Re: [PHP] Help with msql_fetch_array() FIXED ! Now cookie problems :(

2002-07-25 Thread Matthew Bielecki

Hello again,

I got the fetch_array problems fixed.  I was using the actual server name, 
when I switched back to localhost everything worked!!

Now I have a question about how to make cookies work on a Windows machine. 
 This is what I have these parameters set to but it's not working.  Do I 
have slashes/backslashes wrong, or is there something else I have to do in 
Apache??:

session.save_path = C:/Program Files/Apache 
Group/Apache/web/php/dir/files/temp

; Whether to use cookies.
session.use_cookies = 1


; Name of the session (used as cookie name).
session.name = PHPSESSID

; Initialize session on request startup.
session.auto_start = 0

; Lifetime in seconds of cookie or, if 0, until browser is restarted.
session.cookie_lifetime = 0

; The path for which the cookie is valid.
session.cookie_path = c:/Program Files/Apache 
Group/Apache/web/php/dir/files/temp

; The domain for which the cookie is valid.
session.cookie_domain = www.mydomain.com



As always, thanks for your help.








PHPCoder [EMAIL PROTECTED]
07/24/02 01:50 PM

 
To: Matthew Bielecki [EMAIL PROTECTED]
cc: php-general [EMAIL PROTECTED]
Subject:Re: [PHP] Help with msql_fetch_array()


I can almost guarantee that it's not the second line that is failing, 
the problem here is that $result is not containing naything, and that is 
normally due to the fact that you are not connecting to the db, or the 
table tablename is not there.

I use the following format as my standard MySQL connect and query 
snippet:

$link = @mysql_connect(localhost,$username,$password) or die ('Could 
not connect!'); //@ suppresses the default error message generated by 
this function and the or die() bit kills the script right then and 
there should it not be able to connect.
mysql_select_db(YOUR_DB_NAME,$link);
$sql = select * from your_table_name;
if ( $result = mysql_query($sql)) {  // checks to see if $result 
contains anything before it even tries to fetch an associative array 
from it.
 $row = mysql_fetch_assoc($result);
} else {
echo Empty result set!;

Note also that I use mysql_fetch_assoc and NOT mysql_fecth_array, as 9 
out of 10 times, you don't need the array element id's that is returned 
by mysql_fetch_array.

Matthew Bielecki wrote:

I have a couple of scripts that fail with the error of:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL 
result 
resource in...

I'm new to both SQL and PHP and I'm wondering if I have some setting 
turned off or what.

Here's the piece of code that is failing (the second line fails):

$result = mysql_db_query($dbname, SELECT * FROM tablename ORDER BY id);
$row = mysql_fetch_array($result);


Thanks for your help in advance!!








Re: [PHP] Help with msql_fetch_array() FIXED ! Now cookie problems:(

2002-07-25 Thread Scott

On Thu, 25 Jul 2002, Matthew Bielecki wrote:
 session.save_path = C:/Program Files/Apache 
 Group/Apache/web/php/dir/files/temp

Considering using something like:  c:/temp, sure beats typing and 
remembering that path :)

How are you setting the cookie in the code?




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Re: [PHP] Help with msql_fetch_array()

2002-07-24 Thread PHPCoder

I can almost guarantee that it's not the second line that is failing, 
the problem here is that $result is not containing naything, and that is 
normally due to the fact that you are not connecting to the db, or the 
table tablename is not there.

I use the following format as my standard MySQL connect and query snippet:

$link = mysql_connect(localhost,$username,$password) or die ('Could 
not connect!'); // suppresses the default error message generated by 
this function and the or die() bit kills the script right then and 
there should it not be able to connect.
mysql_select_db(YOUR_DB_NAME,$link);
$sql = select * from your_table_name;
if ( $result = mysql_query($sql)) {  // checks to see if $result 
contains anything before it even tries to fetch an associative array 
from it.
 $row = mysql_fetch_assoc($result);
} else {
echo Empty result set!;

Note also that I use mysql_fetch_assoc and NOT mysql_fecth_array, as 9 
out of 10 times, you don't need the array element id's that is returned 
by mysql_fetch_array.

Matthew Bielecki wrote:

I have a couple of scripts that fail with the error of:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result 
resource in...

I'm new to both SQL and PHP and I'm wondering if I have some setting 
turned off or what.

