Retrieving full path of file on upload to php script

How do I retrieve the full path (C:\.....filename) of a file uploaded to a p
hp script on the server.
I think  $_FILES will only provide 'name', which is only the filename itself
.

Thanks, Don


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$path=$_SERVER["DOCUMENT_ROOT"];

"Don" <no@adr.com> schreef in bericht
 news:tr8kn0huoh1lk1uq54aqefjr6ll005vgt1@
4ax.com...
> How do I retrieve the full path (C:\.....filename) of a file uploaded to a
> php script on the server.
> I think  $_FILES will only provide 'name', which is only the filename
> itself.
>
> Thanks, Don
>
>
> ----== Posted via mcse.ms - Unlimited-Uncensored-Secure Usenet
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On Sat, 23 Oct 2004 12:27:19 +0200, "Polaris" <ask4polaris@hotmail.com> wrot
e:

>$path=$_SERVER["DOCUMENT_ROOT"];
>
>"Don" <no@adr.com> schreef in bericht
> news:tr8kn0huoh1lk1uq54aqefjr6ll005vgt1@
4ax.com... 

Thanks for your reply.

Following in client-side <form>
------------------------------------
<p>Picture #1:<br><input type=file name="userfile[]" size=90 maxlength=20000
0 accept="image/jpeg"
tabindex=1>\n';

<p>Picture #2:<br><input type=file name="userfile[]" size=90 maxlength=20000
0 accept="image/jpeg"
tabindex=3>\n';


Tried the following in server-side php script, but it provided the path on t
he server where the file
was uploaded.  What I need is the full path where the file resided on the cl
ient machine.  I need it
for each of the two uploads.  Any ideas?
----------------------------------------------------------------------------
----------------------------------------
print("Document root: = ".$_SERVER['document_root']."\n");



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Try this one. Does it solve your problem?

Save as index.php

<?php

if (isset($_POST["file1"])) $file1=$_POST["file1"];
if (isset($_POST["file2"])) $file2=$_POST["file2"];

print "File 1 = " . $file1;
print "<br />File 2 = " . $file2;


print '<form id="UploadForm" action="index.php" method="post">
<div>
<h1>Select your files</h1>
<strong>1</strong>.      <input type="file" name="file1" size="70"
style="margin-bottom: 3px"/>
<br />
<strong>2</strong>.      <input type="file" name="file2" size="70"
style="margin-bottom: 3px"/>
<br />
<input type=submit name=up value="Upload">
</div>
</form>';

?>

"Don" <no@adr.com> schreef in bericht
 news:17skn01fs2udikt14ac7i822vute9mrkc6@
4ax.com...
> On Sat, 23 Oct 2004 12:27:19 +0200, "Polaris" <ask4polaris@hotmail.com>
> wrote:
> 
>
> Thanks for your reply.
>
> Following in client-side <form>
> ------------------------------------
> <p>Picture #1:<br><input type=file name="userfile[]" size=90
> maxlength=200000 accept="image/jpeg"
> tabindex=1>\n';
>
> <p>Picture #2:<br><input type=file name="userfile[]" size=90
> maxlength=200000 accept="image/jpeg"
> tabindex=3>\n';
>
>
> Tried the following in server-side php script, but it provided the path on
> the server where the file
> was uploaded.  What I need is the full path where the file resided on the
> client machine.  I need it
> for each of the two uploads.  Any ideas?
> --------------------------------------------------------------------------
------------------------------------------
> print("Document root: = ".$_SERVER['document_root']."\n");
>
>
>
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> News==----
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Tried it, but $file1 and $file2 are returned empty.  I think maybe I need to
 be using
$GLOBALS['HTTP_RAW_POST_DATA'] or php://input.  But, I can't seem to get the
m to work either.  Any
ideas?

Thanks, Don


On Sat, 23 Oct 2004 21:01:19 +0200, "Polaris" <ask4polaris@hotmail.com> wrot
e:

>Try this one. Does it solve your problem?
>
>Save as index.php
>
><?php
>
>if (isset($_POST["file1"])) $file1=$_POST["file1"];
>if (isset($_POST["file2"])) $file2=$_POST["file2"];
>
>print "File 1 = " . $file1;
>print "<br />File 2 = " . $file2;
>
>
>print '<form id="UploadForm" action="index.php" method="post">
>    <div>
>      <h1>Select your files</h1>
>      <strong>1</strong>.      <input type="file" name="file1" size="70"
>style="margin-bottom: 3px"/>
>      <br />
>      <strong>2</strong>.      <input type="file" name="file2" size="70"
>style="margin-bottom: 3px"/>
>      <br />
>      <input type=submit name=up value="Upload">
>    </div>
></form>';
>
>?>
>
>"Don" <no@adr.com> schreef in bericht
> news:17skn01fs2udikt14ac7i822vute9mrkc6@
4ax.com... 
>



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"Don" <no@adr.com> schreef in bericht
 news:bvoln05dbb573qkjsrhf91l7n0d9rvbfot@
4ax.com...
> Tried it, but $file1 and $file2 are returned empty.  I think maybe I need
> to be using
> $GLOBALS['HTTP_RAW_POST_DATA'] or php://input.  But, I can't seem to get
> them to work either.  Any
> ideas?
>
> Thanks, Don
>
>
> On Sat, 23 Oct 2004 21:01:19 +0200, "Polaris" <ask4polaris@hotmail.com>
> wrote:
> 

Strange when I run the script here at my local apache server and I fill in
the form $file1 and $file2 do have returning values... I gives complete path
with filename something like K:\\www\\filename.ext Or maybe I don't
understand your problem clearly???

Read up on the $_FILES scope and handling file uploads here >>
(http://us2.php.net/features.file-upload)

This provides alot more information than you think
You can get the client filename, the temp file, as well as size and any
upload errors.

Using the basename() function, you can also trim down the filename to get
the name and extension as well.

V/r,
Charles

"Polaris" <ask4polaris@hotmail.com> wrote in message
news:417b74a1$0$44103$5fc3050@dreader2.news.tiscali.nl...
>
> "Don" <no@adr.com> schreef in bericht
>  news:bvoln05dbb573qkjsrhf91l7n0d9rvbfot@
4ax.com... 
need 
>
> Strange when I run the script here at my local apache server and I fill in
> the form $file1 and $file2 do have returning values... I gives complete
path
> with filename something like K:\\www\\filename.ext Or maybe I don't
> understand your problem clearly???
>
>