RE: [PHP] Oh, the problem

2002-07-21 Thread César Aracena 

Chris. Thanks for living me a hand in such a fast way. It helped me
fine.

César.

 -Original Message-
 From: Chris Earle [mailto:[EMAIL PROTECTED]]
 Sent: Monday, July 22, 2002 12:51 AM
 To: [EMAIL PROTECTED]
 Subject: [PHP] Oh, the problem
 
 It appears that $username isn't being defined within the scope of the
 query.
 You define it only in the function.
 
 Try do this instead:
 
 $username = ;
 
 function pic_upload($userid)
 {
 global $username;
 //... the rest of the code
 }
 // now do the queries and everything else.
 
 I think that will fix your problem.
 
 César aracena [EMAIL PROTECTED] wrote in message
 000301c23129$aebbab30$adc405c8@gateway">news:000301c23129$aebbab30$adc405c8@gateway...
 Hi all. I'm trying to handle a picture upload. So far, I've made my
 script store it where I want with the name I want. The only problem
now,
 is that it should store that given name into a MySQL DB table. I tried
 it by calling a function which stores the picture and returns a
variable
 called $picname then use an UPDATE statement but that variable isn't
 passed. any ideas? The code looks like this:
 
 function pic_upload($userid)
 {
 if (is_uploaded_file($_FILES['devpicture']['tmp_name']))
 {
 $filename = $_FILES['devpicture']['tmp_name'];
 
 $realname = $_FILES['devpicture']['name'];
 
 $username = $userid..jpg;
 
 copy($_FILES['devpicture']['tmp_name'],
 c:/apache/htdocs/os-seek/photos\\.$username);
 
 $username;
 }
 else
 {
 echo Possible file upload attack: filename
 .$_FILES['devpicture']['name']..br;
 }
 }
 
 ---
 and then the updating
 ---
 
 pic_upload($id);
 
 $query = UPDATE os_developers SET devpicture = '$username' WHERE
devid
 = $id;
 $result = mysql_query($query) or die(mysql_error());
 
 Thanks in advance,
 
 Cesar Aracena
 CE / MCSE+I
 Neuquen, Argentina
 +54.299.6356688
 +54.299.4466621
 
 
 
 
 
 
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Re: [PHP] Oh, the problem

2002-07-21 Thread Chris Earle 

Sure thing.  Always love to help.

-Chris


César aracena [EMAIL PROTECTED] wrote in message
001801c23136$6748bb50$adc405c8@gateway">news:001801c23136$6748bb50$adc405c8@gateway...
Chris. Thanks for living me a hand in such a fast way. It helped me
fine.

César.

 -Original Message-
 From: Chris Earle [mailto:[EMAIL PROTECTED]]
 Sent: Monday, July 22, 2002 12:51 AM
 To: [EMAIL PROTECTED]
 Subject: [PHP] Oh, the problem

 It appears that $username isn't being defined within the scope of the
 query.
 You define it only in the function.

 Try do this instead:

 $username = ;

 function pic_upload($userid)
 {
 global $username;
 //... the rest of the code
 }
 // now do the queries and everything else.

 I think that will fix your problem.

 César aracena [EMAIL PROTECTED] wrote in message
 000301c23129$aebbab30$adc405c8@gateway">news:000301c23129$aebbab30$adc405c8@gateway...
 Hi all. I'm trying to handle a picture upload. So far, I've made my
 script store it where I want with the name I want. The only problem
now,
 is that it should store that given name into a MySQL DB table. I tried
 it by calling a function which stores the picture and returns a
variable
 called $picname then use an UPDATE statement but that variable isn't
 passed. any ideas? The code looks like this:

 function pic_upload($userid)
 {
 if (is_uploaded_file($_FILES['devpicture']['tmp_name']))
 {
 $filename = $_FILES['devpicture']['tmp_name'];

 $realname = $_FILES['devpicture']['name'];

 $username = $userid..jpg;

 copy($_FILES['devpicture']['tmp_name'],
 c:/apache/htdocs/os-seek/photos\\.$username);

 $username;
 }
 else
 {
 echo Possible file upload attack: filename
 .$_FILES['devpicture']['name']..br;
 }
 }

 ---
 and then the updating
 ---

 pic_upload($id);

 $query = UPDATE os_developers SET devpicture = '$username' WHERE
devid
 = $id;
 $result = mysql_query($query) or die(mysql_error());

 Thanks in advance,

 Cesar Aracena
 CE / MCSE+I
 Neuquen, Argentina
 +54.299.6356688
 +54.299.4466621






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