[SOLVED] SQL syntax error

[Armantas]

Hello, i`m not good at english but i will try to explain my problem.
When i trying to submit new story on step 3 i get this error:


How can i fix it?

Sincerily,
Armantas

[AnAlienHolakres3]

I suppose you did some customization didnt you? If yes, backup your customized file and repalce with the original, then refresh page and see effect. I usually got this error when link_id was 0 or empty string.

If you didnt make any changes with original pligg check links table,check auto_increment value etc. If anything is ok...then....I dont know

[Armantas]

I didnt make any changes. I installed pligg 9.7, login and try to submit news but i get this syntax error :/

[AshDigg]

Please provide us with more information so we can help you.

Are you (or your webhost) using Linux or Windows for the server.
What PHP / MySQL versions are you using?

[Armantas]

This is my host information:

OS: Linux Debian 4.0 (Etch)
PHP: 5.2.0
MySQL: 5.0.45

[AshDigg]

in /libs/links.php look around line 150 for

 $db->query("UPDATE " . table_links . " set link_summary='$link_summary', link_title_url='$link_title_url', link_url='$link_url', link_url_title='$link_url_title', link_title='$link_title', link_content='$link_content', link_tags='$link_tags', link_field1='$link_field1', link_field2='$link_field2', link_field3='$link_field3', link_field4='$link_field4', link_field5='$link_field5', link_field6='$link_field6', link_field7='$link_field7', link_field8='$link_field8', link_field9='$link_field9', link_field10='$link_field10', link_field11='$link_field11', link_field12='$link_field12', link_field13='$link_field13', link_field14='$link_field14', link_field15='$link_field15' WHERE link_id=$this->id"); 
  

replace with

 $sql = "UPDATE " . table_links . " set link_summary='$link_summary', link_title_url='$link_title_url', link_url='$link_url', link_url_title='$link_url_title', link_title='$link_title', link_content='$link_content', link_tags='$link_tags', link_field1='$link_field1', link_field2='$link_field2', link_field3='$link_field3', link_field4='$link_field4', link_field5='$link_field5', link_field6='$link_field6', link_field7='$link_field7', link_field8='$link_field8', link_field9='$link_field9', link_field10='$link_field10', link_field11='$link_field11', link_field12='$link_field12', link_field13='$link_field13', link_field14='$link_field14', link_field15='$link_field15' WHERE link_id=$this->id";
echo $sql;
$db->query($sql); 
  

When you get to step 3, you'll see the MySQL statement. Copy and paste that for us to see.

thanks

[Armantas]

This is what i get:

[AshDigg]

I'm sorry, that was the wrong query!

Around line 178 - 179 you'll see this...

             $sql = "UPDATE " . table_links . " set `link_reports`=$link_reports, `link_comments`=$link_comments, link_author=$link_author, link_status='$link_status', link_randkey=$link_randkey, link_category=$link_category, link_modified=NULL, link_date=FROM_UNIXTIME($link_date), link_published_date=FROM_UNIXTIME($link_published_date), link_votes=$link_votes, link_karma=$link_karma WHERE link_id=$this->id";
            $db->query($sql); 
  

just put a

 echo $sql; 
  

in between those two lines and paste the results.

thanks!

[Armantas]

Result:

[AshDigg]

Thanks.

Try replacing your /libs/link.php file with this one. If it doesn't work, change line 54 to 'true' and post the results.

thanks