find and Exec statment in PHP not working

[ajd335]

Hi All,
I have a page in which user can enter first name,last name , date and according to that My script should search for the records which matches the above last n first names and print it.
I have the below code,
But seems not working..

<?php
$title = " Form";
echo "Hello";
 if(isset($_POST['fname'])) { $f1 =  $_POST['fname']; }
 
if (isset($_POST['lname'])) { $ll =  $_POST['lname']; }
if (isset($_POST['add'])) { $a1 =  $_POST['add']; }
if (isset($_POST['date'])) { $d1 =  $_POST['date']; }
 
$fullname = $l1.","f1;
echo "$fullname;
$dirname = "/home/gr/ereports/O1"  ;
 
$handle = opendir($dirname) or print "cant open";
 
$ans = shell_exec("find /home/gr/ereports/O1/$d1 -name \*.ps  -exec grep -i -e "$fullname" {} \;  -print; ");
 
?>
 
 

If i go on shell to home/gr/ereports/O1 and then write the code as below

find dirname -name "*.ps" -exec grep -i -e "CAK, SO" {} \; -print ;
 

Here in above "CAK , SO" in which CAK is the lastname n SO is the first name
its giving me the result perfectly.
I am thinking that the way I used the above statment in PHP is wrong..
Can anybody help me...
Thanks

[Atli]

Hi.

There are a lot of syntax errors in that code.
Try turning on the debug messages and fix the errors. See if that helps.

[ajd335]

Hi..
Oh m sorry..have posted that with making some changes..And thus , it has so many syntax errors..
When I execute the below on shell it works fine...

find /home/gr/ereports/O1/2008-06-17 -name "*.ps" -exec grep -i -e "LISA" {} \; -print ;
 

And in PHP i am writing

$ans = shell_exec("find /home/gr/ereports/O1/$d1 -name /"/*.ps/"  -exec grep -i -e $l1  {} \;  -print; ");
 

How can I write "*.ps" in shell_exec ..here I have used /"/*.ps/" back slashes..

[pbmods]

Heya, AJD.

Use these slashes: \ to escape quotes.