Here's the piece of code that is failing (the second line fails):

$result = mysql_db_query($dbname, SELECT * FROM tablename ORDER BY id);
$row = mysql_fetch_array($result);


Thanks for your help in advance!!




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Re: [PHP] Help with msql_fetch_array()

2002-07-24 Thread Kevin Stone

Try using mysql_query(); instead of mysql_db_query();  The SQL query is the
first parameter in this function.  The second parameter is a pointer
connecting to the mysql server.  The pointer is generated by mysql_connect()
and you'll also need to select the database with mysql_select_db().  But
it's the standard way..

$db = mysql_connect(localhost, username, password);// connect to
mysql server
mysql_select_db(mytable, $db);
// select database by name
$query = SELECT * FROM mytable ORDER BY id; // define query to
submit
$result = mysql_query($query, $db);  //
submit query
if (mysql_num_rows($result)  0)
// skip if no rows were found
{
while ($row = mysql_fetch_array($result))
{
// .. do whtever..
}
}
mysql_close($db);

It's also a common practice to put the first couple of lines in a file to
include() back into your main script so that you can protect your useranme
and password.

Good luck.
-Kevin

- Original Message -
From: Matthew Bielecki [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Wednesday, July 24, 2002 10:34 AM
Subject: [PHP] Help with msql_fetch_array()


 I have a couple of scripts that fail with the error of:
 Warning: mysql_fetch_array(): supplied argument is not a valid MySQL
result
 resource in...

 I'm new to both SQL and PHP and I'm wondering if I have some setting
 turned off or what.

 Here's the piece of code that is failing (the second line fails):

 $result = mysql_db_query($dbname, SELECT * FROM tablename ORDER BY id);
 $row = mysql_fetch_array($result);


 Thanks for your help in advance!!


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Re: [PHP] Help with msql_fetch_array()

2002-07-24 Thread Kevin Stone

You beat me too the punch and I think you explained it better than me, but
just one minor little thing to note.  Where you said..

if ( $result = mysql_query($sql)) 

This is not a valid way to check if the query has returned anything.
mysql_query() returns FALSE on error.  So if there was no error but there
also wasn't anything returned then the object stored in $result wiill more
than likely evaluate to TRUE.  For the determining factor you should count
the number of rows with mysql_num_rows($result).  If the returned value is
zero then you know it hasn't returned anything.

-Kevin

- Original Message -
From: PHPCoder [EMAIL PROTECTED]
To: Matthew Bielecki [EMAIL PROTECTED]
Cc: php-general [EMAIL PROTECTED]
Sent: Wednesday, July 24, 2002 11:50 AM
Subject: Re: [PHP] Help with msql_fetch_array()


 I can almost guarantee that it's not the second line that is failing,
 the problem here is that $result is not containing naything, and that is
 normally due to the fact that you are not connecting to the db, or the
 table tablename is not there.

 I use the following format as my standard MySQL connect and query
snippet:

 $link = @mysql_connect(localhost,$username,$password) or die ('Could
 not connect!'); //@ suppresses the default error message generated by
 this function and the or die() bit kills the script right then and
 there should it not be able to connect.
 mysql_select_db(YOUR_DB_NAME,$link);
 $sql = select * from your_table_name;
 if ( $result = mysql_query($sql)) {  // checks to see if $result
 contains anything before it even tries to fetch an associative array
 from it.
  $row = mysql_fetch_assoc($result);
 } else {
 echo Empty result set!;

 Note also that I use mysql_fetch_assoc and NOT mysql_fecth_array, as 9
 out of 10 times, you don't need the array element id's that is returned
 by mysql_fetch_array.

 Matthew Bielecki wrote:

 I have a couple of scripts that fail with the error of:
 Warning: mysql_fetch_array(): supplied argument is not a valid MySQL
result
 resource in...
 
 I'm new to both SQL and PHP and I'm wondering if I have some setting
 turned off or what.
 
 Here's the piece of code that is failing (the second line fails):
 
 $result = mysql_db_query($dbname, SELECT * FROM tablename ORDER BY id);
 $row = mysql_fetch_array($result);
 
 
 Thanks for your help in advance!!
 



