Re: [PHP] Re: [PHP General] RE: Problems with MySQL-Link [3:14370:14592]

2002-02-01 Thread Berthold

Hi again!

If it would be so simple I would not ask...

OK, once more:
#!/usr/local/bin/php -q
?php
$link1 = mysql_connect(localhost,user,passwd)
  || die (Fehler! Zeile .__LINE__.:\n.$php_errormsg.\n.mysql_error());
echo Link 1: $link1\n;
if (mysql_select_db(test, $link1)) {
   echo Successful select of test\n;
} else {
   echo Failed to select test\n;
   die (Fehler! Zeile .__LINE__.:\n.mysql_error().\n);
}
?
gives this output:
Link1: 1
Warning: Supplied argument is not a valid MySQL-Link resource in
/home/bt/bin/test.php on line 5
Failed to select db
Fehler! Zeile 9:

And once again I describe it:
I've tested all settings for MySQL in the php.ini.




Jason Wong wrote:

 On Friday 01 February 2002 15:38, Berthold wrote:
 
Hi!

Thank you.
But I cannot use that $link1 and/or $link2 to select two databases on
different servers...
If I use $link1 the first time to select the first database on the first
server I got that error. But without that $link1-variable AND only one
connection in the script it works.

And it makes no sense if I use it as shell-prog or with Apache-Builtin-PHP.

eg:
#!/usr/local/bin/php -q
?php
$link1 = mysql_connect(localhost,a);
echo 'Link 1: '.$link1 .'BR';
if (mysql_select_db(test, $link1)) {
  echo 'Successful select of testBR';
} else {
  echo 'Failed to select testBR';
}
?
gives this output:
Warning: Supplied argument is not a valid MySQL-Link resource in
/home/bt/bin/test.php on line 4

 
 Try:
 
  $link1 = mysql_connect(localhost,a) or die (Cannot connect to db!); 
 
 If it dies, then you know you have a problem connecting to the db.
 
 If you follow the examples in the code with regards to the DB functions they 
 all have a die() statement on the end for debugging purposes. It's there to 
 help you.
 
 In this case your real problem is most likely that you're not giving the 
 correct number of parameters to mysql_connect($host, $user, $password).
 
 Again, proper use of die() would have quickly helped you track down where the 
 real problem was. Instead of trying to figure out what is wrong with line 4!
 
 hth
 


-- 
Berthold


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Re: [PHP] Re: [PHP General] RE: Problems with MySQL-Link [3:14370:14592]

2002-02-01 Thread Jason Wong

On Friday 01 February 2002 16:23, Berthold wrote:
 Hi again!

 If it would be so simple I would not ask...

 OK, once more:
 #!/usr/local/bin/php -q


1) Has the mysql functions ever worked for you on that system?

2) Are you only get this problem when using php from the command line?


-- 
Jason Wong - Gremlins Associates - www.gremlins.com.hk

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Re: [PHP] Re: [PHP General] RE: Problems with MySQL-Link [3:14370:14592]

2002-02-01 Thread Berthold

Hi!

1) Yes of course. I only have to delete the $link-variables...
But then - of course - I can only use the last connection.

2) It made no sense. Shell or Apache.
And the settings in php.ini are '-1' for 'unlimited'.
And I use the right php.ini, even with the shell-version.

Jason Wong wrote:

 On Friday 01 February 2002 16:23, Berthold wrote:
 
Hi again!

If it would be so simple I would not ask...

OK, once more:
#!/usr/local/bin/php -q

 
 
 1) Has the mysql functions ever worked for you on that system?
 
 2) Are you only get this problem when using php from the command line?
 
 
 


-- 
Berthold


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Re: [PHP] Re: [PHP General] RE: Problems with MySQL-Link [3:14370:14592]

2002-01-31 Thread Berthold

Hi!

mysql_error returns nothing...

The error arrives while selecting the database via:

$link = mysql_connect(host,user,passwd);
mysql_select_db(bla, $link);

echo $link returns always '1'!




