error ...it doesn't recognise code ending - php

Question

<?php
$file_handle=fopen("normal.txt","rb");
$i=0;
while (!feof($file_handle)){
$fline=gets($file_handle);
$fparts=preg_split("/:|;\");
$x_values[$i]=(float)$parts[1];
$y_values[$i]=(float)$parts[2];
$i=$i+1;
}
include 'libchart/libchart/classes/libchart.php';
$data_length=sizeof($x_values);
for(i=0;$i< $data_length; $i++)
$dataset -> addPoint(newPoint('',$y_values[$i]));
$chart -> render('figs/text.png');
?>
//this part is not recognised...so it is thhrowing an error.
I think the rest of the code is ok....but that thing keeps throwing that error at the end of my code.I just don't know why...could please someone help me? I'd appreciate that much .That is the only error i find....if there are others too please tell me...i am just a beginner at website designing...
So your help is precious to me

Answer 1

You have delimted your closing quote in this line:
preg_split("/:|;\");
You should put in a second \ to counter it:
preg_split("/:|;\\", $string);
Normally when you put a \ before a quotation, it means "treat the quotation as part of the string". As you don't want it to mean that you need to delimit the delimiter heh. The compiler thinks that the rest of your code is still part of the string - even the syntax highlight here on SO is showing it.
See the difference:
$fline=gets($file_handle);
$fparts=preg_split("/:|;\\", $string);
$x_values[$i]=(float)$parts[1];
$y_values[$i]=(float)$parts[2];
$i=$i+1;
Edit: The elephant in the room was pointed out by KarelG. Good show. preg_split does indeed need to params. Without the extra \ it would have still not worked though :)