PHP Language - Retrieving full path of file on upload to php script

Author

Retrieving full path of file on upload to php script

Don

2004-10-23, 8:55 am

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

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Polaris

2004-10-23, 8:55 am

$path=$_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 Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet
> News==----
> http://www.newsfeeds.com The #1 Newsgroup Service in the World! >100,000
> Newsgroups
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Don

2004-10-23, 3:55 pm

On Sat, 23 Oct 2004 12:27:19 +0200, "Polaris" <ask4polaris@hotmail.com> wrote:
[color=darkred]
>$path=$_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=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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Polaris

2004-10-23, 3:55 pm

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");
>
>
>
> ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet
> News==----
> http://www.newsfeeds.com The #1 Newsgroup Service in the World! >100,000
> Newsgroups
> ---= East/West-Coast Server Farms - Total Privacy via Encryption =---

Don

2004-10-23, 8:55 pm

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:

>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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Polaris

2004-10-24, 8:55 am

"Don" <no@adr.com> schreef in bericht
news:bvoln05dbb573qkjsrhf91l7n0d9rvbfot@
4ax.com...[color=darkred]
> 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???

Charles Pelkey

2004-10-24, 3:55 pm

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[color=darkred]
>
> 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???
>
>

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