David Robley wrote:
  In article [EMAIL PROTECTED], [EMAIL PROTECTED] says...
 
 So,
 
 I've testet all possible I found in my brain...
 CGI or Apache module: same error
 mysql_connect or mysql_pconnect: same error
 Variing settings in php.ini: same error
 
 How can I solve that problem?
 BTW: Here I describe it again:
 Connecting to a database works well: ok
 Working with that database: ok
 Using the additional $link-Parameter: false
 Not a valid resource id ...
 
 And it does not matter how many different links with different servers I
 use! The resource-id is always '1'.
 
 Strange...
 
 
  Not a valid resource ID usually means a problem with the query - in fact
  this might be in the FAQ. Use mysql_error() after the database query to
  see if mysql returns a useful error.
 
 


-- 
Berthold



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Re: [PHP] Re: [PHP General] RE: Problems with MySQL-Link [3:14370:14592]

2002-01-31 Thread David Robley

In article [EMAIL PROTECTED], [EMAIL PROTECTED] says...
 Hi!
 
 mysql_error returns nothing...
 
 The error arrives while selecting the database via:
 
 $link = mysql_connect(host,user,passwd);
 mysql_select_db(bla, $link);
 
 echo $link returns always '1'!
 
 
 
 
 David Robley wrote:
   In article [EMAIL PROTECTED], [EMAIL PROTECTED] says...
  
  So,
  
  I've testet all possible I found in my brain...
  CGI or Apache module: same error
  mysql_connect or mysql_pconnect: same error
  Variing settings in php.ini: same error
  
  How can I solve that problem?
  BTW: Here I describe it again:
  Connecting to a database works well: ok
  Working with that database: ok
  Using the additional $link-Parameter: false
  Not a valid resource id ...
  
  And it does not matter how many different links with different servers I
  use! The resource-id is always '1'.
  
  Strange...
  
  
   Not a valid resource ID usually means a problem with the query - in fact
   this might be in the FAQ. Use mysql_error() after the database query to
   see if mysql returns a useful error.

Guess I misread your question. However, a litle playing around with this 
script

?php
$link1 = mysql_connect(localhost,a);
$link2 = mysql_connect(localhost);
echo 'Link 1: '.$link1 .'BR';
echo 'Link 2: '.$link2 .'BR';
if(mysql_select_db(test, $link1)) {
echo 'Successful select of testBR';
}else{
echo 'Failed to select testBR';
}
if(mysql_select_db(test2, $link2)) {
echo 'Successful select of test2BR';
}else{
echo 'Failed to select test2BR';
}

?
which gives this output
Link 1: Resource id #1
Link 2: Resource id #2
Successful select of test
Successful select of test2

seems to indicate that php is smart enough not to open a new link even if 
requested if there is already a link for the user! Removing the a in 
the first mysql_connect gives me two links with a resource ID of 1.

I can force errors of course by feeding a wrong username/password which 
eventually leads to a 'Not a valid resource id' error but has other error 
messages as well. You aren't by chance hiding other error responses by 
using @ or 'or die'?

-- 
David Robley
Temporary Kiwi!

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Re: [PHP] Re: [PHP General] RE: Problems with MySQL-Link [3:14370:14592]

2002-01-31 Thread Berthold

Hi!

Thank you.
But I cannot use that $link1 and/or $link2 to select two databases on 
different servers...
If I use $link1 the first time to select the first database on the first 
server I got that error. But without that $link1-variable AND only one 
connection in the script it works.

And it makes no sense if I use it as shell-prog or with Apache-Builtin-PHP.

eg:
#!/usr/local/bin/php -q
?php
$link1 = mysql_connect(localhost,a);
echo 'Link 1: '.$link1 .'BR';
if (mysql_select_db(test, $link1)) {
echo 'Successful select of testBR';
} else {
echo 'Failed to select testBR';
}
?
gives this output:
Warning: Supplied argument is not a valid MySQL-Link resource in 
/home/bt/bin/test.php on line 4

I don't know what to do...



