Upload an image to a folder, via mysql

Anon

Hello.

How do we upload an image via Php and Mysql, "but" upload it to a directory,
and only manage and keep its location [like folders/images/one.gif] or
[http://www.domain.com/images/one.gif] in the mysql database and then be
able to call back the location of the url again when needed so it can be
displayed again on a php page as we wish.

In short, a way to upload a file to a folder and record its url in a mysql database.

Thanks.

Winnie

[deleted]

SO what's the problem exactly?

Anon

Hello.

At this moment I only know how to upload the whole binary image into the mysql database, and then call it there.

I want it to be uploaded it to a directory via a form that an admin can us, and only the location of that image be stores in the mysql database and not the image itself.

Thanks.

vincent wrote:

SO what's the problem exactly?

[deleted]

There seems to be this misconception that a file upload puts the file in memory somewhere then it goes by magic to the sql database.

What happens is the file is uploaded to the /tmp directory (unless you changed the defaults), then the php script gives you variables that start with the name you specified in the form.

For example, if you said "<input type="file" name="blah"> you will have:

$blah which will be equal to the file in the /tmp directory

$blah_name which will be equal to the name of the file as it was on the PC.

So you could create a unique ID for that file (the way you create the ID is your call), then copy($blah, /newdir/newsubdir/$newid), insert this ID to your sql database, possibly with the name of the file that was stored on the user pc ($blah_name), and now you have:

1) the sql record that contains an ID and the file name.
2) the picture or file in the directory you want.

Anon

The suggestion you received is correct.
To facilitate the users that enter the data, I have used a text box with a link "Upload" next to it;this opens a window for your upload script, which should output a variable with the url that the photo was saved.
Then with a javascript like:

the url of the photo will be directly copied to the textbox.

Anon

Hello.

I am using the following piece of code to upload the current file:

"
<body>
<form method="post" action="upload.php" enctype="multipart/form-data">
<input name="userfile[]" type="file">

<input type="submit" value="Upload!!!" > 

</form>

</body>
"

and the following php code to upload the files into a directory "file" in the folder that upload script[below] and html files are in:

<?php
for($i=0;$i<sizeof($userfile);$i++)
{
if(!$userfile_size[$i])
continue;

$UPLOAD = fopen( $userfile[$i], "r" ); 
$contents = fread( $UPLOAD,$userfile_size[$i]);
fclose( $UPLOAD );
$SAVEFILE = fopen( "file//".$userfile_name[$i], "wb" ); 
fwrite( $SAVEFILE, $contents,$userfile_size[$i] );
fclose( $SAVEFILE );
} 
echo "Server HaD Receive the Upload Files!"; 
?>

Now, where do I embed the javascript? I am lost here, since I tried it on the php script, and the html file itself and it didn't work but gives a javascript error.

I am not good in javascript, so maybe that's one reason.

Any ideas?

Anon

The javascript should reflect the names of the initial data form (i suppose the main form has other data fields also),
dataentryform should be the name of your form where the text box with the url should be copied and photo is the name of this same field.

You should use the javascript in the output of the upload script and replace the variable to print with the variable that outputs the location of the file the way you want.