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 PHP General Mailing List (http://www.php.net/)
 To unsubscribe, visit: http://www.php.net/unsub.php



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Re: [PHP] Help with msql_fetch_array()

2002-07-24 Thread PHPCoder

Yes, what on earth was I thinking!
should be:
...
$result = mysql_query($sql);
if (mysql_num_rows($result)) {
 $row = mysql_fetch_assoc($result);
 } else {
 echo whatever;
 }
...


Kevin Stone wrote:

You beat me too the punch and I think you explained it better than me, but
just one minor little thing to note.  Where you said..

if ( $result = mysql_query($sql)) 

This is not a valid way to check if the query has returned anything.
mysql_query() returns FALSE on error.  So if there was no error but there
also wasn't anything returned then the object stored in $result wiill more
than likely evaluate to TRUE.  For the determining factor you should count
the number of rows with mysql_num_rows($result).  If the returned value is
zero then you know it hasn't returned anything.

-Kevin

- Original Message -
From: PHPCoder [EMAIL PROTECTED]
To: Matthew Bielecki [EMAIL PROTECTED]
Cc: php-general [EMAIL PROTECTED]
Sent: Wednesday, July 24, 2002 11:50 AM
Subject: Re: [PHP] Help with msql_fetch_array()


I can almost guarantee that it's not the second line that is failing,
the problem here is that $result is not containing naything, and that is
normally due to the fact that you are not connecting to the db, or the
table tablename is not there.

I use the following format as my standard MySQL connect and query

snippet:

$link = @mysql_connect(localhost,$username,$password) or die ('Could
not connect!'); //@ suppresses the default error message generated by
this function and the or die() bit kills the script right then and
there should it not be able to connect.
mysql_select_db(YOUR_DB_NAME,$link);
$sql = select * from your_table_name;
if ( $result = mysql_query($sql)) {  // checks to see if $result
contains anything before it even tries to fetch an associative array
from it.
 $row = mysql_fetch_assoc($result);
} else {
echo Empty result set!;

Note also that I use mysql_fetch_assoc and NOT mysql_fecth_array, as 9
out of 10 times, you don't need the array element id's that is returned
by mysql_fetch_array.

Matthew Bielecki wrote:

I have a couple of scripts that fail with the error of:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL

result

resource in...

I'm new to both SQL and PHP and I'm wondering if I have some setting
turned off or what.

Here's the piece of code that is failing (the second line fails):

$result = mysql_db_query($dbname, SELECT * FROM tablename ORDER BY id);
   $row = mysql_fetch_array($result);


Thanks for your help in advance!!



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Re: [PHP] Help with msql_fetch_array()

2002-07-24 Thread Martin Clifford

Shouldn't it be:

$result = mysql_query($sql, $link_id);

PHP will be quick to tell you that there is a missing link identifier in the 
mysql_query() call.  It's happened to me plenty of times.  HTH!

Martin Clifford
Homepage: http://www.completesource.net
Developer's Forums: http://www.completesource.net/forums/


 PHPCoder [EMAIL PROTECTED] 07/24/02 02:16PM 
Yes, what on earth was I thinking!
should be:
...
$result = mysql_query($sql);
if (mysql_num_rows($result)) {
 $row = mysql_fetch_assoc($result);
 } else {
 echo whatever;
 }
...


Kevin Stone wrote:

You beat me too the punch and I think you explained it better than me, but
just one minor little thing to note.  Where you said..

if ( $result = mysql_query($sql)) 

This is not a valid way to check if the query has returned anything.
mysql_query() returns FALSE on error.  So if there was no error but there
also wasn't anything returned then the object stored in $result wiill more
than likely evaluate to TRUE.  For the determining factor you should count
the number of rows with mysql_num_rows($result).  If the returned value is
zero then you know it hasn't returned anything.

-Kevin

- Original Message -
From: PHPCoder [EMAIL PROTECTED]
To: Matthew Bielecki [EMAIL PROTECTED]
Cc: php-general [EMAIL PROTECTED]
Sent: Wednesday, July 24, 2002 11:50 AM
Subject: Re: [PHP] Help with msql_fetch_array()


I can almost guarantee that it's not the second line that is failing,
the problem here is that $result is not containing naything, and that is
normally due to the fact that you are not connecting to the db, or the
table tablename is not there.

