[PHP] Showing an image outside of the web folder

[James Nunnerley]

I'm creating a file manager application, from which I want the user to be
able to edit/view files.

On the text side of things, it's pretty easy, however from the image side of
things, I'm not sure how to allow the user to view files outside of the web
folder.

Doing a readfile ($file_location) outputs the binary... can someone point
me in the direction of being able to translate that binary into a viewable
image file?

Cheers
Nunners

 

Re: [PHP] Showing an image outside of the web folder

[John Nichel]

James Nunnerley wrote:

I'm creating a file manager application, from which I want the user to be
able to edit/view files.

On the text side of things, it's pretty easy, however from the image side of
things, I'm not sure how to allow the user to view files outside of the web
folder.

Doing a readfile ($file_location) outputs the binary... can someone point
me in the direction of being able to translate that binary into a viewable
image file?

Cheers
Nunners



PHP Manual - Function Reference - Image Functions

Google - php display image

IMO
It's easier to say RTFM and/or STFW
/IMO

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John C. Nichel IV
Programmer/System Admin (ÜberGeek)
Dot Com Holdings of Buffalo
716.856.9675
[EMAIL PROTECTED]

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Re: [PHP] Showing an image outside of the web folder

[Rabin Vincent]

On 5/5/06, James Nunnerley [EMAIL PROTECTED] wrote:

Doing a readfile ($file_location) outputs the binary... can someone point
me in the direction of being able to translate that binary into a viewable
image file?


Drop the appropriate content-type header:

header('Content-type: image/png');
readfile('/home/foo/bar.png');

Rabin

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[PHP] showing an image with php

[Ross Hulford]

I want to asssign an image to a variable and show it if certain conditions 
are met. E.g


If condition is met

{
$myimage= retrieve image (images/myimage.jpg)
 echo $myimage
}

Have not used image creation before with php just dipping my toe in the 
water.

Thanks


R. 

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Re: [PHP] showing an image with php

[Richard Lynch]

 I want to asssign an image to a variable and show it if certain conditions
 are met. E.g


 If condition is met

 {
 $myimage= retrieve image (images/myimage.jpg)
  echo $myimage
 }

 Have not used image creation before with php just dipping my toe in the
 water.

Not sure you need to use image creation now.

All you seem to really want to do is display an IMG tag, or not, based on
the if.

if (...){
  echo img src=images/myimage.jpg;
}

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[PHP] Showing an Image

[kachaloo]

HI,
I want to show an image from a database but to do that I am using :
 $result = @mysql_query($sql);
 $data = @mysql_result($result, 0, PICTURE);
 $type = @mysql_result($result, 0, FILETYPE);
 Header(Content-type: $type);
 echo $data;


but after this I also want to show some text info. How can I do this ? I
will have to use something else besides header. Please Help.
Thanks,
Vishal



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Re: [PHP] Showing an Image

[mike cullerton]

how about something like,

 $result = @mysql_query($sql);
 $data = @mysql_result($result, 0, PICTURE);
 $type = @mysql_result($result, 0, FILETYPE);
 $comment = @mysql_result($result, 0, COMMENT);
 Header(Content-type: $type);
 echo $data . br . $comment;

or, instead of using Header, create a page and inside it put
 printf(%sbr%s,$image,$comment)

mike

on 7/2/01 1:39 AM, kachaloo at [EMAIL PROTECTED] wrote:

 HI,
 I want to show an image from a database but to do that I am using :
 $result = @mysql_query($sql);
 $data = @mysql_result($result, 0, PICTURE);
 $type = @mysql_result($result, 0, FILETYPE);
 Header(Content-type: $type);
 echo $data;
 
 
 but after this I also want to show some text info. How can I do this ? I
 will have to use something else besides header. Please Help.
 Thanks,
 Vishal
 
 


-- mike cullerton   [EMAIL PROTECTED]



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