fetch_assoc() error

[Daimler]

I installed a script and when i go to administration > languages i receive the fallowing error:

 

Fatal error: Call to a member function fetch_assoc() on a non-object in /home/site/public_html/admin/inc/fnc.lang_search.php on line 83

 

on line 83 i have this:  while ($row = $res->fetch_assoc())

 

Any ideeas please ?

[bh]

Hi,

 

Post your code. The error message says that your $res is not an object.

[Daimler]

This is the all code:

<?php

 

function searchLangKeys($phrase, $search_key, $search_value) {

 

$_search = array( '%', '_', '*' );

$_replace = array( '\\%', '\\_', '%' );

 

$_phrase = str_replace( $_search, $_replace, ' '.mysql_real_escape_string( trim( $phrase ) ).' ' );

 

$_phrase = str_replace( ' ', '%', $_phrase );

 

$_where_definition = array();

if( $search_key )

$_where[] = '`key` LIKE ?phrase';

if( $search_value )

$_where[] = '`value` LIKE ?phrase';

 

$_where_definition = empty( $_where ) ? '1' : implode( ' OR ', $_where );

 

$def_lang = SK_Config::section('languages')->get('default_lang_id');

 

$query = sql_placeholder( "SELECT `section` , `parent_section_id`, `lang_section_id`, `".TBL_LANG_KEY."`.`lang_key_id`, `key`, `value`

FROM `".TBL_LANG_SECTION."` LEFT JOIN `".TBL_LANG_KEY."` USING(`lang_section_id`)

LEFT JOIN `".TBL_LANG_VALUE."` USING(`lang_key_id`) WHERE (".$_where_definition.")",

array( 'phrase' => $_phrase )  );

 

$keys = array();

 

$res = SK_MySQL::query($query);

 

while ($row = $res->fetch_assoc()) {

$path = '';

if ($row['parent_section_id'] != 0) {

$path = $row['section'];

$parent_sect = $row['parent_section_id'];

do {

$query = sql_placeholder( "SELECT `parent_section_id`, `section`, `lang_section_id`,

`".TBL_LANG_KEY."`.`lang_key_id`, `key`

FROM `".TBL_LANG_SECTION."`

LEFT JOIN `".TBL_LANG_KEY."` USING(`lang_section_id`)

WHERE `lang_section_id` = ?", $parent_sect );

 

$parent = SK_MySQL::query($query)->fetch_assoc();

$parent_sect = $parent['parent_section_id'];

 

$end = strlen($path) == 0 ? "" : "." . $path;

$path = $parent['section'] . $end;

}

while ( $parent_sect != 0);

 

$row['path'] = $path;

}

else $row['path'] = $row['section'];

 

$keys[] = $row;

}

 

return $keys;

}

 

 

function getMissingLabels($lang_id)

{

$languages_num = count(SK_LanguageEdit::getLanguages());

 

$missing_labels = MySQL::FetchArray( "SELECT `lang_key_id`

FROM `".TBL_LANG_KEY."`

WHERE `lang_key_id` NOT IN

(SELECT `lang_key_id` from `".TBL_LANG_VALUE."` WHERE `lang_id`=".$lang_id." AND `value` != '')");

 

if (is_array($missing_labels)) {

 

foreach( $missing_labels as $id => $num )

{

$query = "SELECT `section` , `parent_section_id`, `lang_section_id`,

`".TBL_LANG_KEY."`.`lang_key_id`, `key`, `value`

FROM `".TBL_LANG_SECTION."` LEFT JOIN `".TBL_LANG_KEY."` USING(`lang_section_id`)

LEFT JOIN `".TBL_LANG_VALUE."` USING(`lang_key_id`)

WHERE `lang_key_id`=".$num['lang_key_id']." GROUP BY `lang_key_id`";

 

$res = SK_MySQL::query($query);

 

while ($row = $res->fetch_assoc())

{

$path = '';

if ($row['parent_section_id'] != 0) {

$path = $row['section'];

$parent_sect = $row['parent_section_id'];

do {

$query = "SELECT `parent_section_id`, `section`, `lang_section_id`,

`".TBL_LANG_KEY."`.`lang_key_id`, `key`

FROM `".TBL_LANG_SECTION."`

LEFT JOIN `".TBL_LANG_KEY."` USING(`lang_section_id`)

WHERE `lang_section_id` =" . $parent_sect ;

 

$parent = SK_MySQL::query($query)->fetch_assoc();

$parent_sect = $parent['parent_section_id'];

 

$end = strlen($path) == 0 ? "" : "." . $path;

$path = $parent['section'] . $end;

}

while ( $parent_sect != 0);

 

$row['path'] = $path;

}

else $row['path'] = $row['section'];

 

$keys['labels'][] = $row;

}

}

$keys['total'] = count($keys['labels']);

}

else {

$keys = array();

$keys['total'] = 0;

}

 

return $keys;

}

 

[bh]

Check your $res variable, probably your query(*) has an error.

 

*

$res = SK_MySQL::query($query);

 

Debug your query with simple constants example in phpMyAdmin.

