variable inside a like?

[Manat]

if someone types in a name in a search field, I would like them to go to a page with more info on that name. The below doesn't work, any idea why?

  

  $iName=$HTTP_GET_VARS['iName'];
  $strSQL = "SELECT * FROM tableName WHERE Name LIKE '%$iName%';

Also, some sites allow you to, if you type in a wrong name it gives suggestions, i'm sure most sites dont code this but rather has software that does it, but is there a simple way in php for this?

If not, then what's wrong with the above code, how do I put a $var inside %%.

thx

[LordShryku]

You never put and ending quote on your SQL string. Here's what I usually use

$strSQL = "SELECT * 
           FROM   tableName 
           WHERE  Name LIKE '%".$iName."%'";

[root1590]

just do this

select * from tblHealth where hltp_subject like '%$nRid%' or hltp_detail like '%$nRid%'";

where "nRid" is posted to this file and i use with only $sign

[Manat]

Sorry,

what is hltp_detail ?

Thanks