[PHP] Re: Problems with MySQL-Link [3:14370:14592] (Solved)
Hi all, Do not use || instead of or !!! Example: $result = mysql_query(select now()) OR die (mysql_error()); -- Berthold -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Re: [PHP General] RE: Problems with MySQL-Link [3:14370:14592]
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]
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 -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
[PHP] Re: [PHP General] RE: Problems with MySQL-Link [3:14370:14592]
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... Metin Arman wrote: http://www.zend.com/phorum//read.php?num=3id=14592thread=14370 Sorry, Berthold! I can't think of anything else either. And now, I saw also in your quot;phpinfoquot;-page that you are running Php in CGI mode. And that explains that quot;mysql.allow_persistentquot; shows quot;Offquot;. When Php is running in CGI mode - persistent connections are not possible. Anyhow, tell me if you solve your problem! It's interesting what causes it. Regards Metin -- Berthold -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP] Re: [PHP General] RE: Problems with MySQL-Link [3:14370:14592]
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 -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP] Re: [PHP General] RE: Problems with MySQL-Link [3:14370:14592]
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] Link-Identifyer MySQL Problem
Hello! I have a problem with the connection-identifyer, comming from mysql_connect(). Everything worked well until I add a second MySQL-Server. To do that I have to add the link-identifyer. OK. This must be a bug in PHP: Warning: Supplied argument is not a valid MySQL-Link resource... because without that link it works well! I use php 4.1.1 on Linux. The value of link is '1'. The second link is '1' too! Is there a php.ini-setting for allowing only one connection??? -- Berthold -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP] Link-Identifyer MySQL Problem
Hello again! Yes, there is a setting! But the value is '-1' for 'unlimited' It seems that mysql_connect returns always 'false' or '1'. Strange... Berthold wrote: Hello! I have a problem with the connection-identifyer, comming from mysql_connect(). Everything worked well until I add a second MySQL-Server. To do that I have to add the link-identifyer. OK. This must be a bug in PHP: Warning: Supplied argument is not a valid MySQL-Link resource... because without that link it works well! I use php 4.1.1 on Linux. The value of link is '1'. The second link is '1' too! Is there a php.ini-setting for allowing only one connection??? -- Berthold -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
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