 

[Daimler]

can you please be more specific ? i am a very beginner in php thank you very much for your time 

[bh]

One of your query probably wrong:

 

$query = sql_placeholder( "SELECT `section` , `parent_section_id`, `lang_section_id`, `".TBL_LANG_KEY."`.`lang_key_id`, `key`, `value`

      FROM `".TBL_LANG_SECTION."` LEFT JOIN `".TBL_LANG_KEY."` USING(`lang_section_id`)

      LEFT JOIN `".TBL_LANG_VALUE."` USING(`lang_key_id`) WHERE (".$_where_definition.")",

      array( 'phrase' => $_phrase )  );

 

$query = "SELECT `section` , `parent_section_id`, `lang_section_id`,

            `".TBL_LANG_KEY."`.`lang_key_id`, `key`, `value`

            FROM `".TBL_LANG_SECTION."` LEFT JOIN `".TBL_LANG_KEY."` USING(`lang_section_id`)

            LEFT JOIN `".TBL_LANG_VALUE."` USING(`lang_key_id`)

            WHERE `lang_key_id`=".$num['lang_key_id']." GROUP BY `lang_key_id`";

 

Try to test it with simple constants. If its good with sample values, than your dynamic php values wrong.

 

[Daimler]

Can you please tell me how to test this ? and what is a simple constant ? 

[bh]

Yeah, subtitue your dynamic variables with constants.

 

E.g change "TBL_LANG_KEY" to "lang_key"... and check in PHP your values, whether good/wrong. E.g your $_where_definition variable is a correct mysql where statement and etc...

[Daimler]

Hmm i dont think this problem is in script: i installed thos script on my localhost (xampp) and all works fine but not on my webserver.  what could this be ?

[Daimler]

Hey, i changed that TBL_LANG_KEY and others TBL * with lang_key and etc. and i receive this errors now:

Warning: MySQL error: Table 'gayland_gay.lang_key' doesn't exist

in query: SELECT `lang_key_id` FROM `lang_key` WHERE `lang_key_id` NOT IN (SELECT `lang_key_id` from `lang_value` WHERE `lang_id`=1 AND `value` != '')

in /home/gayland/public_html/internals/SK6_inc/class.mysql.php on line 299

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/gayland/public_html/internals/SK6_inc/class.mysql.php on line 50

 

Warning: MySQL error: Table 'gayland_gay.lang_key' doesn't exist

in query: SELECT `lang_key_id` FROM `lang_key` WHERE `lang_key_id` NOT IN (SELECT `lang_key_id` from `lang_value` WHERE `lang_id`=8 AND `value` != '')

in /home/gayland/public_html/internals/SK6_inc/class.mysql.php on line 299

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/gayland/public_html/internals/SK6_inc/class.mysql.php on line 50

 

Warning: MySQL error: Table 'gayland_gay.lang_key' doesn't exist

in query: SELECT `lang_key_id` FROM `lang_key` WHERE `lang_key_id` NOT IN (SELECT `lang_key_id` from `lang_value` WHERE `lang_id`=13 AND `value` != '')

in /home/gayland/public_html/internals/SK6_inc/class.mysql.php on line 299

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/gayland/public_html/internals/SK6_inc/class.mysql.php on line 50

 

Warning: MySQL error: Table 'gayland_gay.lang_key' doesn't exist

in query: SELECT `lang_key_id` FROM `lang_key` WHERE `lang_key_id` NOT IN (SELECT `lang_key_id` from `lang_value` WHERE `lang_id`=14 AND `value` != '')

in /home/gayland/public_html/internals/SK6_inc/class.mysql.php on line 299

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/gayland/public_html/internals/SK6_inc/class.mysql.php on line 50

 

Warning: MySQL error: Table 'gayland_gay.lang_key' doesn't exist

in query: SELECT `lang_key_id` FROM `lang_key` WHERE `lang_key_id` NOT IN (SELECT `lang_key_id` from `lang_value` WHERE `lang_id`=15 AND `value` != '')

in /home/gayland/public_html/internals/SK6_inc/class.mysql.php on line 299

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/gayland/public_html/internals/SK6_inc/class.mysql.php on line 50

 

Warning: MySQL error: Table 'gayland_gay.lang_key' doesn't exist

in query: SELECT `lang_key_id` FROM `lang_key` WHERE `lang_key_id` NOT IN (SELECT `lang_key_id` from `lang_value` WHERE `lang_id`=16 AND `value` != '')

in /home/gayland/public_html/internals/SK6_inc/class.mysql.php on line 299

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/gayland/public_html/internals/SK6_inc/class.mysql.php on line 50

 

Warning: MySQL error: Table 'gayland_gay.lang_key' doesn't exist

in query: SELECT `lang_key_id` FROM `lang_key` WHERE `lang_key_id` NOT IN (SELECT `lang_key_id` from `lang_value` WHERE `lang_id`=19 AND `value` != '')

in /home/gayland/public_html/internals/SK6_inc/class.mysql.php on line 299

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/gayland/public_html/internals/SK6_inc/class.mysql.php on line 50

 

[bh]

You dont understood me Change those field names and etc to constants that are correct at all.

[timvdalen]

Hmm i dont think this problem is in script: i installed thos script on my localhost (xampp) and all works fine but not on my webserver.  what could this be ?

Did you also move the database? Because you can't host the php files somewhere else if you don't change the address from localhost to your ip and open some ports

[Daimler]

ah no..the script is installed separated. on localhost and on webserver

[Daimler]

bh

 

sure i dont understood you i am a beginner.  thanks for being so pactient with me, but were can i see the correct constants? in the phpmyadmin tables ? or ?

[timvdalen]

bh

 

sure i dont understood you i am a beginner.  thanks for being so pactient with me, but were can i see the correct constants? in the phpmyadmin tables ? or ?

Your configuration is wrong. Localhost is your own computer.

You installed your site on a webserver, somewhere else. If you tell php to go look for a database on localhost, it will look on the pc that php is installed on, your webserver.

 

You have to put your database on the same pc as the webserver or edit the host to match your own pc (believe me, this can be very hard en frustrating the first few times, do the first thing)