Warning: mysql_query():supplied argument is not a valid MySQL-Link resource

sidekick · #1 07-Jul-2008, 20:12

Hi everyone
Please help me. I'm trying to install a script but it comes up with this error message:
"Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in D:\www\www.test4.co.za\....\lang\lang.php on line 35"

This is on lines 35 and 57 which are identical, in red below:

PHP Code:


<?php

//if(!empty($_GET)) extract($_GET);
//if(!empty($_POST)) extract($_POST);

if (isset($_REQUEST['lang'])) {
    $sql = "SELECT * FROM lang WHERE `lang_code`='".$_REQUEST['lang']."'";
    $result = mysql_query($sql, $jb_mysql_link) or die (mysql_error());
    if (mysql_num_rows($result)>0) {
        $_SESSION["LANG"] = strtoupper($_REQUEST["lang"]);
        // save the requested language
        setcookie("SAVED_LANG", strtoupper($_REQUEST["lang"]), 2147483647);
        
    } else {

        $sql = "SELECT * FROM lang WHERE `is_default`='Y'";
        $result = mysql_query($sql, $jb_mysql_link) or die (mysql_error());
        $row = mysql_fetch_array($result, MYSQL_ASSOC);
        $_SESSION["LANG"] = strtoupper($row["lang_code"]);
        // save the requested language
        setcookie("SAVED_LANG", strtoupper($row["lang_code"]), 2147483647);
        echo "Invalid language. Reverting to default language.";
    }
} elseif (!isset($_SESSION["LANG"])) {
    // get the default language, or saved language

    if ($_COOKIE['SAVED_LANG']!='') {
        $_SESSION["LANG"] = strtoupper($_COOKIE['SAVED_LANG']);

    } else {

        $sql = "SELECT * FROM lang WHERE `is_default`='Y' ";
        [b][color="Red"]if ($result = mysql_query ($sql, $jb_mysql_link)) {[/color][/b]
            $row = mysql_fetch_array($result, MYSQL_ASSOC);
            $_SESSION["LANG"] = strtoupper($row['lang_code']);
            if ($row['charset']!='') {
                setlocale(LC_TIME, $row['charset']);
            }
        } else {
            $_SESSION["LANG"] = 'EN';

        }
    }
    
}

global $AVAILABLE_LANGS;
global $LANG_FILES;
global $FCK_LANG_FILES;

// load languages into array.. map the language code to the filename
// if mapping didn't work, default to english..

$sql = "SELECT * FROM lang ";
[b][color="Red"]if ($result = mysql_query ($sql, $jb_mysql_link)) {[/color][/b]
    while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
        $lang_code = strtoupper($row['lang_code']);
        $AVAILABLE_LANGS [$lang_code] = $row['name'];
        $LANG_FILES[$lang_code] = $row['lang_filename'];
        $FCK_LANG_FILES[$lang_code] = $row['fckeditor_lang'];
    }
    
    if (($_SESSION["LANG"] != '') ) {
        include ($LANG_FILES[$_SESSION["LANG"]]);
        
    } else {
        include ("english.php");
    }

} else {
    $DB_ERROR = mysql_error();

}


?>

PLEASE help me

Thanking you in advance

Howard_L · #2 08-Jul-2008, 01:21

...So , the error says that on line 35 a "supplied argument is not a valid MySQL-Link resource in D:\www\www.test4.co.za\...."
The line is: $row = mysql_fetch_array($result, MYSQL_ASSOC);

...the arguments are: '$result' and 'MYSQL_ASSOC'

I have only seen mysql_fetch_array( $result ); run without the 'MYSQL_ASSOC' constant in there.
From this: http://www.php.net/mysql_fetch_array I see that it specifies the array type to be returned.
Did you put the MYSQL_ASSOC in there or did it come like that? That may be a problem , but I don't think so.

Something's comming up with that: D:\www\www.test4.co.za we see in the warning.
You can check the values leading up to those arguments to see if they all are as expected.
Try inserting the echo statement below into the corresponding location of your code and see what output you get:

PHP Code:


{      $sql = "SELECT * FROM lang WHERE `is_default`='Y' ";      if ($result = mysql_query ($sql, $jb_mysql_link))        {          echo "\sql='". $sql ."', $jb_mysql_link='". $jb_mysql_link ."', \$result='". $result ."' MYSQL_ASSOC='". MYSQL_ASSOC ."' <br />";         $row = mysql_fetch_array($result, MYSQL_ASSOC);  /*<--problem L35 ****/

post back what you get.

sidekick · #3 19-Jul-2008, 10:36

Hi
Thanx for replying. Was away for a while, so sorry for taking so long to post back.
As you may have noticed, this error occurred on my local pc server (windows). I decided to upload the script into a subdomain of one of my live sites, and all worked 100%. It installed perfectly and I could further configure and use it without any issues. The site is linux.
I'm thinking that it maybe configuration of the local server. I have no clue....I usually do everything locally, then upload to live site once I'm ready to launch the site.
Could it be windows?

Howard_L · #4 19-Jul-2008, 14:05

Your home server is a Windows? Which one?
Can you run other php scripts?
...ones that can manipulate mysql?
I think you can but as you say something is not set up right.

Code:

The error came from something invalid comming back from:
        $result = mysql_query($sql, $jb_mysql_link) or die (mysql_error());

...so then the value in $result was invalid and when this line executed:
        $row = mysql_fetch_array($result, MYSQL_ASSOC);

You got an error.
To see what it is try printing the value of '$result
echo "result= '". $result ." <br /> ";

I suspect it will have something sto do with:
  D:\www\www.test4.co.za\....

Does that file exsist?
What is the value in:  $jb_mysql_link ? That might be another one to print...

I do not have a home server (especially windows) and so can not be of much help.
But I imagine there are some default file searching parameters which might be a cause... what do you think?

sidekick · #5 19-Jul-2008, 18:28

Hi Howard,
My home server is windows xp. All php scripts run, this is the first to give me problems. Mysql works fine with php.
I tried what you suggested, but results still the same: "Warning: mysql_query():supplied argument is not a valid MySQL-Link resource"
I have no clue what's the issue is. I've just about checked everything I suspected.
So frustrating...
Thanks so much for your time and input.

Howard_L · #6 19-Jul-2008, 20:27

You didn't get any values for either of those variables?
Try putting this echo before the section that errors.

HTML Code:

        echo "sql=  jb_'". $sql ."' , mysql_link= '". $jb_mysql_link ."' <br /> ";
        $result = mysql_query($sql, $jb_mysql_link) or die (mysql_error());

...you need to find out what $jb_mysql_link holds,, I don't see it assigned in the code.
What does it hold on the server where you get this script to work?

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Warning: mysql_query():supplied argument is not a valid MySQL-Link resource

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Re: Warning: mysql_query():supplied argument is not a valid MySQL-Link resource

Re: Warning: mysql_query():supplied argument is not a valid MySQL-Link resource

Re: Warning: mysql_query():supplied argument is not a valid MySQL-Link resource