I use the following format as my standard MySQL connect and query

snippet:

$link = @mysql_connect(localhost,$username,$password) or die ('Could
not connect!'); //@ suppresses the default error message generated by
this function and the or die() bit kills the script right then and
there should it not be able to connect.
mysql_select_db(YOUR_DB_NAME,$link);
$sql = select * from your_table_name;
if ( $result = mysql_query($sql)) {  // checks to see if $result
contains anything before it even tries to fetch an associative array
from it.
 $row = mysql_fetch_assoc($result);
} else {
echo Empty result set!;

Note also that I use mysql_fetch_assoc and NOT mysql_fecth_array, as 9
out of 10 times, you don't need the array element id's that is returned
by mysql_fetch_array.

Matthew Bielecki wrote:

I have a couple of scripts that fail with the error of:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL

result

resource in...

I'm new to both SQL and PHP and I'm wondering if I have some setting
turned off or what.

Here's the piece of code that is failing (the second line fails):

$result = mysql_db_query($dbname, SELECT * FROM tablename ORDER BY id);
   $row = mysql_fetch_array($result);


Thanks for your help in advance!!



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To unsubscribe, visit: http://www.php.net/unsub.php 






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Re: [PHP] Help with msql_fetch_array()

2002-07-24 Thread Phillip S. Baker

At 11:26 AM 7/24/2002 Wednesday, Martin Clifford wrote:
Shouldn't it be:

$result = mysql_query($sql, $link_id);

Actually mysql_query should default to the last database connection opened 
if no link identifier has been specified.
So the link identifier to not absolutely required.

Phillip


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Re: [PHP] Help with msql_fetch_array()

2002-07-24 Thread Martin Clifford

It is on whatever installation my host is running :o)  I know it's 4.x, though I 
should really make them upgrade.

 Phillip S. Baker [EMAIL PROTECTED] 07/24/02 02:34PM 
At 11:26 AM 7/24/2002 Wednesday, Martin Clifford wrote:
Shouldn't it be:

$result = mysql_query($sql, $link_id);

Actually mysql_query should default to the last database connection opened 
if no link identifier has been specified.
So the link identifier to not absolutely required.

Phillip


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Re: [PHP] Help with msql_fetch_array()

2002-07-24 Thread Jason Wong

On Thursday 25 July 2002 00:34, Matthew Bielecki wrote:
 I have a couple of scripts that fail with the error of:
 Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result
 resource in...

 I'm new to both SQL and PHP and I'm wondering if I have some setting
 turned off or what.

 Here's the piece of code that is failing (the second line fails):

 $result = mysql_db_query($dbname, SELECT * FROM tablename ORDER BY id);
 $row = mysql_fetch_array($result);

It's a very good idea to add some error checking code it'll save you a lot of 
grief. See examples in manual for details.

-- 
Jason Wong - Gremlins Associates - www.gremlins.com.hk
Open Source Software Systems Integrators
* Web Design  Hosting * Internet  Intranet Applications Development *

/*
Nine years of ballet, asshole.
-- Shelly Long, to the bad guy after making a jump over a gorge that he
   couldn't quite, in Outrageous Fortune
*/


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Re: [PHP] Help with msql_fetch_array()

2002-07-24 Thread Matthew Bielecki

Well I think you were correct about not connecting to the db, but I don't 
understand why.  I wrote another little script just to test the 
connection.

?php

$link = mysql_connect(servername,username,password)
  or die(Couldn't make connection.);
$diditwork = mysql_select_db(news, $link);

if ($diditwork  FALSE)
   {
  echo got the db..yea!;
  echo $link;
   }
   else
   {
  echo didnt get db...bo;
  echo $link;
   }
?

This returns didnt get db...booResource id #1

I don't understand how I can get a resource ID but then not be able to use 
the news database.  Like I described earlier, this happens with my other 
db as well.  I can connect to the db through the console or any other 
client I have on the physical server and do whatever I want with the db's, 
I'm just having problems with php.

Thanks again for your help!!









PHPCoder [EMAIL PROTECTED]
07/24/02 01:50 PM

 
To: Matthew Bielecki [EMAIL PROTECTED]
cc: php-general [EMAIL PROTECTED]
Subject:Re: [PHP] Help with msql_fetch_array()


I can almost guarantee that it's not the second line that is failing, 
the problem here is that $result is not containing naything, and that is 
normally due to the fact that you are not connecting to the db, or the 
table tablename is not there.

