squeezing file into variable...

[Anon]

Greetings!
I am sorry to repost this message, but I've goofed... I should sent it into coding help first, but I have made a mistake.. Sorry...

I have one minor question, but I don't know, how to make it go thru...

What is what I want o actually do is t put the entire source of the file into one variable, something like:
$variable=readfile("/path/file.txt"), but that's a FAR way from that it could work, as readfile opens the file WITHOUT sending it to variable...

I wish for something like
$variable=readfile("/path/file.txt");
mysql_query="insert int database '$var1', '$var2', '$var3', '$variable');

Is there anyway to actually DO that?

Please help.

Best regards.
zer0Kode, Linux developer.

[dstockto]

I think that is the wrong function to do what you want to do. According to the PHP manual, the readfile function reads the file and outputs to standard output. The value returned is the number of bytes read from the file. So, your $variable is probably an int containing the filesize of the file, and you are probably getting the contents of the file output to the screen.

http://www.php.net/manual/en/function.readfile.php

I would suggest taking a look into fopen() and fread() which look like they will do more what you are trying to do.

http://www.php.net/manual/en/function.fopen.php

http://www.php.net/manual/en/function.fread.php

[Anon]

$file_array = file('filename');

which puts it into an array, however if you want it into a variable use -

$file_data = join('',file('filename'));