not a valid MySQL-Link resource
This is a discussion on not a valid MySQL-Link resource within the PHP Language forums, part of the PHP Programming Forums category; Hi everybody, I cannot understand the following thinks. The last line of the fillowing code produces a message about mistake (...
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04-26-2007
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not a valid MySQL-Link resource
Hi everybody,
I cannot understand the following thinks. The last line of the fillowing code produces a message about mistake (not a valid MySQL- Link resource): $link = mysql_connect( "localhost","tmp_user","tmpxxx" ); mysql_select_db( $db_name, $link); $link = mysql_connect( "localhost","tmp_user","tmpxxx" ); mysql_close( $link ); mysql_select_db( "sss", $link ); It is clear why it is happens. Because I use $link to the database, which has been closed in the previouse line. To remove this problem I have to change the code in the following way: $link = mysql_connect( "localhost","tmp_user","tmpxxx" ); mysql_select_db( $db_name, $link); $link_new = mysql_connect( "localhost","tmp_user","tmpxxx" ); mysql_close( $link_new ); mysql_select_db( "sss", $link ); Now I do the same but with the usage of a function. As it is expected, the following code produce the message about mistake: $link = mysql_connect( "localhost","tmp_user","tmpxxx" ); mysql_select_db( $db_name, $link); function fff() { $link = mysql_connect( "localhost","tmp_user","tmpxxx" ); mysql_close( $link ); } fff(); mysql_select_db( "sss", $link ); But what is strange and what I cannot understand is why the previosly used solution does not work. The following code alse generate the same message! $link = mysql_connect( "localhost","tmp_user","tmpxxx" ); mysql_select_db( $db_name, $link); function fff() { $link_new = mysql_connect( "localhost","tmp_user","tmpxxx" ); mysql_close( $link_new ); } fff(); mysql_select_db( "sss", $link ); Thanks. 
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04-26-2007
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Re: not a valid MySQL-Link resource
"Kurda Yon" <kurdayon@yahoo.com> wrote in message news:1177613411.140647.31800@n35g2000prd.googlegro ups.com... | Hi everybody, | | I cannot understand the following thinks. The last line of the | fillowing code produces a message about mistake (not a valid MySQL- | Link resource): | | $link = mysql_connect( "localhost","tmp_user","tmpxxx" ); | mysql_select_db( $db_name, $link); | $link = mysql_connect( "localhost","tmp_user","tmpxxx" ); | mysql_close( $link ); | mysql_select_db( "sss", $link ); | | It is clear why it is happens. Because I use $link to the database, | which has been closed in the previouse line. To remove this problem I | have to change the code in the following way: | | $link = mysql_connect( "localhost","tmp_user","tmpxxx" ); | mysql_select_db( $db_name, $link); | $link_new = mysql_connect( "localhost","tmp_user","tmpxxx" ); | mysql_close( $link_new ); | mysql_select_db( "sss", $link ); | | Now I do the same but with the usage of a function. As it is expected, | the following code produce the message about mistake: | | $link = mysql_connect( "localhost","tmp_user","tmpxxx" ); | mysql_select_db( $db_name, $link); | function fff() | { | $link = mysql_connect( "localhost","tmp_user","tmpxxx" ); | mysql_close( $link ); | } | fff(); | mysql_select_db( "sss", $link ); | | But what is strange and what I cannot understand is why the previosly | used solution does not work. The following code alse generate the same | message! | | $link = mysql_connect( "localhost","tmp_user","tmpxxx" ); | mysql_select_db( $db_name, $link); | function fff() | { | $link_new = mysql_connect( "localhost","tmp_user","tmpxxx" ); | mysql_close( $link_new ); | } | fff(); | mysql_select_db( "sss", $link ); hard to understand you. is this your *real* code? you realize that $link INSIDE the function is not $link OUTSIDE your function, right? why are you worrying about closing the connection anyway? the connection is dropped after the script is run/exited.
