Array Prolems

Christian_F

Hi folks,

i have a problem with an array.

First i list out all files wich are in a directory called data and save them into an array like this:

$count = 0;
$ordner = opendir("data");

while($datei = readdir($ordner))
{

if(! is_file("data/$datei"))

continue;

$array[$count] = $datei;
$count++;

}

closedir($ordner);

foreach($array as $wert)
{
print "$wert<p>";
}

It works Fine.

But now i want to open all files with fopen and i wnat to give the $filename the value of ech filenames wich are in the directory. I think the $array isnt fitting to the $filename cause i get an error of non existing file and an endless loop at the fopen point.

here is my try:

while($count > 0)
{

 $filename = $array[$count];

 $fp = fopen ($filename, "r");
 $inhalt = fread ($fp, filesize ($filename));
 fclose ($fp);


   echo "<HTML><HEAD></HEAD><BODY>";

   $data = explode (" : ", $inhalt);

   for ($i = 0; $i >= 0; $i++) {
   echo "$data[$i]<br>";
   }

   echo "<HTML><HEAD></HEAD><BODY>";

   $count--;

}

error: warning supplied argument is not a valid stream source

im thankfull 4 any help !

PS: sorry 4 bad english, im french

Weedpacket

A couple of things: The most important is that you're reading the filenames from the 'data' directory, but trying to open the files in the current directory.

.
|
+-data
| |
| +-file1
| +-file2
| +-you're getting the filenames from here
|
+-but you're trying to open them here

In short, you should be trying to open "data/$filename".

The other points are stylistic: [man]file_get_contents[/man] may be more straightforward; using a [man]for[/man] loop would make the code simpler than trying to simulate one with a while statement; and if there is doubt about the content of a variable (including if that variable is an array), [man]print_r[/man] or [man]var_dump[/man] can give information about its content.

Christian_F

thnx for your help, i really didnt saw that i tried to read files in wrong dir.

but its still not working....

can some one explain to me, how i can fill in the $array[] into the $filename for each filename wich in in the array?

for ($i = 0; $i >= $filenamenmbr; $i++) {

$filename = $array[$filenamenmbr];

}

this didnt seems to be right

thnx again

Weedpacket

 for ($i = 0; $i < $filenamenmbr; $i++) {

$filename = $array[$i];

}

Christian_F

thank you very much 4 your serious help weedpacket !

i made a little workaround and changed everything a bit

auslesen.php:

<?php

$count = 0;
$ordner = opendir("data");

while($datei = readdir($ordner))
{

if(! is_file("data/$datei"))

continue;



$array[$count] = $datei;
$count++;

}

closedir($ordner);

foreach($array as $wert)
{
print "$wert<p>";
}



include 'data/lesen.php';

?>

lesen.php:

<?php
for ($e = 0; $e < $count; $e++) { 


$filename = $array[$e]; 

$fp = fopen ($filename, "r");
$inhalt = fread ($fp, filesize ($filename));
fclose ($fp);


   echo "<HTML><HEAD></HEAD><BODY>";

   $data = explode (" : ", $inhalt);

   for ($i = 0; $i <= 3; $i++) {
   echo "$data[$i]<br>";
   }

   echo "<HTML><HEAD></HEAD><BODY>";


}
?>

Browser output:

lesen.php
testuser.txt

dsfdfs.txt

", $inhalt); for ($i = 0; $i >= 0; $i++) { echo "$data[$i]
"; } echo ""; ?>

Warning: fopen(testuser.txt): failed to open stream: No such file or directory in /srv/www/htdocs/test/data/lesen.php on line 7

Warning: filesize(): Stat failed for testuser.txt (errno=2 - No such file or directory) in /srv/www/htdocs/test/data/lesen.php on line 8

Warning: fread(): supplied argument is not a valid stream resource in /srv/www/htdocs/test/data/lesen.php on line 8

Warning: fclose(): supplied argument is not a valid stream resource in /srv/www/htdocs/test/data/lesen.php on line 9

Warning: fopen(dsfdfs.txt): failed to open stream: No such file or directory in /srv/www/htdocs/test/data/lesen.php on line 7

Warning: filesize(): Stat failed for dsfdfs.txt (errno=2 - No such file or directory) in /srv/www/htdocs/test/data/lesen.php on line 8

Warning: fread(): supplied argument is not a valid stream resource in /srv/www/htdocs/test/data/lesen.php on line 8

Warning: fclose(): supplied argument is not a valid stream resource in /srv/www/htdocs/test/data/lesen.php on line 9

Its so nice that you help me, cause i just started this week with php, and the language realy makes me intereted in.

I really cant see the failure by now

MFG

Chris

alvinkoh

Hi Chris,

The error is on line 7.

You need to include the full path of the file when using fopen.
In your case, it's something like:

fopen("$your_directory/$filename","r");

The path can also be absolute, e.g.
fopen("$_SERVER[DOCUMENT_ROOT]/$your_directory/$filename");

An alternative is to specify the working directory of your script to recurse into the "data1" folder, e.g. chdir()

chdir("$your_directory");

Cheers.