www.chiccantine.com
--
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infoshop.com
I think the problem here has been that this is such a basic operation
and most of us just are too busy with more complicated stuff...that we
didn't catch it...
?
//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();
?
anyway, I am learning a lot...
thanks, guys... you're all great...
I have lots more coming... :-D
--
Phil Jourdan --- p...@ptahhotep.com
http://www.ptahhotep.com
http://www.chiccantine.com
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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
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Re: catch the error
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
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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
a Free CD of Ubuntu mailed to your door without any cost. Visit :
www.ubuntu.com http://www.ubuntu.com
--
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
--
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To unsubscribe, visit: http://www.php.net/unsub.php
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
--
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RE: [PHP] Re: catch the error
-Original Message-
From: PJ [mailto:af.gour...@videotron.ca]
Sent: Thursday, February 26, 2009 1:16 PM
To: a...@ashleysheridan.co.uk
Cc: Darryle Steplight; Ricardo Dias Marques;
php-general@lists.php.net;
MySql
Subject: 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?
Jesus Christ... everyone on this list must've had a long week, because
you guys are going blind. :D
In examples sent to you, people foolishly replaced your $db var with
$db_connect ONLY FOR PART OF THE SCRIPT. You've defined your database
connection as $db_connect in some versions of the source, but then you
reference $db (without _connect) in your mysql_select call in that same
source.
$db = mysql_connect([option list here]); # -- this code instantiates a
connection
mysql_select_db([some name], $db); # notice how $db is here?
$result = mysql_query([some query], $db); # it's here, too!
$db becomes your resource link when you use mysql_connect. That resource
link must
Re: [PHP] Re: catch the error
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.
Ash
www.ashleysheridan.co.uk
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Re: catch the error
Boyd, Todd M. wrote: Jesus Christ... everyone on this list must've had a long week, because you guys are going blind. :D In examples sent to you, people foolishly replaced your $db var with $db_connect ONLY FOR PART OF THE SCRIPT. You've defined your database connection as $db_connect in some versions of the source, but then you reference $db (without _connect) in your mysql_select call in that same source. $db = mysql_connect([option list here]); # -- this code instantiates a connection mysql_select_db([some name], $db); # notice how $db is here? $result = mysql_query([some query], $db); # it's here, too! $db becomes your resource link when you use mysql_connect. That resource link must then be passed to your mysql functions. Otherwise, they have no idea which database connection you are attempting to use. # BAD: $db = mysql_connect([options]); $result = mysql_query([some query], $db_connect); # what's $db_connect? who knows! # GOOD: $db = mysql_connect([options]); $result = mysql_query([some query], $db); # same resource link from mysql_connect See? HTH, // Todd Yes, and I would also strongly urge (scream at) PJ, to turn on E_ALL error reporting in any script he has problems with. #1 you'll figure out the problem because the errors will tell you exactly what the problem is undefined variable db_connect, etc... #2 if you can't figure it out, someone here can easily if you give the errors. Virtually every problem you've posted throws an error(s) that will tell you what the F*** is the problem! -- Thanks! -Shawn http://www.spidean.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Re: catch the error
In examples sent to you, people foolishly replaced your $db var with $db_connect ONLY FOR PART OF THE SCRIPT. You've defined your database connection as $db_connect in some versions of the source, but then you reference $db (without _connect) in your mysql_select call in that same source. $db = mysql_connect([option list here]); # -- this code instantiates a connection mysql_select_db([some name], $db); # notice how $db is here? $result = mysql_query([some query], $db); # it's here, too! $db becomes your resource link when you use mysql_connect. That resource link must then be passed to your mysql functions. Otherwise, they have no idea which database connection you are attempting to use. RTFM? If no connection is specified, the last one is used. It is an optional argument (only *really* needed when you have multiple connections in the same script). -- Postgresql php tutorials http://www.designmagick.com/ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] Re: catch the error
-Original Message- From: Chris [mailto:dmag...@gmail.com] Sent: Thursday, February 26, 2009 4:16 PM To: Boyd, Todd M. Cc: PJ; PHP General list Subject: Re: [PHP] Re: catch the error In examples sent to you, people foolishly replaced your $db var with $db_connect ONLY FOR PART OF THE SCRIPT. You've defined your database connection as $db_connect in some versions of the source, but then you reference $db (without _connect) in your mysql_select call in that same source. $db = mysql_connect([option list here]); # -- this code instantiates a connection mysql_select_db([some name], $db); # notice how $db is here? $result = mysql_query([some query], $db); # it's here, too! $db becomes your resource link when you use mysql_connect. That resource link must then be passed to your mysql functions. Otherwise, they have no idea which database connection you are attempting to use. RTFM? If no connection is specified, the last one is used. It is an optional argument (only *really* needed when you have multiple connections in the same script). RTF E-mail I sent? He had used $db_connect instead of $db. $db_connect hadn't been set to anything. He was specifying a connection, but it was null. Unless it falls back to the last connection used in the case of an empty variable, then this was most likely (read: proven to be) the problem. :p -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Re: catch the error