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04-27-2007
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Re: not a valid MySQL-Link resource
On Apr 26, 9:20 pm, "Steve" <no....@example.com> wrote:
> "Kurda Yon" <kurda...@yahoo.com> wrote in message > > news:1177613411.140647.31800@n35g2000prd.googlegro ups.com... > | Hi everybody, > | > | I cannot understand the following thinks. The last line of the > | fillowing code produces a message about mistake (not a valid MySQL- > | Link resource): > | > | $link = mysql_connect( "localhost","tmp_user","tmpxxx" ); > | mysql_select_db( $db_name, $link); > | $link = mysql_connect( "localhost","tmp_user","tmpxxx" ); > | mysql_close( $link ); > | mysql_select_db( "sss", $link ); > | > | It is clear why it is happens. Because I use $link to the database, > | which has been closed in the previouse line. To remove this problem I > | have to change the code in the following way: > | > | $link = mysql_connect( "localhost","tmp_user","tmpxxx" ); > | mysql_select_db( $db_name, $link); > | $link_new = mysql_connect( "localhost","tmp_user","tmpxxx" ); > | mysql_close( $link_new ); > | mysql_select_db( "sss", $link ); > | > | Now I do the same but with the usage of a function. As it is expected, > | the following code produce the message about mistake: > | > | $link = mysql_connect( "localhost","tmp_user","tmpxxx" ); > | mysql_select_db( $db_name, $link); > | function fff() > | { > | $link = mysql_connect( "localhost","tmp_user","tmpxxx" ); > | mysql_close( $link ); > | } > | fff(); > | mysql_select_db( "sss", $link ); > | > | But what is strange and what I cannot understand is why the previosly > | used solution does not work. The following code alse generate the same > | message! > | > | $link = mysql_connect( "localhost","tmp_user","tmpxxx" ); > | mysql_select_db( $db_name, $link); > | function fff() > | { > | $link_new = mysql_connect( "localhost","tmp_user","tmpxxx" ); > | mysql_close( $link_new ); > | } > | fff(); > | mysql_select_db( "sss", $link ); > > hard to understand you. > > is this your *real* code? you realize that $link INSIDE the function is not > $link OUTSIDE your function, right? why are you worrying about closing the > connection anyway? the connection is dropped after the script is run/exited. Sorry, may be I explained it in a bad way. The problem is that the following code produce a message about mistake: $link = mysql_connect( "localhost","tmp_user","tmpxxx" ); mysql_select_db( $db_name, $link); function fff() { $link_new = mysql_connect( "localhost","tmp_user","tmpxxx" ); mysql_close( $link_new ); } fff(); mysql_select_db( "sss", $link ); Or more precisely the last line causes the following message: Warning: mysql_select_db(): 2 is not a valid MySQL-Link resource And it happens only if I execute the fff() function before. I suppose that origin of the problem is that I close connection to the database in the function. But I do not understand why. I do give another name to the connection.
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04-27-2007
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Re: not a valid MySQL-Link resource
Kurda Yon wrote:
> On Apr 26, 9:20 pm, "Steve" <no....@example.com> wrote: >> "Kurda Yon" <kurda...@yahoo.com> wrote in message >> >> news:1177613411.140647.31800@n35g2000prd.googlegro ups.com... >> | Hi everybody, >> | >> | I cannot understand the following thinks. The last line of the >> | fillowing code produces a message about mistake (not a valid MySQL- >> | Link resource): >> | >> | $link = mysql_connect( "localhost","tmp_user","tmpxxx" ); >> | mysql_select_db( $db_name, $link); >> | $link = mysql_connect( "localhost","tmp_user","tmpxxx" ); >> | mysql_close( $link ); >> | mysql_select_db( "sss", $link ); >> | >> | It is clear why it is happens. Because I use $link to the database, >> | which has been closed in the previouse line. To remove this problem I >> | have to change the code in the following way: >> | >> | $link = mysql_connect( "localhost","tmp_user","tmpxxx" ); >> | mysql_select_db( $db_name, $link); >> | $link_new = mysql_connect( "localhost","tmp_user","tmpxxx" ); >> | mysql_close( $link_new ); >> | mysql_select_db( "sss", $link ); >> | >> | Now I do the same but with the usage of a function. As it is expected, >> | the following code produce the message about mistake: >> | >> | $link = mysql_connect( "localhost","tmp_user","tmpxxx" ); >> | mysql_select_db( $db_name, $link); >> | function fff() >> | { >> | $link = mysql_connect( "localhost","tmp_user","tmpxxx" ); >> | mysql_close( $link ); >> | } >> | fff(); >> | mysql_select_db( "sss", $link ); >> | >> | But what is strange and what I cannot understand is why the previosly >> | used solution does not work. The following code alse generate the same >> | message! >> | >> | $link = mysql_connect( "localhost","tmp_user","tmpxxx" ); >> | mysql_select_db( $db_name, $link); >> | function fff() >> | { >> | $link_new = mysql_connect( "localhost","tmp_user","tmpxxx" ); >> | mysql_close( $link_new ); >> | } >> | fff(); >> | mysql_select_db( "sss", $link ); >> >> hard to understand you. >> >> is this your *real* code? you realize that $link INSIDE the function is not >> $link OUTSIDE your function, right? why are you worrying about closing the >> connection anyway? the connection is dropped after the script is run/exited. > > Sorry, may be I explained it in a bad way. The problem is that the > following code produce a message about mistake: > > $link = mysql_connect( "localhost","tmp_user","tmpxxx" ); > mysql_select_db( $db_name, $link); > function fff() > { > $link_new = mysql_connect( "localhost","tmp_user","tmpxxx" ); > mysql_close( $link_new ); > } > fff(); > mysql_select_db( "sss", $link ); > > Or more precisely the last line causes the following message: > Warning: mysql_select_db(): 2 is not a valid MySQL-Link resource > > And it happens only if I execute the fff() function before. I suppose > that origin of the problem is that I close connection to the database > in the function. But I do not understand why. I do give another name > to the connection. > > From the mysql_connect() manual page: resource mysql_connect ( [string $server [, string $username [, string $password [, bool $new_link [, int $client_flags]]]]] ) [...] If a second call is made to mysql_connect() with the same arguments, no new link will be established, but instead, the link identifier of the already opened link will be returned. The new_link parameter modifies this behavior and makes mysql_connect() always open a new link, even if mysql_connect() was called before with the same parameters. In SQL safe mode, this parameter is ignored. This might be what causes your error. HTH Sh.
