upload file and rename

[anfetienne]

ive got an upload file form and it can have as many fields as a user wants to add images but say the user accidently pushes submit and the images upload (they are renamed as 1.jpg, 2.jpg, 3.jpg etc etc) but the user still wants to upload more images.....is it possible for php to read a directory and check what the filenames are and then renamed all extra images that are uploaded starting from the next digit?

so if there is already a 1.jpg and a 2.jpg once more pictures are uploaded it starts from 3.jpg and so forth....ive tried looking for something like this but dont know where to start.

thanks

[Markus]

Here's something to get you started:

readdir() - read to documentation!
You'll probably also need to explode() the file name to get the name minus it's extension.

Theory:

1. Loop through the directory's files using readdir(),
2. Determine whether you are at the end of the directory (I'll let you figure this out as a little brain-workout),
3. find the number in the files name,
4. increment the number to use as your next file name.

- mark.

[anfetienne]

ok thanx markus....once ive gotta tester ill come back for more advice......thanks again

[Markus]

@anfetienne
No problem, anfetienne.

- mark.

[anfetienne]

ok ive got to the point where i can open a directory and list all files in the directory and strip off the .jpg......i used a similar code to rename files and move them.....i haven't finished building it as I am restructure all my coding to suit this. but based on what you showed me the cide i need to use for starters is this

<?php
if ($handle = opendir('.')) {
    while (false !== ($file = readdir($handle))) {
        if ($file != "." && $file != "..") {
            echo "$file\n";
        }
    }
    closedir($handle);
}
?> 
 

from here i have no idea on how to get the php to detect what number the last uploaded image is on and then how to get that into a variable

[anfetienne]

my guess was to try get it into an array so its read to the last value and then add onto that but that attempt failed.....i was nowhere close to it lol

[Markus]

Assuming your files are in the format x.jpg (x being a number), you can explode() the file name using '.jpg' as your needle.

$file = "12.jpg";
 
$file = explode(".jpg", $file);
 
// $file is now an array.
// $file[0] holds the number 12
 

[anfetienne]

so to add the extra numbers ontop would i use?

$i=1;

$file[0].$i++

[Markus]

Something like below:

 
$num;
 
while( /* read directory */ )
{
    // Do explode.
    $num = $file[0];
}
 
// Loop has ended. Increment $num by 1 for new number.
++$num;
 
 

[anfetienne]

ahhhhhh i see what you are getting at......

so lastly could i use ++$num like this?

(rename($pathA.$filename, $pathB.++$num.'.jpg')

[Markus]

@anfetienne
Sure, but be careful: ++ increments the variable, so multiple calls to ++$num may confuse you on the output.

 
$i = 9;
 
echo ++$i; // ouput: 10.
echo ++$i; // output: 11.
 
 

[anfetienne]

would you recommend me using echo to make sure that ++$num outputs as it should?

[anfetienne]

this is the code i have made to test this....i add to images and it increments fine to show the number 3 but if i add anymore images it doesn't detect them and continue incrementing.

<?
$path="upload/test/";
$num;
 
if ($handle = opendir($path)) {
   while (false !== ($file = readdir($handle))) {
    if ($file != "." && $file != "..") {
 
 
            $file = explode(".jpg", $file);
            $num = $file[0];
 
       }
   }
 
}
closedir($handle);
 
++$num;
 
echo $num;
?>
 

[anfetienne]

any ideas where ive gone wrong anyone?

[Atli]

Hi.

Before you take this any further, let me offer an alternate solution (or two)...

First, if you want to name the files "1.jpg", "2.jpg", etc...
Rather than count the actual files, it would be much much easier to simply store the last used number in a file (or preferably, a database), so that you could just read that number and increment it.

The file_get_contents and file_put_contents can make that very easy.

Second...
Unless you actually have a good reason to do so, storing the images as "1.jpg", "2.jpg", etc... isn't all that efficient.
It requires that you go out of your way to count the images and make sure there are no collisions.

I would recommend using a timestamp instead, which would require minimal tampering with to ensure that there are no collisions.

PHP has the function time, or even better, microtime, which will give you the current UNIX timestamp in seconds or milliseconds, respectively.