I use the following format as my standard MySQL connect and query 
snippet:

$link = @mysql_connect(localhost,$username,$password) or die ('Could 
not connect!'); //@ suppresses the default error message generated by 
this function and the or die() bit kills the script right then and 
there should it not be able to connect.
mysql_select_db(YOUR_DB_NAME,$link);
$sql = select * from your_table_name;
if ( $result = mysql_query($sql)) {  // checks to see if $result 
contains anything before it even tries to fetch an associative array 
from it.
 $row = mysql_fetch_assoc($result);
} else {
echo Empty result set!;

Note also that I use mysql_fetch_assoc and NOT mysql_fecth_array, as 9 
out of 10 times, you don't need the array element id's that is returned 
by mysql_fetch_array.

Matthew Bielecki wrote:

I have a couple of scripts that fail with the error of:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL 
result 
resource in...

I'm new to both SQL and PHP and I'm wondering if I have some setting 
turned off or what.

Here's the piece of code that is failing (the second line fails):

$result = mysql_db_query($dbname, SELECT * FROM tablename ORDER BY id);
$row = mysql_fetch_array($result);


Thanks for your help in advance!!








Re: [PHP] Help with msql_fetch_array()

2002-07-24 Thread Kevin Stone

That should definitely be working.  Only thing I can think of is that you're
database name isn't correct.  You're certain news is the full name of the
database.. I mean it's not news_com or something like that?  The $link
must be valid becuase the script isn't ending with Couldn't make
connection.  So that just leaves the mysql_select_db() function.  And the
only thing that will make that function return FALSE is if it can't find the
database name for that user on the specified server.
-Kevin

- Original Message -
From: Matthew Bielecki [EMAIL PROTECTED]
To: PHPCoder [EMAIL PROTECTED]; [EMAIL PROTECTED]
Cc: php-general [EMAIL PROTECTED]
Sent: Wednesday, July 24, 2002 2:07 PM
Subject: Re: [PHP] Help with msql_fetch_array()


 Well I think you were correct about not connecting to the db, but I don't
 understand why.  I wrote another little script just to test the
 connection.

 ?php

 $link = mysql_connect(servername,username,password)
   or die(Couldn't make connection.);
 $diditwork = mysql_select_db(news, $link);

 if ($diditwork  FALSE)
{
   echo got the db..yea!;
   echo $link;
}
else
{
   echo didnt get db...bo;
   echo $link;
}
 ?

 This returns didnt get db...booResource id #1

 I don't understand how I can get a resource ID but then not be able to use
 the news database.  Like I described earlier, this happens with my other
 db as well.  I can connect to the db through the console or any other
 client I have on the physical server and do whatever I want with the db's,
 I'm just having problems with php.

 Thanks again for your help!!









 PHPCoder [EMAIL PROTECTED]
 07/24/02 01:50 PM


 To: Matthew Bielecki [EMAIL PROTECTED]
 cc: php-general [EMAIL PROTECTED]
 Subject:Re: [PHP] Help with msql_fetch_array()


 I can almost guarantee that it's not the second line that is failing,
 the problem here is that $result is not containing naything, and that is
 normally due to the fact that you are not connecting to the db, or the
 table tablename is not there.

 I use the following format as my standard MySQL connect and query
 snippet:

 $link = @mysql_connect(localhost,$username,$password) or die ('Could
 not connect!'); //@ suppresses the default error message generated by
 this function and the or die() bit kills the script right then and
 there should it not be able to connect.
 mysql_select_db(YOUR_DB_NAME,$link);
 $sql = select * from your_table_name;
 if ( $result = mysql_query($sql)) {  // checks to see if $result
 contains anything before it even tries to fetch an associative array
 from it.
  $row = mysql_fetch_assoc($result);
 } else {
 echo Empty result set!;

 Note also that I use mysql_fetch_assoc and NOT mysql_fecth_array, as 9
 out of 10 times, you don't need the array element id's that is returned
 by mysql_fetch_array.

 Matthew Bielecki wrote:

 I have a couple of scripts that fail with the error of:
 Warning: mysql_fetch_array(): supplied argument is not a valid MySQL
 result
 resource in...
 
 I'm new to both SQL and PHP and I'm wondering if I have some setting
 turned off or what.
 
 Here's the piece of code that is failing (the second line fails):
 
 $result = mysql_db_query($dbname, SELECT * FROM tablename ORDER BY id);
 $row = mysql_fetch_array($result);
 
 
 Thanks for your help in advance!!
 







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