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04-27-2007
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Re: not a valid MySQL-Link resource
On Apr 27, 12:07 pm, Schraalhans Keukenmeester
<bitbuc...@invalid.spam> wrote: > Kurda Yon wrote: > > On Apr 26, 9:20 pm, "Steve" <no....@example.com> wrote: > >> "Kurda Yon" <kurda...@yahoo.com> wrote in message > > >>news:1177613411.140647.31800@n35g2000prd.googleg roups.com... > >> | Hi everybody, > >> | > >> | I cannot understand the following thinks. The last line of the > >> | fillowing code produces a message about mistake (not a valid MySQL- > >> | Link resource): > >> | > >> | $link = mysql_connect( "localhost","tmp_user","tmpxxx" ); > >> | mysql_select_db( $db_name, $link); > >> | $link = mysql_connect( "localhost","tmp_user","tmpxxx" ); > >> | mysql_close( $link ); > >> | mysql_select_db( "sss", $link ); > >> | > >> | It is clear why it is happens. Because I use $link to the database, > >> | which has been closed in the previouse line. To remove this problem I > >> | have to change the code in the following way: > >> | > >> | $link = mysql_connect( "localhost","tmp_user","tmpxxx" ); > >> | mysql_select_db( $db_name, $link); > >> | $link_new = mysql_connect( "localhost","tmp_user","tmpxxx" ); > >> | mysql_close( $link_new ); > >> | mysql_select_db( "sss", $link ); > >> | > >> | Now I do the same but with the usage of a function. As it is expected, > >> | the following code produce the message about mistake: > >> | > >> | $link = mysql_connect( "localhost","tmp_user","tmpxxx" ); > >> | mysql_select_db( $db_name, $link); > >> | function fff() > >> | { > >> | $link = mysql_connect( "localhost","tmp_user","tmpxxx" ); > >> | mysql_close( $link ); > >> | } > >> | fff(); > >> | mysql_select_db( "sss", $link ); > >> | > >> | But what is strange and what I cannot understand is why the previosly > >> | used solution does not work. The following code alse generate the same > >> | message! > >> | > >> | $link = mysql_connect( "localhost","tmp_user","tmpxxx" ); > >> | mysql_select_db( $db_name, $link); > >> | function fff() > >> | { > >> | $link_new = mysql_connect( "localhost","tmp_user","tmpxxx" ); > >> | mysql_close( $link_new ); > >> | } > >> | fff(); > >> | mysql_select_db( "sss", $link ); > > >> hard to understand you. > > >> is this your *real* code? you realize that $link INSIDE the function is not > >> $link OUTSIDE your function, right? why are you worrying about closing the > >> connection anyway? the connection is dropped after the script is run/exited. > > > Sorry, may be I explained it in a bad way. The problem is that the > > following code produce a message about mistake: > > > $link = mysql_connect( "localhost","tmp_user","tmpxxx" ); > > mysql_select_db( $db_name, $link); > > function fff() > > { > > $link_new = mysql_connect( "localhost","tmp_user","tmpxxx" ); > > mysql_close( $link_new ); > > } > > fff(); > > mysql_select_db( "sss", $link ); > > > Or more precisely the last line causes the following message: > > Warning: mysql_select_db(): 2 is not a valid MySQL-Link resource > > > And it happens only if I execute the fff() function before. I suppose > > that origin of the problem is that I close connection to the database > > in the function. But I do not understand why. I do give another name > > to the connection. > > From the mysql_connect() manual page: > > resource mysql_connect ( [string $server [, string $username [, string > $password [, bool $new_link [, int $client_flags]]]]] ) > > [...] > > If a second call is made to mysql_connect() with the same arguments, no > new link will be established, but instead, the link identifier of the > already opened link will be returned. The new_link parameter modifies > this behavior and makes mysql_connect() always open a new link, even if > mysql_connect() was called before with the same parameters. In SQL safe > mode, this parameter is ignored. > > This might be what causes your error. > > HTH > > Sh. Thanks! It helps.
