ajax php mysql search problem

[Vizonz]

found a little ajax script to display search results so i changed it up a little  but when searching getting an error. anyone give me a hand finding out why..

 

 

$qry = "SELECT  DISTINCT artist,label,genre,composer FROM songlist GROUP BY artist ORDER BY
  songlist.date_added";
$searchText = "";
if($_REQUEST['search_text']!=""){
$searchText = $_REQUEST['search_text'];
$qry .=" where artist like '$searchText%'";

 

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'where artist like 'Artist%'' at line 2

 

 

if you need more info i will post more  but maybe its just a error in that above code  didnt know where to post this in ajax mysql or php but it seems more a mysql problem then any of the others

[radar]

Have you tried echoing out your $qry and running that through phpmyadmin to see what it says?  Whenever I have an SQL issue, thats the first thing I do, as even if i cant figure it out it gives me some further diagnostic information to post here along with the query itself.

[Vizonz]

the query works its displaying all the results. but when i use the search box  to find a specific thing thats where i get the error

[radar]

My query is a bit different than yours, but it works for my search.  note that i use the smarty template engine, and smarty paginate...  but perhaps this will help you on your way too.

 

$sql = "SELECT SQL_CALC_FOUND_ROWS * FROM employs WHERE l_name LIKE '".$_pid."%' LIMIT ".
								SmartyPaginate::getCurrentIndex().",".SmartyPaginate::getLimit();

 

perhaps its something so simple as making WHERE and LIKE capitals...  or as a test you could get rid of the % and see if that works...  if it doesnt work then try it like '".$searchText."%' and if that doesn't work i have no clue ;(

[Vizonz]

thanks  for trying   tried all of the above and no result

[Vizonz]

Anyone this is kinda important its set up and running besides the search. The results are in ajax pagination so thats working  just the search dont function

[PFMaBiSmAd]

You are building the query incorrectly. The specific parts of the query must be in a certain order - http://www.phpfreaks.com/forums/index.php/topic,305028.msg1442575.html#msg1442575

[dezkit]

$qry = "SELECT  DISTINCT artist,label,genre,composer FROM songlist ";
$searchText = "";
if($_REQUEST['search_text']!=""){
$searchText = $_REQUEST['search_text'];
$qry .=" WHERE artist LIKE '$searchText%'";
}
$qry .=" GROUP BY artist ORDER BY songlist.date_added";

try this

[Vizonz]

thanks so much both of you   that led to it working  

[radar]

I should have realized that fix the first go around...  the where clause has to go directly after the table  your selecting as it is now..  bah!