So, to generate a unique name, you could simply do:

<?php
$name = microtime(true) . mt_rand(1,999) . ".jpg";
?>

Which would give you:
- 1239162720.1566.jpg

And even if you happen to have two requests processed at the very same millisecond, there is only 1:999 chance that you get a collision. (Which can easily be increased)

[anfetienne]

it has to be incremented as 1.jpg 2.jpg etc...

if using a database to store the last is an easier way to do this then i shall but how is it possible to store the last number into a database?

also how can i get readdir to only check for the last file within a directory?

[anfetienne]

ahhhh i guess i wouldn't need to get it to check the directory......if i get it to check the database against the corresponding ID and check the value that is there and go forth from there......

thanks for the idea atli....i will work on that now and get back to you with the results and my final code

[anfetienne]

i have worked on the coding and i got it to work the way i need it to using the database as suggested but i have a problem with the sql coding

i use this code to recall what i need it to from the database

"SELECT * FROM imgUpload WHERE tempID='{$random_digit}'"
 

and i get this error

Error in query:You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHERE tempID='1371669417'' at line 5

[Markus]

@anfetienne
What data type is tempID? If it's an integer, you cannot use quotes.

[anfetienne]

ok so i should set it to text.....thanks

[Markus]

@anfetienne
No, you should just remove the quotes from around the variable in your SQL.

[anfetienne]

ok so
"SELECT * FROM imgUpload WHERE tempID='{$random_digit}'"

becomes
"SELECT * FROM imgUpload WHERE tempID={$random_digit}"

[Markus]

@anfetienne
That's correct, anfetienne.

- mark.

[anfetienne]

thanks for the advice

[waqasahmed996]

Sir i think we can use quotes even if data type is interger.

we can write like this

SELECT * FROM `table` WHERE abc='1'
 

even if abc has interger data type

am i right?

[Dormilich]

that depends on the database driver. best practice is to quote it accordingly.

if you're using Prepared Statements the quoting is done internally.

[anfetienne]

well i have tried

"SELECT * FROM imgUpload WHERE tempID= '{$random_digit}' "

"SELECT * FROM imgUpload WHERE tempID={$random_digit}"

and

"SELECT * FROM imgUpload WHERE tempID= $random_digit "

and i still get the same error message. it has completely got me stumped

[Dormilich]

is there any error number given? you could look this up to get an idea what kind of error it is.

[Markus]

What is the table structure?

[anfetienne]

two columns .... tempID and lastDigit

it's really simple

[Dormilich]

what data type have these columns? (like VARCHAR, INT(5), …)

[anfetienne]

tempID INT(15), lastDigit INT(5)

[Dormilich]

has there been a MySQL Error Number?

[anfetienne]

it inserts data to the db perfectly fine but when it checks if there is already a row that has the tempID it comes up with that error message

[anfetienne]

no mysql error numbers just the message of "Error in query:You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHERE tempID='1371669417'' at line 5"

[Dormilich]

if you want to overwrite data, use the UPDATE keyword.

[anfetienne]

i have......

$query="
 
UPDATE imgUpload SET 
 
            lastDigit='{$i}', 
 
            WHERE tempID='$random_digit'";
 
}

[Dormilich]

does it work now?
...

[anfetienne]

its always been there from the beginning......i've been getting the error even though i have used this......the only thing i can think of is im using the wrong sql to update a table that has only two columns i.e.

"SELECT * FROM imgUpload WHERE tempID= '{$random_digit}' "

[Markus]

Strange strange.

For some reason, PHP does not show the error number associated with the error (security reasons, maybe?).

Anyway, if you have phpMyAdmin installed, run the same query through it's SQL interface. It will give you the error number (eg #1064) and the error text.

[anfetienne]

#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '"SELECT * FROM imgUpload WHERE tempID= '{$random_digit}' "' at line 1

[Dormilich]

1064 is a syntax error like parenthesis or quotation mark mismatch.

[anfetienne]

@anfetienne
as stated in a previous post of mine

[Markus]

@anfetienne
Can you show how you're doing this via PHP.