Shawn McKenzie wrote: Boyd, Todd M. wrote: Jesus Christ... everyone on this list must've had a long week, because you guys are going blind. :D In examples sent to you, people foolishly replaced your $db var with $db_connect ONLY FOR PART OF THE SCRIPT. You've defined your database connection as $db_connect in some versions of the source, but then you reference $db (without _connect) in your mysql_select call in that same source. $db = mysql_connect([option list here]); # -- this code instantiates a connection mysql_select_db([some name], $db); # notice how $db is here? $result = mysql_query([some query], $db); # it's here, too! $db becomes your resource link when you use mysql_connect. That resource link must then be passed to your mysql functions. Otherwise, they have no idea which database connection you are attempting to use. # BAD: $db = mysql_connect([options]); $result = mysql_query([some query], $db_connect); # what's $db_connect? who knows! # GOOD: $db = mysql_connect([options]); $result = mysql_query([some query], $db); # same resource link from mysql_connect See? HTH, // Todd Yes, and I would also strongly urge (scream at) PJ, to turn on E_ALL error reporting in any script he has problems with. #1 you'll figure out the problem because the errors will tell you exactly what the problem is undefined variable db_connect, etc... #2 if you can't figure it out, someone here can easily if you give the errors. Virtually every problem you've posted throws an error(s) that will tell you what the F*** is the problem! Ok, ok, I understant the frustrations But, yes, I think that some of the suggestions were rather hastily made as most of the guys must be very busy and as we are all somewhat normal, we do make mistakes. I mixed up some suggestions with other errors of mine and so we had spaghetti. So, now let's kiss and make up... or as the Italians would say, Amici, come prima. :-) I'm glad you mention the error reporting and I admit i haven't been doing much of that. I do not understand how it should be set up or where I should look for the errors. I did try ini_set(display_errors,true); error_reporting(E_STRICT|E_ALL); but it meant absolutely nothing. I could not find any logs for mysql or php errors and none appeared magically in the page. Perhaps you could steer me to an explanation somewhere. I was able to save some grief for all as I dif manage to find some guidelines on the mysql manual site that were pretty helpful. Thanks for your tolerance of my ignorance - love you all! PJ -- Phil Jourdan --- p...@ptahhotep.com http://www.ptahhotep.com http://www.chiccantine.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
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
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Re: catch the error
Boyd, Todd M. wrote: -Original Message- From: Chris [mailto:dmag...@gmail.com] Sent: Thursday, February 26, 2009 4:16 PM To: Boyd, Todd M. Cc: PJ; PHP General list Subject: Re: [PHP] Re: catch the error In examples sent to you, people foolishly replaced your $db var with $db_connect ONLY FOR PART OF THE SCRIPT. You've defined your database connection as $db_connect in some versions of the source, but then you reference $db (without _connect) in your mysql_select call in that same source. $db = mysql_connect([option list here]); # -- this code instantiates a connection mysql_select_db([some name], $db); # notice how $db is here? $result = mysql_query([some query], $db); # it's here, too! $db becomes your resource link when you use mysql_connect. That resource link must then be passed to your mysql functions. Otherwise, they have no idea which database connection you are attempting to use. RTFM? If no connection is specified, the last one is used. It is an optional argument (only *really* needed when you have multiple connections in the same script). RTF E-mail I sent? He had used $db_connect instead of $db. $db_connect hadn't been set to anything. He was specifying a connection, but it was null. Unless it falls back to the last connection used in the case of an empty variable, then this was most likely (read: proven to be) the problem. The last two emails I saw (no I haven't read the whole thread) were: $db = mysql_connect($db_host, $db_user, $db_pass); mysql_select_db($db_name,$db); snip $result1 = mysql_query($sql1,$db); and $db = mysql_connect($db_host, $db_user, $db_pass); mysql_select_db($db_name,$db); which have the right variables. Plus I was picking on the you must do this - using the link identifier is an optional thing as I already said. -- Postgresql php tutorials http://www.designmagick.com/ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Re: catch the error
PJ you should be getting Warning errors the way you had the code. You definitely is a fun lover and enjoy moments. But while coding dont only think of fastness of coding but also code with logic. I was going to answer you about the $db but later I found lots response already came so didn't really dig to get the errors. Look back, see, I asked if you got rid of the errors :) Now, if you dont use the variable in the parameter it will look for the immediate opened database resource but if you use a variable. THEN it must contain that required resource thats just OBVIOUS reason. And, your error reporting should be reporting that. I haven't checked the code. A funny thing is you are making others think for the things you should be thinking. I dont say its bad. Its actually good for all of us having some drill on the basics :D Regards Lenin www.twitter.com/nine_L www.lenin9l.wordpress.com On Fri, Feb 27, 2009 at 6:37 AM, Chris dmag...@gmail.com wrote: Boyd, Todd M. wrote: -Original Message- From: Chris [mailto:dmag...@gmail.com] Sent: Thursday, February 26, 2009 4:16 PM To: Boyd, Todd M. Cc: PJ; PHP General list Subject: Re: [PHP] Re: catch the error In examples sent to you, people foolishly replaced your $db var with $db_connect ONLY FOR PART OF THE SCRIPT. You've defined your database connection as $db_connect in some versions of the source, but then you reference $db (without _connect) in your mysql_select call in that same source. $db = mysql_connect([option list here]); # -- this code instantiates a connection mysql_select_db([some name], $db); # notice how $db is here? $result = mysql_query([some query], $db); # it's here, too! $db becomes your resource link when you use mysql_connect. That resource link must then be passed to your mysql functions. Otherwise, they have no idea which database connection you are attempting to use. RTFM? If no connection is specified, the last one is used. It is an optional argument (only *really* needed when you have multiple connections in the same script). RTF E-mail I sent? He had used $db_connect instead of $db. $db_connect hadn't been set to anything. He was specifying a connection, but it was null. Unless it falls back to the last connection used in the case of an empty variable, then this was most likely (read: proven to be) the problem. The last two emails I saw (no I haven't read the whole thread) were: $db = mysql_connect($db_host, $db_user, $db_pass); mysql_select_db($db_name,$db); snip $result1 = mysql_query($sql1,$db); and $db = mysql_connect($db_host, $db_user, $db_pass); mysql_select_db($db_name,$db); which have the right variables. Plus I was picking on the you must do this - using the link identifier is an optional thing as I already said. -- Postgresql php tutorials http://www.designmagick.com/ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
