filename variable

[nadirah81]

Hi...

Is this possible??? I have created a file, e.g., 1.txt using fopen(). The next time a file is been created, I wanted to create 2.txt and it follows with 3.txt and so on....

I know I need to use variable, something like that. How should I do it???

[Shrike]

You could parse the filename to figure out which number is next. Check out basename().

[nadirah81]

This is what i have done but something is not right. I have done it in two ways so please take a look at it. I am not good in php so, could you make corrections to it? Which one is better?

Thanks.

P.S.
Just to tell you that I am trying to create a text file in a folder called input. The filename is automatically created using variable, e.g., if there are no input, it will create 1.txt, if there are 1.txt, 2.txt and 3.txt, it will create 4.txt and so on.

First:

<?php
//select text file in the Input directory
$path = "C:\Input\";
$folder = dirname ($path);
$file = basename ($path, ".txt");

//create file

$array = array
(if (i=>1)
{
$fp = fopen($folder.$file.$path, "w");
$string = "$qseq";
$write = fputs($fp, $string);
fclose($fp);
i++
}
)
?>

Second:

<?php
$dir = "C:\Input";
$dh = opendir ($dir);
while (false !== ($filename = readdir ($dh)))
{
$files = array_sum ($filename);
}
sort ($files);
count ($files);

if ($files = i)
//create a text file with filename (i+1).txt

$filename = '(i+1).txt';
$fp = fopen($filename, "w");
$string = "$qseq";
$write = fputs($fp, $string);
fclose($fp)

?>

[PallaviDalvi]

i have a code snippet that returns the next file number.

$handle=opendir("C:/Input/");
$new_name=0;
while ($contents=readdir($handle))
{
	if (is_file($contents))  //we need to work only if its a file
	{
		$info=pathinfo($contents);
		if ($info['extension'] == "txt")  //check ifits a text file...we can omit all other file types.
		{
			$flname=basename($contents,".txt");  // get only the name...omit the extension
			if ($new_name < intval($flname) )  // store the last file (the number is the greatest)
			{
				$new_name = $flname;
			}

	}
}
}
echo ++$new_name; // the new file name wud be $new_name.txt

hope this helps u

however in ur code-... i didnt see the second approach but in the first one...
$fp = fopen($folder.$file.$path, "w");
creates a path like C:\InputC:/Input/
moreover i wud suggest u use
$path = "C:/Input / ";
as ur approach created probs with me....maybe u have no probs.

[nadirah81]

Thanks for your help. I have used the functions that you gave and i have made some changes as i do not want and "echo" to the file name. Here are the whole functions:

<?php
$handle=opendir("C:\Input\");
$new_name=0;
while ($contents=readdir($handle))
{
if (is_file($contents)) //we need to work only if its a file
{
$info=pathinfo($contents);
if ($info['extension'] == "txt") //check if it is a text file...we can omit all other file types.
{
$flname=basename($contents,".txt"); // get only the name...omit the extension
if ($new_name < intval($flname) ) // store the last file (the number is the greatest)
{
$new_name = $flname;
++$new_name;
}

    } 
} 

}
$fp = fopen($flname, "a");
$string = "$qseq";
$write = fputs($fp, $string);
fclose($fp);
?>

But when I execute it, I've got an error message:
Parse error: parse error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in filename.php on line 9

FYI, the register_global=Off in my php.ini file. I am using php 4.3.3.

[PallaviDalvi]

dont know exactly where the error is occuring(not sure of where line 9 is) but i did spot some thing...

$fp = fopen($flname, "a"); //$flname=4; assume

shud be

$new_name .= ".txt" ; //$new_name="5.txt"; the name of the file
$fp = fopen($new_name, "a");

remember...all the code on top..... that creates $new_name which is the new name for the file... u need to append an extension to the file number and then open that file.

get back with a pointer towards line no 9 in the code. maybe things become clearer to me.

[nadirah81]

This is line no. 9:

if ($info['extension'] == "txt")

Another thing that comes to my mind is... is it something to do with $_REQUEST? because this php file is actually a process that was taken from a data in a html form. the data is "qseq", stated in the php file.

[PallaviDalvi]

ok... i suggest u echo the value of $info['extension'] and see whats wrong. coz at my end this peice of code worked perfectly fine. so i think there is some prob with the values being used at ur end. $_REQUEST has nothing to do here ... coz irrespective of whether the form is posted or not... the file name is generated from files on the disk. so the part of code posted by me has to work. unless ofcourse there is a prob with the values picked up from the directory. try printing out the values and u can see for urself.
get back if there are any further probs.

[nadirah81]

I have used the snippet that you gave me but, I have changed this:

while ($contents=readdir($handle))

to this:

while (false !== ($file = readdir($handle)))
and also I added the closedir() function at the end.

and now, the parse error is gone. But, now i got errors stating that

Warning: fopen(0.txt): failed to open stream: Permission denied in c:\inetpub\wwwroot\openclosedir.php on line 21

which is this line:
$fp = fopen($new_file, "a");

Warning: fputs(): supplied argument is not a valid stream resource in c:\inetpub\wwwroot\openclosedir.php on line 23

which is this line:
$write = fputs($fp, $string);

and,

Warning: fclose(): supplied argument is not a valid stream resource in c:\inetpub\wwwroot\openclosedir.php on line 24

which is this line:
fclose($fp);

I guess something is wrong when it tries to execute the fopen().
Any idea how I could make this work??

This is the whole part of it...

$new_file = "$new_name.txt";
$fp = fopen($new_file, "a");
$string = "$qseq";
$write = fputs($fp, $string);
fclose($fp);

[PallaviDalvi]

Warning: fopen(0.txt): failed to open stream: Permission denied in c:\inetpub\wwwroot\openclosedir.php on line 21

Check the permissions on the directory.

[nadirah81]

hahaha... it's working....!!!!!😉 thanks a lot for your help. Thank you.

[nadirah81]

hiii....

I am not so sure what i have done, or ......?! it did not work anymore.

I tried copying the one that u gave me (as above), but it still wouldn't work. If there is no file in the input folder, it will create one, 0.txt. Then, when the next time I try to post another input, i just overwrites whatever in the existing file.

Pls help!!!!!

I do not know what is going on.