[anfetienne]

$username="tempsUser";
$password="auction";
$database="auctionTemps";
 
 
 
// OPEN CONNECTION ---> 
mysql_connect(localhost,$username,$password);
 
@mysql_select_db($database) or die( "Unable to select database");
 
// Select column 1 from table name where column name = $your_var.
 
// If mysql_query returns false, we'll die with the error.
$result = mysql_query("SELECT * FROM imgUpload WHERE tempID = {$random_digit}");
 
// If a there is a match
if ( mysql_num_rows( $result ) > 0 )
{
$row = mysql_fetch_array( $result );
 
//fill in details about the path
$path="upload/imgtemp/m/$random_digit/";
$pathA="upload/$random_digit/images/";
 
//variable used for the name of each file
$i=$row['lastDigit'];
 
$od = opendir($path);
while (false !== ($filename = readdir($od)))
{
    //you can add any type of filenames you wish to skip (for instance Thumbs.db on windows)
    if($filename != '.' && $filename != '..' && !is_dir($path.$filename))
    {
        //we give files a name - here we use increasing numbers for jpg files:
        if(rename($path.$filename, $pathA.$i++.'.jpg'))
        echo 'Renamed file '.$filename.'<br />';
    }
}
 
closedir($od);
 
 
//fill in details about the path
$pathT="upload/imgtemp/t/$random_digit/";
$pathF="upload/$random_digit/images/";
//variable used for the name of each file
$i=$row['lastDigit'];
 
$odT = opendir($pathT);
while (false !== ($filename = readdir($odT)))
{
    //you can add any type of filenames you wish to skip (for instance Thumbs.db on windows)
    if($filename != '.' && $filename != '..' && !is_dir($pathT.$filename))
    {
        //we give files a name - here we use increasing numbers for jpg files:
        if(rename($pathT.$filename, $pathF.$i++.'b.jpg'))
        echo 'Renamed file '.$filename.'<br />';
 
 
    }
}
closedir($odT);
 
 
 
$query="
 
UPDATE imgUpload SET 
 
            lastDigit='{$i}', 
 
            WHERE tempID= '{$random_digit}' ";
 
}
 
else
{
//fill in details about the path
$path="upload/imgtemp/m/$random_digit/";
$pathA="upload/$random_digit/images/";
//variable used for the name of each file
$nw=1;
 
$od = opendir($path);
while (false !== ($filename = readdir($od)))
{
    //you can add any type of filenames you wish to skip (for instance Thumbs.db on windows)
    if($filename != '.' && $filename != '..' && !is_dir($path.$filename))
    {
        //we give files a name - here we use increasing numbers for jpg files:
        if(rename($path.$filename, $pathA.$nw++.'.jpg'))
        echo 'Renamed file '.$filename.'<br />';
    }
}
 
closedir($od);
 
 
//fill in details about the path
$pathT="upload/imgtemp/t/$random_digit/";
$pathF="upload/$random_digit/images/";
//variable used for the name of each file
$nw=1;
 
$odT = opendir($pathT);
while (false !== ($filename = readdir($odT)))
{
    //you can add any type of filenames you wish to skip (for instance Thumbs.db on windows)
    if($filename != '.' && $filename != '..' && !is_dir($pathT.$filename))
    {
        //we give files a name - here we use increasing numbers for jpg files:
        if(rename($pathT.$filename, $pathF.$nw++.'b.jpg'))
        echo 'Renamed file '.$filename.'<br />';
    }
}
closedir($odT);
 
$query="INSERT imgUpload (tempID,lastDigit)
 
        VALUES(    
            '$random_digit',
            '$nw')";
 
}
 
//////-----> 
$result=mysql_query($query) or die("Error in query:".mysql_error()); 
//if ($result) 
    //echo mysql_affected_rows()." row inserted into the database effectively."; 
 
//  CLOSE CONNECTION ---> 
mysql_close(); 
 
/////////////////////////////////////////////////////////////////////////////////// 
 

[Markus]

There shouldn't be a comma on line 71 :D

Is that what is giving the error, or the query at the beginning of the file?

[anfetienne]

just like to say thank you for the help......always appreciated

[Markus]

@anfetienne
You have returned.

You're welcome.

- mark.

[anfetienne]

yeah not been too well.....only just recovered.....thanks again though