David Robley wrote:

 In article [EMAIL PROTECTED], [EMAIL PROTECTED] says...
 
Hi!

mysql_error returns nothing...

The error arrives while selecting the database via:

$link = mysql_connect(host,user,passwd);
mysql_select_db(bla, $link);

echo $link returns always '1'!




David Robley wrote:
  In article [EMAIL PROTECTED], [EMAIL PROTECTED] says...
 
 So,
 
 I've testet all possible I found in my brain...
 CGI or Apache module: same error
 mysql_connect or mysql_pconnect: same error
 Variing settings in php.ini: same error
 
 How can I solve that problem?
 BTW: Here I describe it again:
 Connecting to a database works well: ok
 Working with that database: ok
 Using the additional $link-Parameter: false
 Not a valid resource id ...
 
 And it does not matter how many different links with different servers I
 use! The resource-id is always '1'.
 
 Strange...
 
 
  Not a valid resource ID usually means a problem with the query - in fact
  this might be in the FAQ. Use mysql_error() after the database query to
  see if mysql returns a useful error.

 
 Guess I misread your question. However, a litle playing around with this 
 script
 
 ?php
 $link1 = mysql_connect(localhost,a);
 $link2 = mysql_connect(localhost);
 echo 'Link 1: '.$link1 .'BR';
 echo 'Link 2: '.$link2 .'BR';
 if(mysql_select_db(test, $link1)) {
   echo 'Successful select of testBR';
   }else{
   echo 'Failed to select testBR';
   }
 if(mysql_select_db(test2, $link2)) {
   echo 'Successful select of test2BR';
   }else{
   echo 'Failed to select test2BR';
   }
 
 ?
 which gives this output
 Link 1: Resource id #1
 Link 2: Resource id #2
 Successful select of test
 Successful select of test2
 
 seems to indicate that php is smart enough not to open a new link even if 
 requested if there is already a link for the user! Removing the a in 
 the first mysql_connect gives me two links with a resource ID of 1.
 
 I can force errors of course by feeding a wrong username/password which 
 eventually leads to a 'Not a valid resource id' error but has other error 
 messages as well. You aren't by chance hiding other error responses by 
 using @ or 'or die'?
 
 


-- 
Berthold


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PHP General Mailing List (http://www.php.net/)
To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
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Re: [PHP] Re: [PHP General] RE: Problems with MySQL-Link [3:14370:14592]

2002-01-31 Thread Jason Wong

On Friday 01 February 2002 15:38, Berthold wrote:
 Hi!

 Thank you.
 But I cannot use that $link1 and/or $link2 to select two databases on
 different servers...
 If I use $link1 the first time to select the first database on the first
 server I got that error. But without that $link1-variable AND only one
 connection in the script it works.

 And it makes no sense if I use it as shell-prog or with Apache-Builtin-PHP.

 eg:
 #!/usr/local/bin/php -q
 ?php
 $link1 = mysql_connect(localhost,a);
 echo 'Link 1: '.$link1 .'BR';
 if (mysql_select_db(test, $link1)) {
   echo 'Successful select of testBR';
 } else {
   echo 'Failed to select testBR';
 }
 ?
 gives this output:
 Warning: Supplied argument is not a valid MySQL-Link resource in
 /home/bt/bin/test.php on line 4

Try:

 $link1 = mysql_connect(localhost,a) or die (Cannot connect to db!); 

If it dies, then you know you have a problem connecting to the db.

If you follow the examples in the code with regards to the DB functions they 
all have a die() statement on the end for debugging purposes. It's there to 
help you.

In this case your real problem is most likely that you're not giving the 
correct number of parameters to mysql_connect($host, $user, $password).

Again, proper use of die() would have quickly helped you track down where the 
real problem was. Instead of trying to figure out what is wrong with line 4!

hth
-- 
Jason Wong - Gremlins Associates - www.gremlins.com.hk

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