Question

<?php

$top = "../top.txt";
$middle = "../middle.txt";
$bottom = "../bottom.txt";
$end = "/st.txt";
$data = "/dt.txt";

$handle1 = fopen($top, "r"); 
$contents1 = fread($handle1, filesize($top)); 
fclose($handle1); 

$handle2 = fopen($end, "r"); 
$contents2 = fread($handle2, filesize($end)); 
fclose($handle2);

$handle3 = fopen($middle, "r"); 
$contents3 = fread($handle3, filesize($middle)); 
fclose($handle3);

$handle4 = fopen($data, "r"); 
$contents4 = fread($handle4, filesize($data)); 
fclose($handle4);

$handle5 = fopen($bottom, "r"); 
$contents5 = fread($handle5, filesize($bottom)); 
fclose($handle5);

echo $contents1;
echo $contents2;
echo $contents3;
echo $contents4;
echo $contents5;

?>

I get these errors for every one of them:

Warning: fopen(../top.txt) [function.fopen]: failed to open stream: No such file or directory

Warning: filesize() [function.filesize]: stat failed for ../top.txt

Warning: fread(): supplied argument is not a valid stream resource

Warning: fclose(): supplied argument is not a valid stream resource

CHMOD in all files and folders and is set to 777

All the files exist on the server

PHP5 is installed on the server

What am i doing wrong?

Answer 1

$top = "../top.txt";

means that from your current position it moves one folder up, and searches for the file top.txt. It’s same for $middle = "../middle.txt"; and $bottom = "../bottom.txt";. For these two:

$end = "/st.txt";
$data = "/dt.txt"

it’s searching from the root directory. Do make sure you have the files located at the places you refer them to. I may think you wanted this in the last two variables:

$end = "./st.txt";
$data = "./dt.txt"

Where ./ represents the current directory.

So, to give you more detailed example. Lets say you have your PHP file located at /var/www/httpdocs/project/phpFile.php then your files would be linked like:

$top = "/var/www/httpdocs/top.txt";
$middle = "/var/www/httpdocs/middle.txt";
$bottom = "/var/www/httpdocs/bottom.txt";

$end = "/st.txt";
$data = "/dt.txt"

So now notice the difference in the last two variables – they stay the same, because you’ve set it to be looking from root directory (/). Where as if you would use the linking with ./, then the paths would be:

$end = "/var/www/httpdocs/project/st.txt";
$data = "/var/www/httpdocs/project/dt.txt"

Hope this explains it. And yes, do look into the functions other people mentioned in the answers.