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04-27-2007
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Re: not a valid MySQL-Link resource
It is me again. It seems that I solve the previous problem (with your
help) but get a new one (which seems to be very related to the previous one). This part of code produce a message about mistake: $link = mysql_connect( "localhost","tmp_user","pswxxx" ); function fff() { $link_new = mysql_connect( "localhost","tmp_user","pswxxx", true ); mysql_close( $link_new ); } fff(); mysql_select_db( "sss", $link ); $result = mysql_query( "select * from root" ); The message is: Access denied for user: 'nobody@localhost' (Using password: NO) A link to the server could not be established It complains about the last line. And this message disappears if I do not coll the function fff (I replace fff(); by //fff();). Can anybody explain this behavior? It seems that connection made to the database in the function somehow influences on other connection.
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04-27-2007
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Re: not a valid MySQL-Link resource
On Fri, 27 Apr 2007 04:53:04 -0700, Kurda Yon wrote:
> It is me again. It seems that I solve the previous problem (with your > help) but get a new one (which seems to be very related to the > previous one). > > This part of code produce a message about mistake: > > $link = mysql_connect( "localhost","tmp_user","pswxxx" ); > function fff() > { > $link_new = mysql_connect( "localhost","tmp_user","pswxxx", true ); > mysql_close( $link_new ); > } > fff(); > mysql_select_db( "sss", $link ); > $result = mysql_query( "select * from root" ); > > The message is: > Access denied for user: 'nobody@localhost' (Using password: NO) > A link to the server could not be established > > It complains about the last line. And this message disappears if I do > not coll the function fff (I replace fff(); by //fff();). > > Can anybody explain this behavior? It seems that connection made to > the database in the function somehow influences on other connection. I'm guessing here: You use literal strings here in your post making the mysql connection, but do you perhaps use variables in your real code? If so, (and assuming globals are off) the function doesn't see these variables unless you specify them global in the function. Other than that, I can't think of anything right now. I cannot reproduce the error anyway with my code (below). <?PHP require 'dbdata.inc.php'; $link = mysql_connect($dbhost,$dbuser,$dbpasswd); function dbopen() { global $dbhost, $dbuser, $dbpasswd; $new_link = mysql_connect($dbhost,$dbuser,$dbpasswd,true); mysql_close($new_link); } dbopen(); mysql_select_db($dbname,$link); $result= mysql_query('select count(*) from customers',$link); $row = mysql_fetch_row($result); echo $row['0']; // outputs 13, oughtta be closer to 50 if you ask me;-) ?>
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04-27-2007
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Re: not a valid MySQL-Link resource
On Apr 27, 3:02 pm, Schraalhans Keukenmeester <inva...@invalid.spam>
wrote: > On Fri, 27 Apr 2007 04:53:04 -0700, Kurda Yon wrote: > > It is me again. It seems that I solve the previous problem (with your > > help) but get a new one (which seems to be very related to the > > previous one). > > > This part of code produce a message about mistake: > > > $link = mysql_connect( "localhost","tmp_user","pswxxx" ); > > function fff() > > { > > $link_new = mysql_connect( "localhost","tmp_user","pswxxx", true ); > > mysql_close( $link_new ); > > } > > fff(); > > mysql_select_db( "sss", $link ); > > $result = mysql_query( "select * from root" ); > > > The message is: > > Access denied for user: 'nobody@localhost' (Using password: NO) > > A link to the server could not be established > > > It complains about the last line. And this message disappears if I do > > not coll the function fff (I replace fff(); by //fff();). > > > Can anybody explain this behavior? It seems that connection made to > > the database in the function somehow influences on other connection. > > I'm guessing here: You use literal strings here in your post making the > mysql connection, but do you perhaps use variables in your real code? If > so, (and assuming globals are off) the function doesn't see these > variables unless you specify them global in the function. > > Other than that, I can't think of anything right now. I cannot reproduce > the error anyway with my code (below). > > <?PHP > require 'dbdata.inc.php'; > $link = mysql_connect($dbhost,$dbuser,$dbpasswd); > function dbopen() { > global $dbhost, $dbuser, $dbpasswd; > $new_link = mysql_connect($dbhost,$dbuser,$dbpasswd,true); > mysql_close($new_link);} > > dbopen(); > mysql_select_db($dbname,$link); > $result= mysql_query('select count(*) from customers',$link); > $row = mysql_fetch_row($result); > echo $row['0']; // outputs 13, oughtta be closer to 50 if you ask me;-) > ?> Thank you for your code. I have compared it with my code (line by line) and found out that the origin of the mistake was that I did not put $link, as the second argument of the function mysql_query (in the last line). Thanks!
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